MATHEMATICS FORM TWO SUMMARIZED

Chapter One: Rates and Variations

Introduction

Understanding rates and variations is essential for grasping how quantities change in relation to each other. Some variables are interdependent, meaning a change in one affects the others. Many real-life situations involve such relationships. In this chapter, you will describe the concepts of rates and variations, explain the types of variations, and solve problems on rates and variations. The competence developed will enable you to solve real-life challenges such as finding shopping deals, assessing fuel efficiency, or predicting data trends in fields like economics, science, engineering, and many other applications.

Rates

Rates show how one quantity is related to another quantity either increasing or decreasing its quantity. For example, speed of moving object is the ratio between the distance covered and time taken.

Activity 1.1: Exploring rates in real life

Engage in Activity 1.1 to explore the concept of rates in real life.

Example 1.1

If a car travels 150 kilometres in 3 hours. What is its speed?

Solution

Given distance = 150 km and time = 3 hrs.

But, speed is the rate of change of distance with respect to time. That is,

speed = distance/time

Thus, speed = 150 km / 3 hrs = 50 km per hour

Therefore, the rate of travel of the car is 50 km in every hour.

Example 1.2

If a water tank fills up with 200 litres of water in 5 minutes, what is the rate of flow of water?

Solution

Given the volume of water is 200 L and the time the tank takes to be filled is 5 minutes.

Rate of flow = Volume / Time

Rate of flow = 200 L / 5 min = 40 L/min

Therefore, water enters the tank at the rate of 40 litres per minute.

Example 1.3

A student had two plant seedlings. She measured the rate at which the seedlings were growing. Seedling A grew 5 cm in 10 days and seedling B grew 8 cm in 12 days. Which seedling was growing more quickly?

Solution

The rates of growth of the two seedlings is computed as follows:

Rate of growth of seedling A = 5 cm / 10 days = 0.5 cm per day

Rate of growth of seedling B = 8 cm / 12 days = 0.67 cm per day

The growth rate of seedling B is higher than that of seedling A.

Therefore, seedling B was growing more quickly than seedling A.

Example 1.4

Two pipes, A and B are used to fill a water tank. Pipe A can fill the tank in 6 hours, while pipe B can fill the same tank in 4 hours. If both pipes are opened at the same time, how long will it take to fill the tank?

Solution

Pipe A fills the tank in 6 hours, so it fills 1/6 of the tank in 1 hour.

Pipe B fills the tank in 4 hours, so it fills 1/4 of the tank in 1 hour.

Combined rate = Rate of A + Rate of B

Combined rate = 1/6 + 1/4 = (2 + 3)/12 = 5/12

Therefore, both pipes together fill 5/12 of the tank in 1 hour.

If 5/12 of the tank is filled in 1 hour, then the time taken to fill the whole tank will be:

Time = Whole tank / Combined rate = 1 / (5/12) = 12/5 = 2.4 hours

Therefore, it will take 2.4 hours to fill the tank when both pipes are opened simultaneously.

Exercise 1.1

1. What rate in metres per second is equivalent to a speed of 45 kilometres per hour?
2. A car covers a distance of 200 kilometres in 60 minutes. What is the speed of the car in metres per second?
3. A water tap takes 10 minutes to fill a 500 litres tank. Find the rate of flow of water in litres per second.
4. If premium petrol costs Tsh 2,500 per litre, how many litres can be purchased for Tshs 10,000?
5. Find the rate in kilometres per hour if a racing car travels 115½ kilometres in 35 minutes.
6. Mwajuma walks 6 kilometres in 1 hour and John walks the same distance in 2 hours. What are their walking speeds? Who walks faster, and by how much?
7. A garden hose fills a 20 litre bucket in 4 minutes. What is the rate of flow in litres per minute?
8. A car travels 400 km using 25 litres of fuel. What is the car's fuel efficiency in kilometres per litre?
9. Mr. Magoda's salary increased from Tsh 800,000 to Tsh 1,000,000 in a year. What is the rate of change of his salary?
10. The temperature in Mrs. Kidunula's room increased from 15°C at 8 am to 25°C at 2 pm. What was the average rate of temperature change per hour?
11. A car accelerates from 0 to 60 km/h in 10 seconds. What is the rate of change of its speed in km/h per second?
12. Mr. Kilenzi's heartbeat increased from 60 beats per minute to 120 beats per minute during exercise over 3 minutes. What was the average rate of change in his heartbeats?
13. A family uses 80 GB of internet data in 30 days. What is the average daily rate of data usage in GB per day?
14. Pipe X can fill a tank in 8 hours, while Pipe Y can fill it in 5 hours. How long will it take to fill the tank if both pipes are opened simultaneously?
15. Pump A can fill a pool in 5 hours, Pump B in 4 hours, and Pump C in 10 hours. How long will it take to fill the pool if all three pumps are operated at the same time?

Exchange Rates

In any country, people expect to do transactions in the currency of their own country. When money from country A is to be used in country B, it is necessary to exchange the currency of country A to the currency of country B. Various currencies in the world are linked together by exchange rates. This enables smooth transfer of money and payments to take place between countries.

Activity 1.2: Performing currency exchange rates

1. Choose a suitable computer Currency Conversion Application.
2. Analyse current exchange rates between the Tanzanian currency and other currencies of your choice using the computer application.
3. Compare the app's rates with market rates and evaluate its accuracy for financial decisions, especially for travel.

Exercise 1.2

Use any computer Currency conversion application to answer questions 1 ~ 10.

1. How much is Tsh 20,600 worth in Indian Rupees?
2. Convert 1 New Zealand Dollar into Saudi Arabia Riyal.
3. How much is Tsh 500,000 worth in Euros?
4. How many Yen are equivalent to Tsh 1?
5. How many Tanzanian shillings can a visitor from Kenya get for exchanging 930 Kenyan shillings?
6. Find the amount in Tanzanian shillings of each of the following:
   1. 30,000 Euros
   2. 4,200 Pula
   3. 640 Rands
   4. 12,000 Riyal
7. Exchange Tsh 300,000 into the following currencies:
   1. Mozambican meticals
   2. Malawian kwachas
   3. Swiss francs
   4. Indian rupees
8. How much is Tsh 6,000,000 worth in Pounds Sterling?
9. Mwanaisha bought story books for 200 AUD (Australian Dollars). How much did she spend in Tanzanian shillings?
10. Mr. Utaligolo wants to exchange 1,000 USD for EUR. If the exchange fee is 2%, how much will he receive after charging the fee?
11. Mrs. Uwemba is shopping in a country where 1 USD = 205 Local Currency (LC). If she spent 205,000 LC, how much did she spend in USD?
12. Ayota Stationery wants to buy items from an online store, which cost 2500 NOK (Norwegian Kroner). If the exchange rate is 1 NOK = 9 Local Currency, how much does the stationery pay in Local Currency?

Variations

Variation is a relationship where a change in one quantity leads to a proportional change in the other. It allows for the exploration of connections between two or more quantities. The four basic types of variations are direct, inverse, joint, and combined.

Direct Variations

Direct variation is a relationship between two variables where one variable is a constant multiple of the other. In other words, if one variable increases or decreases, the other variable changes proportionally in the same direction.

This type of variation is useful for understanding how changes in one variable affect another in a directly proportional manner.

Activity 1.3: Exploring direct variation in real life

1. Learn about direct variation from books or from the internet.
2. Find real-life examples that illustrate direct variations.
3. Demonstrate mathematically how variables in the real-life scenarios relate and use the relationship to solve related problems.

If y is directly proportional to x, it can be written as y ∝ x, where ∝ is a symbol of proportionality. The corresponding mathematical equation connecting x and y is formed by introducing a proportionality constant k, and replaces ∝ with an equal sign to get, y = kx.

For instance, if y varies directly as the square of x, then y ∝ x² and the corresponding equation is y = kx², where k is the constant of proportionality.

For any two pairs of quantities x and y, say (x₁, y₁) and (x₂, y₂), two equations y₁ = kx₁² and y₂ = kx₂² are obtained. This implies that k = y₁/x₁² = y₂/x₂². So, it is said that x and y vary directly if the ratios of the values of y to the values of x are proportional.

If x and y are any two quantities that are in direct variation, then y = kx. The nature of the equation y = kx is a straight line passing through the origin, where k represents the gradient (slope) of the line.

Example 1.5

If y varies directly as x and x = 15 when y = 4, find the value of y when x = 12.

Solution

Given that y ∝ x. This implies that y = kx, where k is the constant of proportionality.

Making k the subject of the equation gives k = y/x

But, for any two pairs of quantities, k = y₁/x₁ and k = y₂/x₂.

Thus, y₁/x₁ = y₂/x₂

Given x₁ = 15, x₂ = 12, and y₁ = 4, the value of y₂ is obtained as follows:

y₂ = (y₁ × x₂)/x₁ = (4 × 12)/15 = 48/15 = 16/5

Therefore, the value of y is 16/5 when x = 12.

Example 1.6

If x varies directly as the square of y and x = 4 when y = 2, find the value of x when y = 8.

Solution

Let x₁ = 4, y₁ = 2 and y₂ = 8. The variation equation is given by:

x₁/x₂ = y₁²/y₂²

Substituting the values of x₁, y₁, and y₂ gives:

4/x₂ = 2²/8² = 4/64

Cross multiplication gives:

4 × x₂ = 4 × 64

x₂ = (4 × 64)/4 = 64

Therefore, the value of x is 64 when y is 8.

Example 1.7

A car travels 60 kilometres using 5 litres of diesel. How many litres of diesel are needed to travel 150 kilometres?

Solution

Let x denote the number of litres of diesel and y denote the number of kilometres.

The equation for the variation becomes x₁/x₂ = y₁/y₂

Given x₁ = 5 litres, y₁ = 60 km, and y₂ = 150 km, the value of x₂ is given by:

x₂ = (x₁ × y₂)/y₁ = (5 litres × 150 km)/60 km = 750/60 = 12.5 litres

Therefore, 12.5 litres of diesel are needed to travel 150 kilometres.

Exercise 1.3

1. If x varies directly as y and x = 16 when y = 10, find the value of y when x = 20.
2. The surface area of a circular object varies directly as the square of its radius. If its surface area is 78.5 cm² when the radius is 5 cm, find the surface area of the circular object when the radius is 7 cm.
3. If x varies directly as y and x = 30 when y = 40, find the value of x when y = 16.
4. A mason can build 100 metres of fence in 20 hours. How long will it take 5 masons with the same ability to build 875 metres of fence?
5. If x varies directly as 2y + 7 and x = 5, when y = 4, find the value of y when x = 6.
6. If 8 men can assemble 16 machines in 12 days, how long will it take 15 men of the same ability to assemble 100 machines?
7. If y varies directly as the square root of x, and y = 12 when x = 4, find the value of y when x = 9.
8. If y varies directly as x and y = 8 when x = 3, find the value of y when x = 18.
9. Two variables x and y that vary directly have corresponding values as shown in the following table.

   x	3	5	6	8
   y	17		34	58

   1. Find the rule connecting x and y.
   2. Fill in the missing values.
   3. Draw a graph which shows that y ∝ x for k = 1.
10. Study the following table and answer the questions that follow.

    Hours worked	2	3	4	5
    Earning (Tsh)	1,150	1,725	2,300	2,875

    1. Do earnings vary directly as the number of hours worked?
    2. If yes, calculate the constant of proportionality and find the equation that describes the relationship.
11. The volume of a sphere varies as the cube of its radius. Three solid spheres of diameters 3/2 m, 2 m, and 5/2 m are melted and combined to form a new solid sphere. Find the diameter of the new sphere.
12. When observing two buildings simultaneously, the length of each building's shadow varies directly with its height. If a 5 floor building has a shadow of length 20 m, how many floors of a building would form a shadow of length 32 m?
13. The resistance of a wire varies as the square of the diameter of its cross-section. Find the percentage change in the resistance when the diameter is (a) doubled (b) reduced by 20%.
14. Two variables A and x are related by the formula A = axⁿ. The following set of data was generated based on this formula.

    x	1	2	3	4
    A	0.5	2	4.5	8

    1. Find the values of a and n.
    2. Find the value of A when x = 5.
15. A precious stone worth Tsh 15,600,000 is accidentally dropped and broken into three pieces. The weights of the pieces are in the proportions of 2:3:5, respectively. If the value of the precious stone varies directly as the cube of its weight, calculate the value of the remaining stone in percentage.

Inverse Variations

A relationship between two or more variables is said to be an inverse variation if the value of one variable increases while the other value decreases, or vice versa.

Activity 1.4: Identifying inverse variations in daily life

1. Identify three real-life activities where increasing one variable decreases the other variable.
2. Perform one of the activities to experience how changes of the variables are related in the activity.
3. Find out through reading books and browsing the internet how to express these inverse relationships mathematically.
4. Use the mathematical expression to explain how changes in one variable affect the other, and present your findings.

Quantities with an inverse variation relationship are said to be inversely proportional to each other. In this case, the quantities vary inversely or in inverse proportion. Inverse proportion is sometimes referred to as indirect proportion.

For example, the number of men employed to cultivate a farm and time taken to complete the work are inversely related. Likewise, the time to travel to a certain place and the speed are inversely related.

Generally, if y has an inverse relationship with x, then y is proportional to the reciprocal of x. This relationship is denoted by:

y ∝ 1/x

The equation relating y and x is formed by introducing a constant of proportionality, k and replacing the symbol of proportionality with an equal sign (=). That is, the inverse variation between y and x becomes:

y = k(1/x)

If y varies inversely as the square of x, the equation connecting x and y is y = k/x² or k = yx².

Example 1.8

If x varies inversely as y and x = 2 when y = 3, find the value of y when x = 18.

Solution

The statement x ∝ 1/y implies that x = k/y, where k is the constant of proportionality.

Making k the subject of the equation gives, xy = k, which implies that x₁y₁ = x₂y₂.

When x₁ = 2, y₁ = 3, and x₂ = 18, the value of y₂ is:

y₂ = (x₁y₁)/x₂ = (2 × 3)/18 = 6/18 = 1/3

Therefore, the value of y is 1/3 when the value of x is 18.

Example 1.9

If it takes 12 days for 10 men to assemble a machine, how long does it take 15 men with the same ability to assemble the same machine?

Solution

Let m be the number of men and d be the number of days. It is obvious that 15 men will take less time to assemble the machine than 10 men.

Thus, m and d vary inversely, that is, m ∝ 1/d, which implies that m = k/d.

Thus, k = md.

For the two pairs of quantities (m₁, d₁) and (m₂, d₂), it follows that, m₁d₁ = m₂d₂.

Thus, d₂ = (m₁d₁)/m₂

Given d₁ = 12 days, m₁ = 10 men and m₂ = 15 men, the value of d₂ is:

d₂ = (10 men × 12 days)/15 men = 120/15 = 8 days

Therefore, it takes 8 days for 15 men to assemble the same machine.

Example 1.10

The intensity of light is inversely proportional to the square of the distance D from the light source. Calculate the percentage change in intensity under the following conditions:

1. The distance is halved.
2. The distance is increased by 30%.

Solution

I ∝ 1/D²

I = k/D²

(a) If the distance D is halved, the new distance is D₁ = D/2.

The new intensity I₁ = k/(D/2)² = k/(D²/4) = 4k/D²

Percentage change = [(I₁ - I)/I] × 100% = [(4k/D² - k/D²)/(k/D²)] × 100% = 300%

Therefore, the intensity increases by 300%.

(b) If the distance is increased by 30%, the new distance is D₁ = 1.3D.

The new intensity I₁ = k/(1.3D)² = k/(1.69D²) = (1/1.69)(k/D²)

Percentage change = [(I₁ - I)/I] × 100% = [((1/1.69)(k/D²) - k/D²)/(k/D²)] × 100% = (1/1.69 - 1) × 100% ≈ -40.83%

Therefore, the intensity decreases by approximately 40.83%.

Joint and Combined Variations

In some activities, one variable can depend on several other variables to operate effectively. Such relationships are described by joint and combined variations.

Activity 1.5: Discovering joint and combined variations in daily life

1. Explore the concepts of joint and combined variations using books and online resources.
2. Find and describe real-life examples of such variations using daily practices such as formulas.
3. Share your final observations and discuss the examples you discovered.

Joint Variation

Consider a formula for finding the area of a triangle. It is given by Area = 1/2 × base × height. In this equation, the area of a triangle varies directly as the product of its base and height. Variable relationships of this nature are known as joint variations. This specific variation can be generally expressed as Area ∝ base × height.

A joint variation occurs when a variable is directly or inversely proportional to the product of two or more variables. Mathematically, if a variable z varies jointly as x and y, the general formula for joint variation can be written as:

z ∝ xy ⇒ z = kxy

Example 1.11

Suppose y varies directly as x and z. Given x = 4, z = 2, and y = 24, find:

1. The variation equation connecting x, y, and z.
2. The value of y when x = 5 and z = 6.

Solution

(a) Since y ∝ x and y ∝ z, it follows that y ∝ xz, y = kxz.

The variation equation is k = y/(xz), where k is a constant of proportionality.

Given x = 4, z = 2, and y = 24, it follows that k = 24/(4 × 2) = 24/8 = 3

Therefore, the variation equation is y = 3xz.

(b) With y = 3xz, if x = 5 and z = 6, then y = 3 × 5 × 6 = 90

Therefore, the value of y is 90.

Example 1.12

A bakery has a project to bake 240 cakes. With 4 bakers working for 5 days, they can complete the task. If the number of cakes is increased to 300 and the bakery decides to use 6 bakers, how many days will it take to bake all 300 cakes?

Solution

Let c be the number of cakes, b be the number of bakers, and d be the number of days.

More cakes can be produced if there are more bakers and more working days, assuming other factors remain constant.

It implies that these variables vary jointly and are directly related. Thus, c ∝ bd.

c = kbd.

Given c = 240, b = 4, and d = 5, it implies that 240 = 4 × 5 × k = 20k

k = 240/20 = 12

From c = kbd, c = 12bd

Given, c = 300, b = 6, 300 = 12 × 6 × d = 72d

d = 300/72 = 4.2

Therefore, 4.2 days will be needed to bake 300 cakes or it requires 4 days, 4 hours, and 48 minutes to bake 300 cakes.

Combined Variation

Combined variation occurs when a variable depends on two or more other variables, with some relationships being direct and others inverse.

Example 1.16

The cost of a certain material varies directly with the quantity purchased and inversely with the number of suppliers. If the cost is Tshs 120,000 when the quantity is 100 units and there are 4 suppliers, find the constant of variation and the cost when the quantity becomes 150 units and there are only 3 suppliers.

Solution

Let C be the cost of materials, q be the quantity purchased, and S be the number of suppliers.

Thus, C ∝ q and C ∝ 1/S. The combined variation is:

C ∝ q/S ⇒ C = k(q/S)

Given C = Tshs 120,000, q = 100, S = 4, then

k = (C × S)/q = (120,000 × 4)/100 = 480,000/100 = 4,800

Now, if S = 3 and q = 150, it follows that

C = (4,800 × 150)/3 = 720,000/3 = 240,000

Therefore, the cost of materials will be Tshs 240,000.

Exercise 1.5

1. If y varies directly as the square of x and inversely as z, find the percentage change in y when x is increased by 10% and z is decreased by 20%.
2. Suppose P varies directly as V and inversely as the square root of R. If P = 180 when R = 25 and V = 9, find the value of P when V = 6 and R = 36.
3. The height h of a cone varies directly as its volume V and inversely as the square of its radius r. Write a formula for the height of the cone.
4. If y² varies directly as x-1 and inversely as x+d and x = 2, d = 4 when y = 1, find the value of x when y = 2 and d = 1.
5. If two typists in a typing pool can type 210 pages in 3 days, how many typists working at the same speed will be needed to type 700 pages in 2 days?
6. Suppose x varies directly as y² and inversely as p. If x = 2, when y = 3 and p = 1, find the value of y when x = 4 and p = 5.
7. If V varies directly as the square of x and inversely as y, and if V = 18 when x = 3 and y = 4, find the value of V when x = 5 and y = 2.
8. Use a mathematical software to draw the following curves. Assume that the constant of proportionality is 1.
   1. y ∝ 1/x
   2. y ∝ 1/x²
   3. y ∝ 1/√x
9. The following table shows the values of y for some selected values of x. The variables x and y are connected by the relation, 'y varies inversely as x'. Calculate the missing values of y.

   x	5	10	15
   y	a	3	b	1.5

10. Express each of the following relations as an equation using k as a constant of proportionality.
    1. c varies directly as p and q, and inversely as s.
    2. d varies jointly as t and r².
    3. d varies directly as y and the square root of z.
11. The heating cost H for a house varies directly with its size S in square metres and inversely with the efficiency rating E of the heating system. If H = 50,000 shillings for a house of 200 square metres with an efficiency rating of 4, find k and the heating cost for a house of 250 square metres with an efficiency rating of 5.
12. The shipping cost C varies directly with the mass W of the package and inversely with the number of packages n. If the cost is TShs 7,000,000 for a package weighing 20 kg, and 6 packages, find k and the cost of a package weighing 30 kg with 5 packages.

Chapter Summary

1. A rate gives the change of one quantity with respect to another quantity.
2. Exchange rate is the conversion rate between different currencies.
3. Variation is the relationship in which the change in one quantity results in a proportional change in the other.
4. If y = kx, then y varies directly with x, or y is directly proportional to x. The constant k is called a constant of proportionality.
5. If y varies as 1/x, then y is inversely proportional to x.
6. If a quantity varies as the product of two or more quantities, then it varies jointly with other quantities.
7. If both direct variation and inverse variation occur at the same time, then it is called a combined variation.

Revision Exercise 1

1. If y = kx and y = 8 when x = 7, find the value of k and the value of y when x = 40.
2. If y is directly proportional to x and y = 10 when x = 4, find the value of y when x = 15 and the value of x when y = 8.4.
3. If y ∝ x and y = 16.5 when x = 3.5, find the equation connecting x and y. Hence, find the value of x when y = 21.
4. If y is proportional to x² and if x = 15 when y = 200, find the equation connecting x and y. Find the value of y when x = 8.5.
5. If y ∝ √x and y = 3.5 when x = 4, express y in terms of x. What is the value of y when x = 25?
6. If y ∝ 1/x, find the values of a, b, and c in the following table.

   x	a	1.2	8
   y	6	b	1.5	0.8

7. Given that y varies directly as x and inversely as z. If y = 10 when x = 8 and z = 5, find the equation connecting x, y, and z. Find the value of y when x = 6 and z = 2.5.
8. If y varies jointly as x and z², and if y = 13⅓ when x = 2.5 and z = 4/3, find the equation connecting the three variables. Find the value of x when z = 3/2 and y = 54.
9. Suppose y varies directly as x² and inversely as √z. If x = 8, y = 16, and z = 25, find the value of y when x = 5 and z = 9.
10. Determine whether the data in the following tables have an inverse variation relationship. If yes, find the missing values.

    1. x	7	9	12	6
       y	10	12	15	12

    2. x	12	6	21	3
       y	4	2	7	3

    3. x	-15	-8	10
       y	-8	-15	10

11. A dairy farm dispenses milk into a container at a rate of 45 litres per minute.
    1. How much milk is dispensed in 2 hours?
    2. How long will it take to empty a 2000 litre tank?
12. A water pump operates at a flow rate of 300 litres per minute.
    1. Calculate the total volume of water pumped in 2 hours.
    2. Calculate the time required to fill a 30,000-litre tank.
13. A truck travels 800 km and uses 90 litres of diesel.
    1. Calculate the fuel consumption rate in litres per 100 km.
    2. How much diesel will the truck use to travel 600 km at the same rate?
    3. If the truck travels 1,300 km, how much diesel will it consume?
14. State whether distance and speed vary directly or inversely. Give reason.
15. If x and y vary inversely, use the given pair of values to find an equation which in each case relate the variables:
    1. x = 6, y = 4
    2. x = 8, y = 12
    3. x = 10, y = 1/3
16. A group of 10 volunteers can pack 400 boxes of food in 5 hours. If the organization decides to pack 600 boxes and increases the number of volunteers to 12, how many hours will it take to pack all 600 boxes?
17. A publishing company needs to print 2,000 copies of a book. With 4 printers, the task is completed in 10 days. If the company decides to print 3,000 copies and uses 6 printers, how many days will it take to complete the printing?
18. A logistics team of 8 members can package and ship 400 orders in 5 days. If the number of team members is increased to 12 and the number of orders to be shipped is raised to 600, calculate the number of days needed to complete the work.
19. The intensity I of light, in lux, on a surface varies directly with the power P in watts of light and inversely with the square of the distance d from the light source. A photographer is setting up lighting for a photoshoot. If the light intensity is 400 lux when a light source of 66 watts is placed 2 m away, what will be the intensity when a light source of 100 watts is placed 3 m away?
20. Machine A can produce 800 parts per hour, while Machine B can produce 1,000 parts per hour. There is an order of 24,000 parts.
    1. How long will the machines take working together to process the order?
    2. Machine B was on preventive maintenance, so Machine A had to work alone for the first 5 hours before Machine B joins in. How many more hours will they have to work together to finish the job?
21. Pipe A can fill a pond in 8 hours, while Pipe B can fill it in 3 hours. However, Pipe B has a defect that causes it to lose water at the rate of 1 litre every 20 minutes while filling. If the pond has a capacity of 200 litres and both pipes are opened simultaneously, how long will it take to fill the pond?

Chapter Two: Congruence

Introduction

In fields like architecture, engineering and design, precise measurements and patterns are crucial. Maintaining such patterns and measurements across objects and designs requires an understanding of the relationship between shapes and their properties. In this chapter, you will learn about postulates, proofs and theorems on congruence as well as the congruence of triangles. The competencies developed will enable you to solve problems such as determining the dimensions of sides of figures without taking actual measurements, designing and making objects of the same shapes and sizes such as worn-out parts.

Think

Without knowledge of congruence industries would take longer time to produce objects of the same shape and size.

The Concept of Congruence

The term congruence is derived from the Latin word congruentia which means agree or fit together exactly. Engage in Activity 2.1 to explore the characteristics of objects from real life situations which fit together.

Activity 2.1: Exploring Congruence of Objects

1. Create a shape of your choice in Ms-Word, copy and paste it, then drag it on top of the original and observe.
2. Enlarge the original shape, overlap it with the copied shape, and note any changes.
3. Use real objects (e.g., coins or books) to repeat task 1 (or 2 where necessary) and compare your observations.

In Activity 2.1, one may have noticed that same objects matched exactly on top of each other. This means that the objects have the same shape and size. Two or more objects with such characteristics are called congruent figures.

Postulates, Theorems and Proofs

Understanding postulates, theorems, and proofs is important for developing a solid foundation in mathematics. These concepts form the basis for logical reasoning and they are used to validate mathematical ideas.

Postulate

A postulate is a statement that is accepted as true without any proof. These statements are universally accepted, self-evident and can form the basis for further reasoning and making arguments. The following are examples of postulates:

• A circle can be drawn with any centre and radius.
• A straight line can be drawn from any point to any other point.
• All right angles are equal to each other.

Theorem

A theorem is an argument that has been proved to be true based on past established results, definitions or postulates. The following are examples of theorems:

• The sum of interior angles of a triangle is 180 degrees.
• In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
• The sum of interior angles of a quadrilateral is 360 degrees.

Proof

A proof is a series of logical statements that are based on definitions, previously established facts, and postulates that may be used to conclude the truth of a mathematical argument. The following are common procedures when undertaking a proof:

1. Draw a clearly labelled diagram to represent a problem. Indicate all the information such as equal angles, parallel lines, and congruent segments.
2. Write down the given information based on the labels of the supporting figures.
3. State the argument which needs to be proved.
4. Where necessary, make additional constructions with dotted lines to make the proof clear.
5. In writing the proof:
   • Refer the figures you have planned to use in the proof.
   • Provide arguments with reasons based on the given information or established facts.
   • Start with statements whose validity are given or are obvious.
   • The final statement is the conclusion about what was supposed to be proved.

Activity 2.2: Exploring Postulates, Proofs, and Theorems

Identify a list of postulates, proofs, and theorems from reliable sources and briefly explain why each is accepted as a postulate, proof or theorem.

Example 2.1

Prove that the sum of interior angles of a triangle is 180°.

Triangle ABC with line XY through C parallel to AB

Proof:

Construction: Draw line XY through C parallel to AB. Thus,

∠ACX = ∠CAB (alternate interior angles as XY ∥ AB) ... (1)

∠YCB = ∠ABC (alternate interior angles as XY ∥ AB) ... (2)

But ∠ACX + ∠BCA + ∠YCB = 180° (degree measure of a straight angle) ... (3)

Substitute (1) and (2) into (3) to get ∠CAB + ∠BCA + ∠ABC = 180°.

Therefore, the sum of interior angles of a triangle is 180°.

Example 2.2

Prove that the sum of two interior angles of a triangle is equal to the exterior angle of the third interior angle.

Triangle ABC with AB extended to D

Proof:

From the figure, it implies that

∠CAB + ∠BCA + ∠ABC = 180° (sum of interior angles of a triangle), and

∠ABC + ∠CBD = 180° (degree measure of a straight angle).

It follows that,

∠CAB + ∠BCA + ∠ABC = ∠ABC + ∠CBD

But ∠ABC is common to both sides of the equation. Thus, ∠CAB + ∠BCA = ∠CBD.

Therefore, the sum of two interior angles of a triangle is equal to the exterior angle of the third interior angle.

Congruence of Triangles

Two triangles ABC and PQR are said to be congruent if pairs of corresponding sides are equal and pairs of corresponding angles are equal. This fact is mathematically represented as ΔABC ≅ ΔPQR. The symbol ≅ means "congruent to".

Figure 2.1: Congruence of triangles

From Figure 2.1, if ΔABC ≅ ΔPQR, the pairs of corresponding sides are AB and PQ, BC and QR, and AC and PR. Therefore, AB = PQ, BC = QR, and AC = PR.

Pairs of corresponding angles are ∠ABC and ∠PQR, ∠BCA and ∠QRP, and ∠BAC and ∠QPR. Therefore, ∠ABC = ∠PQR, ∠BCA = ∠QRP, and ∠BAC = ∠QPR.

Postulates for Congruence of Triangles

Three conditions for congruence are sufficient to prove that two triangles are congruent. The postulates that are commonly used include the Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Angle-Side (AAS), and Right angle-Hypotenuse-Side (RHS).

Side-Side-Side (SSS) Postulate

The SSS postulate states that two triangles are congruent if the pairs of their corresponding sides are equal.

Figure 2.2: Pair of congruent triangles satisfying SSS postulate

From Figure 2.2, it follows that:

• AB = PQ (given)
• BC = QR (given)
• AC = PR (given)

Since the pairs of the corresponding sides of triangles ABC and PQR are equal, then the two triangles are exactly the same. Therefore, it follows that, ΔABC ≅ ΔPQR (by SSS).

Since the two triangles are congruent, it implies that their corresponding angles are also equal, that is, ∠CAB = ∠RPQ, ∠ABC = ∠PQR and ∠BCA = ∠QRP.

Example 2.4

Use the following figure to prove that ΔABC ≅ ΔCDA, hence deduce that ∠DCA = ∠BAC.

Rectangle ABCD with diagonal AC

Solution:

Given, a rectangle ABCD in which AB = DC, and AD = BC.

Construct a line joining A and C.

Proof: From ΔABC and ΔCDA

AB = DC (given)

BC = AD (given)

AC is a common side to both triangles.

Therefore, ΔABC ≅ ΔCDA (By SSS).

Hence ∠BAC = ∠DCA (definition of congruent triangles).

Example 2.5

Triangle ABC is an isosceles triangle in which AB and AC are equal. If D is the midpoint of BC, prove that ΔABD ≅ ΔACD.

Isosceles triangle ABC with D as midpoint of BC

Solution:

Consider the ΔABC such that AB = AC and D is a mid-point of BC as shown in the following figure.

Required to prove that ΔABD ≅ ΔACD.

Construction: Use a dotted line to join the points A and D.

Proof: In ΔABD and ΔACD

AB = AC (given)

BD = DC (given)

AD is a common side to both triangles.

Therefore, ΔABD ≅ ΔACD (by SSS).

Side-Angle-Side (SAS) Postulate

The SAS postulate states that two triangles are congruent if two pairs of corresponding sides and the included angles are equal.

Figure 2.3: Pair of congruent triangles satisfying SAS postulate

From Figure 2.3, it follows that:

• AB = PQ (given)
• AC = PR (given)
• ∠BAC = ∠QPR (given)

Therefore, ΔABC ≅ ΔPQR (by SAS).

Example 2.6

In the following figure, if AB = CD and ∠ABC = ∠DCB, prove that ΔABC ≅ ΔDCB.

Triangles ABC and DCB sharing side BC

Proof:

In ΔABC and ΔDCB:

AB = CD (given)

∠ABC = ∠DCB (given)

BC is a common side to both triangles.

Therefore, ΔABC ≅ ΔDCB (by SAS).

Angle-Angle-Side (AAS) Postulate

Two triangles are congruent if the angles in any two pairs of corresponding angles are equal and the lengths of a pair of corresponding sides are equal.

Figure 2.4: Two congruent figures by AAS postulate

Figure 2.4 shows that:

• BA = QP (given)
• ∠ABC = ∠PQR (given)
• ∠BCA = ∠QRP (given)

Thus, the two triangles satisfy the AAS postulate.

Therefore, ΔABC ≅ ΔPQR (by AAS).

Example 2.7

In the following figure, prove that ΔABC ≅ ΔCDA.

Parallelogram ABCD with diagonal AC

Solution:

Given a parallelogram ABCD where AC is its diagonal.

Required to prove that, ΔABC ≅ ΔCDA.

Proof: In ΔABC and ΔCDA it follows that,

∠CAB = ∠ACD (alternate interior angles as AB ∥ DC).

Similarly, ∠ACB = ∠CAD (alternate interior angles as AD ∥ BC).

AC is a common side.

Therefore, ΔABC ≅ ΔCDA (by AAS).

Right angle-Hypotenuse-Side (RHS) Postulate

Two right-angled triangles are congruent if their hypotenuses have equal length and a pair of the corresponding sides have equal length.

Figure 2.5: Two congruent right-angled triangles with RHS postulate

Figure 2.5 shows that:

• AB = PQ (given)
• AC = PR (given)
• ∠ABC = ∠PQR = 90° (given)

If two triangles satisfy the conditions in the RHS postulate, then the triangles must fit exactly. Thus, Figure 2.5 shows that ΔABC ≅ ΔPQR (by RHS).

Example 2.8

Use the following figure to prove that ΔADB ≅ ΔADC and DB = DC.

Triangle ABC with AD perpendicular to BC

Solution:

Given ΔABC such that AD is perpendicular to BC.

Required to prove that

(a) ΔADB ≅ ΔADC

(b) DB = DC

Proof: From ΔADB and ΔADC, it implies that

AB = AC (given)

∠ADB = ∠ADC = 90° (given)

AD is a common side.

Therefore, ΔADB ≅ ΔADC (by RHS), hence DB = DC (definition of congruence of triangles).

Chapter Summary

1. Two figures are said to be congruent if they have exactly the same size and shape.
2. A postulate is a statement that is accepted without any proof.
3. A theorem is an argument that has been proven to be true based on past established results or definitions or postulates.
4. A proof is a series of logical statements that are based on definitions, previously established facts, and postulates that may be used to conclude the truth of mathematical arguments.
5. Two triangles are congruent if:
   • The sides of one triangle have equal lengths to the corresponding sides of the other triangle (SSS).
   • The lengths of two sides and the included angle of one triangle are respectively equal to the lengths of two corresponding sides and the included angle of other triangle (SAS).
   • Two angles and the included side of one triangle are respectively equal to the corresponding two angles and the included side of the other triangle (ASA).
   • Two angles and non-included side of one triangle are respectively equal to the corresponding two angles and a non-included side of the other triangle (AAS).
6. Two right-angled triangles are congruent if their hypotenuses and a pair of sides have equal length.

Revision Exercise 2

1. In the following figure, BA = BC and KA = KC. Prove that ∠BAK = ∠BCK.
2. A quadrilateral MNOP has the property that MN = OP and MP = NO. Prove that ∠PMN = ∠NOP.
3. Use the following figure to prove that ML = PL.
4. In the following figure, prove that ∠ACD = ∠BDC.
5. Use the following figure to prove that ∠BMA = ∠CMA = 90°, given that AB = AC.
6. If ABCD is a square and AR = BR, prove that R is the midpoint of DC.
7. Prove that the line segment from the vertical angle of an isosceles triangle to the mid-point of its base is perpendicular to the base.
8. In the following figure, prove that BD = DC.
9. Prove that the bisector of the vertical angle of an isosceles triangle is perpendicular to the base at its mid-point.
10. In the following figure AB = HB and RB = BF. Prove that:
    1. ∠RAB = ∠FHB
    2. AM = HM

Chapter Three: Similarity

Introduction

Understanding similar triangles is crucial in scaling, designing structures, even in photographing, where proportions must be maintained. All of these skills are possible for someone with knowledge of similar figures. In this chapter, you will learn the concept of similar figures, recognise the properties of similar triangles and explain postulates, proofs, and theorems of similar triangles. The competencies developed will be applied in constructions and in architectural matters such as finding heights of buildings, bridges and trees, where tape measures cannot be used conveniently and many other applications.

Think

A world without knowledge of scales, proportions, sizes and comparisons.

Similar Figures

The previous chapter discussed congruent figures which are figures with identical shape and size. This chapter introduces the concept of similar figures which advances the concept of congruent figures. Engage in Activity 3.1 to explore the concept of similar figures.

Activity 3.1: Recognizing Similarities Between Figures

1. Cut a rectangle from the edge of a plain paper or a manila sheet as shown in the following figure, then compare the rectangle with the original paper in terms of shape and size.
2. Measure the lengths of sides and angles of both shapes, then find the ratio of the corresponding side lengths.
3. Use your observations to make a general conclusion about the shapes, and explore additional sources such as the internet and books to enrich your competence.
4. Reflect on real objects around you with similar features and characteristics and share your insights.

Similar figures are geometric shapes that have the same shape but differ in size. They maintain the same ratio of corresponding sides and angles. Study Figure 3.1 which illustrates the concept of similar objects.

Figure 3.1: Pictures of similar blackboards

In Figure 3.1, the objects are similar because they have the same shape but differ in size. Based on this experience, a simple method of obtaining similar figures is by uniformly scaling (enlarging or shrinking) the original figure. Engage in Activity 3.2 to explore more about similarity of figures.

Activity 3.2: Using Computer Applications to Demonstrate Similarity of Figures

1. Use a software of your choice (MS Word, Desmos, GeoGebra) to create various shapes.
2. Copy and paste the shapes into a new workspace, then resize by enlarging or reducing the figures.
3. Observe the changes in size in terms of sides and angles and discuss your findings based on the similarity of figures.

Similar Triangles

Triangles are similar when their corresponding angles are equal and their corresponding sides are proportional. Consider the pair of triangles shown in Figure 3.2.

Figure 3.2: Similar triangles

From Figure 3.2, ∠CAB of ΔABC corresponds to ∠RDQ of ΔPQR and each measures 50°, ∠ABC corresponds to ∠PQR and each measures 60°, and ∠BCA corresponds to ∠QRP and each measures 70°.

Since the corresponding angles are equal, then the two triangles are similar. Also, AB corresponds to PQ, BC corresponds to QR, and CA corresponds to RP. The ratios of the corresponding sides are given by:

AB/PQ = 6 cm/4.5 cm = 4/3

BC/QR = 4 cm/3 cm = 4/3

CA/RP = 5 cm/3.75 cm = 4/3

Therefore, AB/PQ = BC/QR = CA/RP = 4/3.

Since the corresponding sides are proportional, the two triangles are similar. Therefore, ΔABC is similar to ΔPQR and is denoted by ΔABC ~ ΔPQR. The symbol ~ means "similar to."

Similar triangles (polygons) are named according to the order of their vertices. For instance, in ΔABC and ΔPQR in Figure 3.2, it can be deduced from the order of the vertices that AB of the first triangle corresponds to PQ of the second triangle. Consequently, BC corresponds to QR, and AC corresponds to PR.

Example 3.1

Given that ΔSLK ~ ΔNFR, identify all the corresponding angles and the corresponding sides.

Solution:

Using the order of vertices of the two similar triangles:

Corresponding angles: ∠SLK corresponds to ∠NFR, ∠KSL corresponds to ∠RNF, ∠LKS corresponds to ∠FRN.

Corresponding sides: LS of ΔSLK corresponds to FN of ΔNFR, SK of ΔSLK corresponds to NR of ΔNFR, and KL of ΔSLK corresponds to RF of ΔNFR.

Example 3.2

Given that ΔABC ~ ΔPQR, find the value of ∠ABC if:

(a) ∠BAC = 120° and ∠PRQ = 25°

(b) ∠QPR + ∠BCA = 145°

Solution:

Consider the following figures of ΔABC and ΔPQR.

Triangles ABC and PQR

(a) Since ∠QRP corresponds to ∠BCA, then ∠QRP = ∠BCA = 25°.

In ΔABC, ∠ABC + ∠BCA + ∠BAC = 180° (sum of interior angles in a triangle)

∠ABC + 25° + 120° = 180°

Thus, ∠ABC = 180° - 145° = 35°.

Therefore, ∠ABC = 35°.

(b) Since ∠QPR corresponds to ∠BAC, then ∠QPR = ∠BAC.

Thus, ∠QPR + ∠BCA = ∠BAC + ∠BCA = 120° + 25° = 145°.

But ∠ABC + ∠BAC + ∠ACB = 180° (sum of interior angles in a triangle). It follows that

145° + ∠ABC = 180°

∠ABC = 180° - 145° = 35°

Therefore, ∠ABC = 35°.

Similarity Theorems

Similarity theorems are used to solve problems based on similar triangles.

If ΔABC ~ ΔXYZ, it means:

• Corresponding angles are equal. That is ∠ABC = ∠XYZ, ∠CAB = ∠ZXY, and ∠BCA = ∠YZX.
• The ratio of corresponding sides is equal. That is, AB/XY = BC/YZ = AC/XZ

Note: The value of the constant ratio obtained from the ratio of corresponding sides of similar figures is called the constant of proportionality or scale factor.

To prove similarity of two triangles, only one of the following conditions is sufficient:

1. Two pairs of corresponding angles are equal. This introduces an Angle-Angle (AA) similarity theorem.
2. Three corresponding sides are proportional. This is described by a Side-Side-Side (SSS) similarity theorem.
3. Two corresponding sides are proportional and one pair of corresponding angles (formed by these sides) are equal is described by the Side-Angle-Side (SAS) similarity theorem.

Angle-Angle (AA) Similarity Theorem

The AA similarity theorem states that, two triangles are similar if two pairs of corresponding angles are equal. This implies that, if two pairs of angles are equal, the third pair of angles will also be equal.

Figure 3.3: Similar triangles by the Angle-Angle theorem

In Figure 3.3, it can be observed that, ∠BAC = ∠QPR, ∠ACB = ∠PRQ.

Thus, ∠ABC = ∠PQR (third pairs of angles of triangles).

Therefore, ΔABC ~ ΔPQR and AB/PQ = BC/QR = AC/PR.

Example 3.3

For each pair of triangles in the following figures, determine whether they are similar or not. Indicate the similarity theorem used to support your argument.

Three pairs of triangles for similarity analysis

Solution:

(a) Required to prove that: ΔOUL ~ ΔMUO

Proof:

∠LOU = ∠OMU = 40° (given)

∠OUL = ∠MUO = 90° (given)

∠ULO = ∠UOM = 50° (third angles in triangles)

Therefore, ΔOUL ~ ΔMUO (by AA-similarity theorem).

(b) ΔCBA and ΔRPQ are not similar because the corresponding angles are not equal.

(c) Required to prove that ΔSVU ~ ΔSWT

Proof:

The ratio of the corresponding sides:

SW/SV = 10 cm/13 cm = 3/4

ST/SU = 9 cm/12 cm = 3/4

TW/UV = 6 cm/8 cm = 3/4

Therefore, ΔSVU ~ ΔSWT (by SSS-Similarity theorem).

Side-Side-Side (SSS) Similarity Theorem

The SSS similarity theorem states that, two triangles are similar if three pairs of corresponding sides are proportional.

Figure 3.4: Similar triangles by the Side-Side-Side theorem

In Figure 3.4, it can be observed that, if the ratio of their sides are such that LM/DE = LN/DF = MN/EF, then ΔLMN ~ ΔDEF.

Therefore, ∠LMN = ∠DEF, ∠MNL = ∠EFD and ∠MLN = ∠EDF.

Side-Angle-Side (SAS) Similarity Theorem

The SAS similarity theorem states that, two triangles are similar if two pairs of their corresponding sides are proportional and one pair of corresponding angles (formed by these sides) are equal.

Figure 3.5: Similar triangles by Side-Angle-Side theorem

In Figure 3.5, if AB/PQ = AC/PR and ∠BAC = ∠QPR, then ΔABC ~ ΔPQR.

Therefore, ∠ABC = ∠PQR and ∠ACB = ∠PRQ.

Example 3.4

In the following figure, find the constant of proportionality needed to obtain a pair of similar triangles, if CE:DE = 1.2:1. Name this pair of similar triangles.

Triangles with proportional sides

Solution:

The ratio of lengths of corresponding sides are given by:

CE/DE = 1.2/1 = 12 cm/10 cm = 6/5

AE/BE = 12 cm/10 cm = 6/5

AC/BD = 18 cm/15 cm = 6/5

Thus, CE/DE = AE/BE = AC/BD = 6/5.

Therefore, ΔACE ~ ΔBDE (corresponding sides are proportional) and 6/5 is the constant of proportionality.

Activity 3.3: Exploring Similarity Through Perpendicular Bisectors

1. Use a geometrical software of your choice (or any other method) to draw a triangle ΔPQR.
2. Construct a line perpendicular to QR through point P, and label the intersection as S.
3. Measure segments PQ, PS, PR, and RS.
4. Calculate the ratios PQ/PS and PR/RS.
5. Drag point C until PQ/PS = PR/RS.
6. For which value of ∠PQR are ΔPQR and ΔPRS similar?
7. Share your findings about the theorem that supports your answer in task 6, and explain how it applies to the similarity observed in this activity.

Chapter Summary

1. AA Similarity theorem: If the correspondence between triangles is such that two pairs of corresponding angles are equal, then the triangles are similar.
2. SAS Similarity theorem: If the correspondence between two triangles is such that the lengths of two pairs of corresponding sides are proportional and the included angles are equal, then the triangles are similar.
3. SSS Similarity theorem: If the correspondence between two triangles is such that the lengths of corresponding sides are proportional, then the triangles are similar.
4. Similar figures have the same shape.
5. In similar figures, the ratios of the lengths of corresponding sides are equal. That is, corresponding sides are proportional. The value of the ratio is called the constant of proportionality or scale factor.
6. The symbol used to indicate similarity between figures is "~".

Revision Exercise 3

1. In the following figure determine the pairs of equal angles and the proportional sides which make ΔAEB ~ ΔDEC.
2. Name two triangles similar to ΔABC in the following figure.
3. If ΔLMN ~ ΔPQR and ΔPQR ~ ΔABC, mention equal angles in ΔABC and ΔLMN, hence mention all proportional sides in ΔPQR and ΔABC. Find the ratio of proportionality for the similarity of ΔPQR and ΔABC.
4. A triangle ΔBC is such that CA is extended to X and BA is extended to Y so that XY ∥ BC. Prove that AX/AC = XY/CB.
5. A trapezium PQRS is such that PQ ∥ RS. If the diagonals intersect at X, prove that PX/RX = QX/SX.
6. Name a triangle which is similar to ΔABC in the following figure.
7. Use similarity to find the value of ED in the following figures.
8. If ΔPQR ~ ΔLMN, PR = 20 dm, NL = 10 dm, NM = 12 dm and LM = 9 dm, find the lengths of the other sides of ΔPQR.
9. Prove that any two equilateral triangles are similar.
10. In ΔPMT and ΔQNS, ∠PMT = ∠QSN = ∠MTP = ∠QNS. Prove that PM × NS = QS × MT.

Chapter Four: Algebra

Introduction

Skills in solving problems systematically, logically, and making connections between variables are essential in daily life. Such skills can be developed and applied by learning algebra. In this chapter, you will learn about binary operations, quadratic expressions, and solving quadratic equations. The competencies developed will enable you to solve daily life problems such as budgeting, comparing prices, planning for a journey and figuring proportions in cooking. Also, algebra is useful in other subjects and fields such as science, engineering and technology and many other applications.

Think

The role algebra plays in real life.

Binary Operations

A binary operation is a rule for combining two quantities to produce a unique quantity from the same set. It makes use of symbols that represent one or more operations.

The symbols used in binary operations are not standard, this means, they do not represent a specific operation. A symbol may be used in a certain operation in an expression, and yet have a different meaning in another expression.

A binary operation may be denoted by symbols such as Δ and *, depending on the instructions given for the operation.

The instructions may be given in words or by symbols. Binary operations may involve the basic arithmetic operations such as addition (+), subtraction (-), multiplication (×), and division (÷) depending on the given definition.

Activity 4.1: Exploring Binary Operations

1. Suppose you have 3 apples and 4 oranges. What is the total number of fruits? How would you represent this operation?
2. How changing the number of apples or oranges affects the total fruit count?
3. If an apple is sold at TShs 950 and an orange is sold for Tshs 200, calculate the total cost of buying all the fruits.

Example 4.1

If a * b = 5a - b, find the value of 6 * 9.

Solution:

From a * b = 5a - b, it implies that,

6 * 9 = 5 × 6 - 9

= 30 - 9

= 21

Therefore, 6 * 9 = 21.

Example 4.2

If a * b = a² - b, find the value of y given that 4 * (2 * y) = 4.

Solution:

Given a * b = a² - b.

Start with the operation in the brackets.

4 * (2 * y) = 4 * (2² - y)

= 4 * (4 - y)

= 4² - (4 - y)

= 16 - 4 + y

= 12 + y

Thus, 12 + y = 4

y = -8

Therefore, y = -8.

Example 4.3

Given that x * y = 4x + 6y, find the value of 6 * 4.

Solution:

Given x * y = 4x + 6y. It follows that,

6 * 4 = (4 × 6) + (6 × 4)

= 24 + 24

= 48

Therefore, 6 * 4 = 48.

Transposition of Formulae

A formula is an equation which shows how variables are related. For example, the formula which is used to find the circumference, C of a circle is given by C = πd, where C is expressed in terms of π and d. In this case, C is the subject of the formula. It is also possible to express d as the subject of the formula, that is, d = C/π.

The procedure of expressing a formula in different ways is called transposition of formula. The following are some examples of formulae:

(a) A = lb

(b) V = πr²h

(c) I = PRT/100

(d) A = ½(a + b)h

(e) T = 2π√(l/g)

(f) y = mx + c

(g) s = ut + ½at²

Example 4.4

Given that y = mx + c, make m the subject of the formula.

Solution:

Given the formula y = mx + c. ...(i)

Subtract c from both sides of equation (i) to get y - c = mx. ...(ii)

Divide equation (ii) by x throughout to get m = (y - c)/x

Therefore, m = (y - c)/x, provided x ≠ 0.

Example 4.5

The volume, V of a cylinder with a base of radius r and height h is given by V = πr²h. Make r the subject of the formula.

Solution:

Given the formula V = πr²h. ...(i)

Divide both sides of equation (i) by πh to get V/(πh) = r². ...(ii)

Apply the square root to both sides of equation (ii) to get √(V/(πh)) = r.

Therefore, r = √(V/(πh)).

Quadratic Expressions

An expression whose highest exponent (degree) of the variable is 2 is called a quadratic expression.

The following are examples of quadratic expressions:

(i) 4z² + 3, the highest exponent (or degree) of z is 2.

(ii) 6y² + y, the highest exponent of y is 2.

(iii) 3n² - 2n + 1, the highest exponent of n is 2.

In general, a quadratic expression has the form ax² + bx + c, where a ≠ 0 and a, b, and c are real numbers. The term ax² is called the quadratic term, where a is the coefficient of x². The term bx is called the linear term (middle term), where b is the coefficient of x and c is the constant term.

For example, in the quadratic expression 4x² - 6x + 7, the coefficient of x² is 4, the coefficient of x is -6 and the constant is 7.

Activity 4.2: Deducing Quadratic Expressions from Real Life

1. A garden designer plans to design different gardens in a compound. During an investigation, he noticed that the length of each garden is 4 m longer than its width. Deduce the simplest possible expression of the area of the garden by expanding the factors.
2. Explore various sources to learn more about how you can expand such results.
3. Make a presentation to demonstrate how you arrived at your conclusion.

From activity 4.2, it can be observed that, quadratic expressions can be derived from the product of two linear expressions. For instance in Figure 4.1, if the width of a rectangle is (y + 1) unit and its length is (2y + 3) unit, then the area is (2y + 3)(y + 1) square units.

Figure 4.1: Rectangle PQRS which demonstrate an area as a product of factors

Thus, 2y² + 5y + 3 is the expanded form of (2y + 3)(y + 1).

Therefore, (2y + 3)(y + 1) = 2y² + 5y + 3.

Example 4.6

Expand (z + 2)(2z - 3).

Solution:

(z + 2)(2z - 3) = z(2z - 3) + 2(2z - 3)

= 2z² - 3z + 4z - 6

= 2z² + z - 6

Therefore, (z + 2)(2z - 3) = 2z² + z - 6.

Factorization of Quadratic Expressions

Factorization is the process of writing an expression as a product of its factors. Numbers and algebraic expressions can be expressed in terms of factors.

For instance, 10 can be expressed as a product of 2 and 5 or 1 and 10.

The expression x² - x - 12 can be expressed as a product of two linear factors x - 4, and x + 3. That is, x² - x - 12 = (x - 4)(x + 3).

Quadratic expressions can be factored using three main methods which are splitting the middle term, difference of two squares and perfect squares.

Factorization by Splitting the Middle Term

Factorization by splitting the middle term involves the splitting of the middle term into two terms. For instance, in the quadratic expression ax² + bx + c, the middle term is bx. To split the middle term, the following steps can be used:

1. Find two numbers whose sum is equal to b and whose product is equal to ac.
2. To find the two numbers in (i), list all the factors of ac and determine a pair whose sum is b.

Example 4.7

Factorize the expression 2x² + 7x + 6 by splitting the middle term.

Solution:

The coefficients are 2, 7 and the constant term is 6. That is, a = 2, b = 7 and c = 6. So ac = 2 × 6 = 12.

Find the pair of factors of 12 whose sum is 7.

The pairs of factors of 12 are:

• The sum of 1 and 12 is 13
• The sum of -1 and -12 is -12
• The sum of 2 and 6 is 8
• The sum of -2 and -6 is -8
• The sum of 3 and 4 is 7
• The sum of -3 and -4 is -7

Therefore, the correct choice is 3 and 4.

So the terms are 3x and 4x.

2x² + 7x + 6 = 2x² + 4x + 3x + 6

= (2x² + 4x) + (3x + 6)

= 2x(x + 2) + 3(x + 2)

= (x + 2)(2x + 3)

Therefore, 2x² + 7x + 6 = (x + 2)(2x + 3).

Quadratic Equations

In the previous section, you learned about quadratic expressions. A quadratic equation is formed when a quadratic expression is equal to a specific value. Quadratic equations take the general form ax² + bx + c = 0. The solutions of such an equation are also known as roots or zeros. There are also several methods for solving quadratic equations, including factorization, completing the square, and the general quadratic formula.

Solving Quadratic Equations by Factorization

To solve quadratic equations by factorization, the following steps can be used:

1. Write the equation in standard form: ax² + bx + c = 0.
2. Rewrite the quadratic expression ax² + bx + c as a product of two linear factors.
3. Set each factor equal to zero and solve for x.

Note: If a quadratic equation is expressed as a product of two linear factors, say (x + a)(x + b) = 0, where a and b are constants, then x + a = 0 or x + b = 0 or both factors are equal to zero. This is called Zero factor theorem which states that, if a product of two or more factors is zero, then at least one of the factors must be zero.

Example 4.8

Solve the equation (x + 4)(x - 3) = 0.

Solution:

If (x + 4)(x - 3) = 0, then either x + 4 = 0 or x - 3 = 0.

Therefore, x = -4 or x = 3.

Example 4.9

Solve the quadratic equation 3x² - x = 0.

Solution:

3x² - x = 0.

x(3x - 1) = 0

Either, x = 0 or 3x - 1 = 0.

Therefore, x = 0 or x = 1/3.

Solving Quadratic Equations by Completing the Square

Completing the square is a technique used to solve quadratic equations by transforming the left-hand side of the equation into a perfect square. The steps involved are demonstrated in the following examples.

Example 4.10

What must be added to x² + 10x to make the expression a perfect square?

Solution:

The term to be added must be the square of half the coefficient of x.

The coefficient of x is 10.

Half of 10 is ½(10) = 5.

The square of 5 is 25.

Therefore, 25 must be added to x² + 10x to make it a perfect square.

Solving Quadratic Equations by Using Quadratic General Formula

The method of solving quadratic equations by completing the square can be used to derive the general formula for solving quadratic equations as follows.

From, ax² + bx + c = 0, a ≠ 0.

Divide each term by a to get, x² + (b/a)x + c/a = 0

Subtract c/a from both sides to obtain, x² + (b/a)x = -c/a

Add (b/(2a))² to both sides of the equation.

That is, x² + (b/a)x + (b/(2a))² = -c/a + (b/(2a))²

Factorize the left-hand side and simplify the right-hand side to obtain,

(x + b/(2a))² = (b² - 4ac)/(4a²)

Take the square root of both sides to get,

x + b/(2a) = ±√(b² - 4ac)/(2a)

Subtract b/(2a) from both sides to get,

x = [-b ± √(b² - 4ac)]/(2a)

Therefore, if ax² + bx + c = 0, where a ≠ 0, then x = [-b ± √(b² - 4ac)]/(2a) provided that b² ≥ 4ac is the general quadratic formula.

Example 4.11

Solve 6x² + 11x + 3 = 0 using the quadratic formula.

Solution:

Comparing ax² + bx + c = 0 with 6x² + 11x + 3 = 0 gives, a = 6, b = 11, and c = 3.

Substitute these values into the quadratic formula x = [-b ± √(b² - 4ac)]/(2a).

x = [-11 ± √((11)² - 4 × 6 × 3)]/(2 × 6)

x = [-11 ± √(121 - 72)]/12

x = [-11 ± √49]/12

= [-11 ± 7]/12

x = -4/12 or x = -18/12

Therefore, x = -1/3 or x = -3/2.

Chapter Summary

1. A binary operation is rule for combining two quantities to produce a unique quantity from the same set.
2. An expression whose highest power of the variable is 2 is called a quadratic expression.
3. Factorization is the process of writing an expression as a product of two or more factors.
4. A quadratic equation is formed when a quadratic expression is equal to a specific value, typically zero.
5. The solutions of quadratic equations are known as roots.
6. Methods for solving quadratic equations include:
   • Factorization method
   • Completing the square method
   • Quadratic formula: x = [-b ± √(b² - 4ac)]/(2a)

Revision Exercise 4

1. Solve each of the following quadratic equations by the factorization method:
   1. x² + 3x = 0
   2. 3x² - 15x = 0
   3. x = 3x²
   4. 2x² = 3x
   5. x(5 - x) = 0
2. Solve each of the following quadratic equations by completing the square:
   1. x² + 6x + 7 = 0
   2. x² - 11x + 1 = 0
   3. x² = 7x - 7
   4. 2x² - 10x + 7 = 0
3. Use the quadratic formula to solve each of the following equations:
   1. 4x² - 4x + 1 = 0
   2. 5x² + 12x + 3 = 0
   3. (3x - 2)(2x - 5) = 5x(x - 2)
4. A man is 4 times as old as his son. In 4 years to come the product of their ages will be 520. Find the son's present age.
5. Find two consecutive numbers such that the sum of their squares is equal to 145.
6. A piece of wire 56 cm long is bent to form a rectangle of area 171 cm². Find the dimensions of the rectangle.

Chapter Five: Exponents and Radicals

Introduction

Numbers can be expressed in different ways, each of which illustrates how mathematical problems can be approached and solved. Simplifying complex problems with challenging equations and solving exponential such as growth and electrical networks may require an understanding of exponents and radicals. In this chapter, you will learn the laws of exponents, perform operations on radicals, and rationalize denominators. The competencies developed will enable you to express numbers in short form and apply the knowledge to describe scientific phenomena such as population growth, magnitude of earthquakes, rates of spread of diseases and rate of growth or decay of bacteria, and many other real-life applications.

Think

Solving mathematical expressions and equations involving powers and roots without the concept of radicals and exponents.

Numbers can be expressed in different ways depending on the context and how they are used. Some of these ways include the use of exponents and radicals. Engage in Activity 5.1 to explore different ways of expressing numbers.

Activity 5.1: Expressing Numbers in Different Ways

1. Choose five whole numbers and express them in different ways you know. You can consult various sources such as the internet to learn such ways.
2. Study any unique pattern and provide reasons regarding your observations.
3. Share your conclusion with other students through a method of your choice.

Exponents

From Activity 5.1, one may have explored various ways used to express numbers. One of the common methods of expressing numbers is by multiplying the same or different numbers several times. Study the following examples.

(i) 3 = 1 × 3

(ii) 4 = 2 × 2

(iii) 8 = 2 × 2 × 2

(iv) 12 = 2 × 2 × 3

(v) 81 = 3 × 3 × 3 × 3

These are expanded form of numbers.

From the given examples, the same numbers that are multiplied several times can be expressed in short form. Some numbers can be expressed in short form by counting the number of times the same number is multiplied and writing the result on the top right hand of the multiplied number. For example:

(i) 4 = 2 × 2 = 2² since 2 is multiplied repeatedly two times.

(ii) 8 = 2 × 2 × 2 = 2³ since 2 is multiplied repeatedly three times.

(iii) 12 = 2 × 2 × 3 = 2² × 3 since 2 is multiplied repeatedly two times, and three is multiplied once.

(iv) 81 = 3 × 3 × 3 × 3 = 3⁴ since 3 is multiplied repeatedly four times.

Numbers expressed in the forms 2², 2³ and 3⁴ are called powers. The numbers which are multiplied several times are called bases, and the number of times any number is multiplied by itself is called exponent. For example, 3⁵:

3⁵ is the power, where

3 is the base and

5 is the exponent.

Likewise, 3⁵ is read as the "fifth power of 3" or "3 raised to exponent five" and 2³ is read as the "third power of 2" or "2 raised to exponent 3".

Activity 5.2: Identifying Features of Powers, Base, and Exponent of the Numbers

1. Choose any counting number greater than 1 and multiply it several times.
2. Write the product in the power form and identify the power, base, and exponent.
3. Share your work and justify your answers.

Generally, aⁿ is a number written in exponential form, where a is the base and n is the exponent. The number aⁿ is called a power.

Example 5.1

Write each of the following expressions in exponential form:

(a) 5 × 5 × 5 × 5

(b) k × k × k × k × k

(c) (-3) × (-3) × (-3) × (-3) × (-3)

(d) m × m × m × ... × m (n times)

Solution:

(a) 5 × 5 × 5 × 5 = 5⁴

(b) k × k × k × k × k = k⁵

(c) (-3) × (-3) × (-3) × (-3) × (-3) = (-3)⁵

(d) m × m × m × ... × m = mⁿ

Laws of Exponents

There are three groups of exponents namely positive, negative, and zero exponents. Simplification of the exponents are usually based on the four laws called multiplication, division, power, and zero power.

Multiplication Law for Exponents

Consider the product of two exponential numbers having the same base with positive exponents such as 5³ × 5². The exponential numbers can be written in expanded form as follows:

5² = 5 × 5 and 5³ = 5 × 5 × 5

Therefore, 5² × 5³ = (5 × 5) × (5 × 5 × 5) = 5 × 5 × 5 × 5 × 5

Count how many times 5 is multiplied by itself. The answer is five times, which gives the exponent. Thus, 5 is multiplied by itself five times. The exponent 5 can also be obtained by adding the exponents from each base. That is, 5² × 5³ = 5²⁺³ = 5⁵.

Similarly, if a is any number, then a⁴ × a² = (a × a × a × a) × (a × a) = a⁴⁺² = a⁶

Therefore, a⁴ × a² = a⁶

Generally, if x is any non-zero number with positive exponents m and n, then xᵐ × xⁿ = xᵐ⁺ⁿ. This is called the multiplication law of exponents.

Example 5.2

Simplify each of the following by using multiplication law of exponents:

(a) 7⁶ × 7⁸

(b) 6⁴ × 6³ × 6⁴

(c) (5/9)³ × (5/9)⁴

(d) (0.45)² × (0.45)⁴ × (0.45)¹²

Solution:

By using multiplication law of exponents, it implies that,

(a) 7⁶ × 7⁸ = 7⁶⁺⁸ = 7¹⁴

(b) 6⁴ × 6³ × 6⁴ = 6⁴⁺³⁺⁴ = 6¹¹

(c) (5/9)³ × (5/9)⁴ = (5/9)³⁺⁴ = (5/9)⁷

(d) (0.45)² × (0.45)⁴ × (0.45)¹² = (0.45)²⁺⁴⁺¹² = (0.45)¹⁸

Division Law of Exponents

Consider the following example of dividing exponential numbers with the same base like 7⁵ ÷ 7³. The solution can be obtained as follows:

7⁵ ÷ 7³ = 7⁵/7³ = (7 × 7 × 7 × 7 × 7)/(7 × 7 × 7) = 7 × 7 = 7²

Alternatively, if the bases of the numerator and denominator are the same, then subtract the denominator exponent from numerator exponent.

That is, 7⁵/7³ = 7⁵⁻³ = 7²

Similarly, a⁷/a⁴ = (a × a × a × a × a × a × a)/(a × a × a × a) = a⁷⁻⁴ = a³ where a ≠ 0

Therefore, a⁷/a⁴ = a⁷⁻⁴ = a³

Generally, xᵐ/xⁿ = xᵐ⁻ⁿ where x ≠ 0. That is when exponential numbers of the same base are divided, subtract the exponent of the divisor from the exponent of the dividend.

Example 5.3

Simplify each of the following expressions:

(a) 3²⁰/3¹²

(b) a²⁸/a¹⁷

(c) (0.47)¹⁰/(0.47)⁶

Solution:

(a) 3²⁰/3¹² = 3²⁰⁻¹² = 3⁸

(b) a²⁸/a¹⁷ = a²⁸⁻¹⁷ = a¹¹

(c) (0.47)¹⁰/(0.47)⁶ = (0.47)¹⁰⁻⁶ = (0.47)⁴

Zero Exponent

An expression 7³/7³ can be simplified by using the division law of exponents as follows:

7³/7³ = (7 × 7 × 7)/(7 × 7 × 7) = 1 ...(1)

Alternatively, 7³/7³ = 7³⁻³ = 7⁰ ...(2)

From equations (1) and (2), it follows that 7⁰ = 1

Similarly, if m ≠ 0, then m⁴/m⁴ = (m × m × m × m)/(m × m × m × m) = 1 ...(3)

Thus, m⁴/m⁴ = m⁴⁻⁴ = m⁰ ...(4)

Therefore, m⁰ = 1 by equations (3) and (4).

Generally, if x is any non-zero number, then x⁰ = 1. In this case 0⁰ is not defined.

Negative Exponents

The expression 8³/8⁵ can be simplified using the division law of exponents as follows:

8³/8⁵ = (8 × 8 × 8)/(8 × 8 × 8 × 8 × 8) = 1/(8 × 8) = 1/8² ...(1)

Therefore, 8³/8⁵ = 1/8².

Also, 8³/8⁵ = 8³⁻⁵ = 8⁻² ...(2)

Comparing equations (1) and (2), gives 8⁻² = 1/8².

Similarly, k⁵/k⁸ = (k × k × k × k × k)/(k × k × k × k × k × k × k × k) = 1/(k × k × k) = 1/k³

Thus, k⁵/k⁸ = k⁻³, for k ≠ 0.

Therefore, k⁻³ = 1/k³.

In general, x⁻ⁿ = 1/xⁿ and xⁿ = 1/x⁻ⁿ, for x ≠ 0.

When n = 1, x⁻ⁿ = x⁻¹ = 1/x¹ = 1/x.

If x ≠ 0, then 1/x is called the reciprocal of x, which can also be written as x⁻¹.

Radicals

A rational exponent, x which can be expressed as ⁿ√x is known as a radical. In ⁿ√x, n is an index, √ is the radical symbol and x is the radicand. The symbol √ is also called a surd.

The expression ⁿ√x is also known as the nᵗʰ root of x, which is a number that, when multiplied n times, gives the original number x. For example:

(a) √4 = (4)¹ᐟ² = (2²)¹ᐟ² = 2

2 is the square root of 4.

(b) ³√343 = (343)¹ᐟ³ = (7³)¹ᐟ³ = 7

7 is the cube root of 343.

Note: The square root of a negative real number does not exist in the set of real numbers.

Example 5.4

Find the square root of 196.

Solution:

Factorize 196 in terms of prime factors:

196 = 2 × 2 × 7 × 7 = 2² × 7²

So, 196 = 2² × 7².

Apply the radical sign in both sides of the equation and simplify to obtain,

√196 = √(2² × 7²) = √2² × √7² = (2²)¹ᐟ² × (7²)¹ᐟ² = 2 × 7 = 14

Therefore, √196 = 14.

Addition and Subtraction of Radicals

Two or more radicals can be added or subtracted if they are alike. Radicals which are alike are those with the same indices and radicand. This means that radicals of the same index can be added or subtracted, just as it is done in algebraic expressions. The radicals √2 and ³√2 cannot be added or subtracted because have unlike indices. Before adding or subtracting radicals, first simplify the terms if possible.

Example 5.5

Simplify each of the following expressions:

(a) √2 + √2

(b) 2√3 + 3√3

(c) 2³√81 + ³√24

(d) ⁴√2 + ⁴√32

Solution:

(a) √2 + √2 = 1√2 + 1√2 = 2√2

(b) 2√3 + 3√3 = 5√3

(c) 2³√81 + ³√24 = 2³√(3×3×3×3) + ³√(2×2×2×3) = 2×3³√3 + 2³√3 = 6³√3 + 2³√3 = 8³√3

(d) ⁴√2 + ⁴√32 = ⁴√2 + ⁴√(2×2×2×2×2) = ⁴√2 + 2×⁴√2 = ⁴√2 + 2⁴√2 = 3⁴√2

Multiplication of Radicals

Multiplication of radicals involves multiplying the numbers inside the radicals and then simplifying if possible. Thus, when multiplying radicals, the indices must be the same. Otherwise, it is impossible.

Generally, ⁿ√a × ⁿ√b = ⁿ√(ab)

Example 5.6

Simplify each of the following expression products:

(a) √3 × √5

(b) √2 × √32

(c) √20 × √28

(d) √12(√3 + √5)

(e) (2√3 - √2) × (√3 + 3√2)

Solution:

(a) √3 × √5 = √15

(b) Given that √2 × √32, it follows, √32 = √(2×2×2×2×2) = 4√2

Thus, √2 × √32 = √2 × 4√2 = 4 × √2 × √2 = 4 × 2 = 8

(c) Given √20 × √28, it follows that, √20 = √(2×2×5) = 2√5, √28 = √(2×2×7) = 2√7

Thus, √20 × √28 = 2√5 × 2√7 = 2×2×√5×√7 = 4×√35 = 4√35

(d) √12(√3 + √5) = √(4×3)(√3 + √5) = 2√3(√3 + √5) = 2√3×√3 + 2√3×√5 = 2×3 + 2√15 = 6 + 2√15

(e) Given (2√3 - √2) × (√3 + 3√2)

Expand as follows:

(2√3 - √2) × (√3 + 3√2) = (2√3×√3) + (2√3×3√2) + (-√2×√3) + (-√2×3√2)

= 2√9 + 6√6 - √6 - 3√4

= 2×3 + 5√6 - 3×2

= 6 + 5√6 - 6

= 5√6

Therefore, (2√3 - √2) × (√3 + 3√2) = 5√6.

Chapter Summary

1. Numbers can be expressed in exponential form as aⁿ, where a is the base and n is the exponent.
2. Laws of exponents include:
   • Multiplication law: xᵐ × xⁿ = xᵐ⁺ⁿ
   • Division law: xᵐ/xⁿ = xᵐ⁻ⁿ
   • Power law: (xᵐ)ⁿ = xᵐⁿ
   • Zero exponent: x⁰ = 1 (x ≠ 0)
   • Negative exponents: x⁻ⁿ = 1/xⁿ
3. A radical is an expression of the form ⁿ√x, where n is the index and x is the radicand.
4. Radicals can be added, subtracted, or multiplied if they have the same index.
5. The relationship between exponents and radicals: x¹ᐟⁿ = ⁿ√x
6. Rationalizing denominators involves eliminating radicals from the denominator of a fraction.

Revision Exercise 5

1. Simplify each of the following expressions:
   1. 5² × 5³
   2. (3²)³
   3. 7⁵/7²
   4. a⁴ × a⁷
   5. (2³ × 3²)²
2. Express with positive exponents:
   1. 5⁻³
   2. x⁻²y³
   3. (2a)⁻²
   4. 3⁻¹ + 2⁻²
3. Simplify each radical expression:
   1. √50
   2. ³√54
   3. √18 + √32
   4. 2√27 - √75
4. Multiply and simplify:
   1. √3 × √12
   2. √8 × √18
   3. (2√5 + √3)(√5 - 3√3)
5. Solve the exponential equations:
   1. 2ˣ = 16
   2. 3ˣ⁺¹ = 27
   3. 4ˣ = 8ˣ⁻¹

Chapter Six: Logarithms

Introduction

Logarithms are powerful mathematical tools that help simplify complex calculations, particularly those involving very large or very small numbers. In this chapter, you will learn about the standard form of numbers, the concept of logarithms, laws of logarithms, and their practical applications. The competencies developed will enable you to solve problems in various fields such as science, engineering, finance, and technology where exponential relationships occur.

Think

How would you calculate 2.5 × 10³⁴ × 3.8 × 10²⁷ without using a calculator? Logarithms provide the answer!

Standard Form of Numbers

Standard form (also known as scientific notation) is a way of writing very large or very small numbers in a compact form. A number is in standard form when it is written as:

A × 10ⁿ

where 1 ≤ A < 10 and n is an integer

Example 6.1

Write the following numbers in standard form:

(a) 450,000,000

(b) 0.00000032

(c) 67,500

Solution:

(a) 450,000,000 = 4.5 × 10⁸

(b) 0.00000032 = 3.2 × 10^-7

(c) 67,500 = 6.75 × 10⁴

Example 6.2

Convert the following from standard form to ordinary numbers:

(a) 2.8 × 10⁶

(b) 5.3 × 10^-4

(c) 7.15 × 10³

Solution:

(a) 2.8 × 10⁶ = 2,800,000

(b) 5.3 × 10^-4 = 0.00053

(c) 7.15 × 10³ = 7,150

Concept of Logarithms

A logarithm is the exponent or power to which a base must be raised to produce a given number.

If a^x = N, then log_aN = x

where:

• a is the base (a > 0, a ≠ 1)
• N is the number (N > 0)
• x is the logarithm

Example 6.3

Express the following in logarithmic form:

(a) 2³ = 8

(b) 10² = 100

(c) 5^-2 = 1/25

Solution:

(a) 2³ = 8 ⇒ log₂8 = 3

(b) 10² = 100 ⇒ log₁₀100 = 2

(c) 5^-2 = 1/25 ⇒ log₅(1/25) = -2

Example 6.4

Express the following in exponential form:

(a) log₃81 = 4

(b) log₁₀0.001 = -3

(c) log₂32 = 5

Solution:

(a) log₃81 = 4 ⇒ 3⁴ = 81

(b) log₁₀0.001 = -3 ⇒ 10^-3 = 0.001

(c) log₂32 = 5 ⇒ 2⁵ = 32

Note: There are two commonly used bases for logarithms:

• Common logarithms: Base 10, written as log₁₀ or simply log
• Natural logarithms: Base e (e ≈ 2.71828), written as log_e or ln

Laws of Logarithms

Logarithms follow specific laws that make calculations easier. These laws are derived from the laws of exponents.

Product Law of Logarithms

log_a(MN) = log_aM + log_aN

Proof:

Let log_aM = x and log_aN = y

Then M = a^x and N = a^y

MN = a^x × a^y = a^x+y

Therefore, log_a(MN) = x + y = log_aM + log_aN

Quotient Law of Logarithms

log_a(M/N) = log_aM - log_aN

Proof:

Let log_aM = x and log_aN = y

Then M = a^x and N = a^y

M/N = a^x/a^y = a^x-y

Therefore, log_a(M/N) = x - y = log_aM - log_aN

Power Law of Logarithms

log_aMⁿ = n log_aM

Proof:

Let log_aM = x

Then M = a^x

Mⁿ = (a^x)ⁿ = a^nx

Therefore, log_aMⁿ = nx = n log_aM

Example 6.5

Use the laws of logarithms to simplify:

(a) log₁₀4 + log₁₀25

(b) log₂48 - log₂3

(c) 3 log₅2

Solution:

(a) log₁₀4 + log₁₀25 = log₁₀(4 × 25) = log₁₀100 = 2

(b) log₂48 - log₂3 = log₂(48/3) = log₂16 = 4

(c) 3 log₅2 = log₅2³ = log₅8

Logarithms of Numbers

To find the logarithm of a number, we can use logarithm tables or calculators. The logarithm consists of two parts:

log₁₀N = characteristic + mantissa

• Characteristic: The integer part (can be positive, negative, or zero)
• Mantissa: The decimal part (always positive)

Example 6.6

Find the characteristic for the following numbers:

(a) 347.5

(b) 0.00456

(c) 25.8

Solution:

(a) 347.5 = 3.475 × 10² ⇒ characteristic = 2

(b) 0.00456 = 4.56 × 10^-3 ⇒ characteristic = -3

(c) 25.8 = 2.58 × 10¹ ⇒ characteristic = 1

Antilogarithms

The antilogarithm is the inverse operation of finding a logarithm. If log₁₀N = x, then antilog(x) = N.

Example 6.7

Find the antilogarithm of:

(a) 2.3010

(b) -1.6990

(c) 0.4771

Solution:

(a) antilog(2.3010) = 200 (approximately)

(b) antilog(-1.6990) = antilog(2.3010) = 0.02 (approximately)

(c) antilog(0.4771) = 3 (approximately)

Applications of Logarithms

Logarithms have numerous practical applications in various fields:

Scientific Applications

• pH scale: pH = -log₁₀[H⁺]
• Richter scale: Measures earthquake magnitude
• Sound intensity: Decibel scale

Financial Applications

• Compound interest: A = P(1 + r/n)^nt
• Population growth: P = P₀e^rt
• Radioactive decay: N = N₀e^-λt

Example 6.8

The population of a town grows according to the formula P = 50,000 × 10^0.02t, where t is time in years. Find:

(a) The population after 10 years

(b) The time when the population will reach 100,000

Solution:

(a) P = 50,000 × 10^0.02×10 = 50,000 × 10^0.2 = 50,000 × 1.5849 = 79,245

(b) 100,000 = 50,000 × 10^0.02t

2 = 10^0.02t

log₁₀2 = 0.02t

0.3010 = 0.02t

t = 0.3010/0.02 = 15.05 years

Example 6.9

The intensity of sound is measured in decibels using the formula dB = 10 log₁₀(I/I₀), where I₀ is the reference intensity. If a sound has intensity 1000 times I₀, what is its decibel level?

Solution:

dB = 10 log₁₀(1000I₀/I₀) = 10 log₁₀1000 = 10 × 3 = 30 dB

Chapter Summary

1. Standard form: A × 10ⁿ where 1 ≤ A < 10 and n is an integer
2. Logarithm definition: If a^x = N, then log_aN = x
3. Laws of logarithms:
   • Product law: log_a(MN) = log_aM + log_aN
   • Quotient law: log_a(M/N) = log_aM - log_aN
   • Power law: log_aMⁿ = n log_aM
4. Common logarithms: Base 10 (log₁₀ or log)
5. Natural logarithms: Base e (log_e or ln)
6. Antilogarithm: The inverse operation of logarithm
7. Applications: Used in science, finance, engineering, and many other fields

Revision Exercise 6

1. Write in standard form:
   1. 256,000,000
   2. 0.0000789
   3. 12,450
2. Express in logarithmic form:
   1. 3⁴ = 81
   2. 10^-2 = 0.01
   3. 2⁵ = 32
3. Express in exponential form:
   1. log₂8 = 3
   2. log₁₀1000 = 3
   3. log₅125 = 3
4. Simplify using logarithm laws:
   1. log₁₀6 + log₁₀5
   2. log₂24 - log₂3
   3. 2 log₃7
5. Solve for x:
   1. log₃x = 4
   2. log_x16 = 2
   3. log₂(x + 1) = 3
6. The population of bacteria doubles every hour. If initially there are 500 bacteria, how long will it take to reach 32,000 bacteria?
7. An investment of Tsh 1,000,000 grows at 8% per year compounded annually. How long will it take to double the investment?
8. The magnitude of an earthquake is given by M = log₁₀(A/A₀), where A is the amplitude. If one earthquake has amplitude 1000 times A₀ and another has amplitude 100,000 times A₀, what is the difference in their magnitudes?

Chapter Seven: Sets

Introduction

Sets are fundamental mathematical concepts that help us organize, classify, and analyze groups of objects. In this chapter, you will learn about the concept of sets, different types of sets, operations on sets, and how to represent sets using Venn diagrams. The competencies developed will enable you to solve problems involving collections of objects, analyze relationships between groups, and apply set theory in various real-life situations such as data analysis, probability, and logical reasoning.

Think

How would you organize and classify different groups of objects around you? Sets provide the mathematical framework for doing this systematically!

Concept of a Set

A set is a well-defined collection of distinct objects. The objects in a set are called elements or members of the set.

Definition: A set is a collection of well-defined, distinct objects.

Elements: The objects that belong to a set are called elements or members.

Notation: Sets are usually denoted by capital letters: A, B, C, ...

Elements are denoted by lowercase letters: a, b, c, ...

Example 7.1

Identify which of the following are sets:

(a) The collection of all students in Form Two

(b) The collection of all tall students in Form Two

(c) The collection of all subjects taught in secondary school

(d) The collection of interesting books

Solution:

(a) This is a set - it is well-defined

(b) This is not a set - "tall" is not well-defined

(c) This is a set - it is well-defined

(d) This is not a set - "interesting" is subjective and not well-defined

Methods of Describing Sets

There are two main methods for describing sets:

1. Roster Method (Tabular Form): Listing all elements within braces { }

Example: A = {1, 2, 3, 4, 5}

2. Set-Builder Notation (Rule Method): Describing the properties that elements must satisfy

Example: A = {x | x is a natural number less than 6}

Example 7.2

Write the following sets using both roster and set-builder notation:

(a) The set of vowels in the English alphabet

(b) The set of even numbers between 1 and 11

(c) The set of days of the week

Solution:

(a) Roster: {a, e, i, o, u}
Set-builder: {x | x is a vowel in the English alphabet}

(b) Roster: {2, 4, 6, 8, 10}
Set-builder: {x | x is an even number and 1 < x < 11}

(c) Roster: {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}
Set-builder: {x | x is a day of the week}

Types of Sets

Empty Set (Null Set)

A set with no elements. Denoted by ∅ or { }.

Example: A = {x | x is a person older than 200 years} = ∅

Finite Set

A set with a countable number of elements.

Example: B = {1, 2, 3, 4, 5} (finite with 5 elements)

Infinite Set

A set with an unlimited number of elements.

Example: C = {1, 2, 3, 4, ...} (set of natural numbers)

Singleton Set

A set with exactly one element.

Example: D = {5} (singleton set)

Equal Sets

Two sets that contain exactly the same elements.

Example: {1, 2, 3} = {3, 2, 1} (order doesn't matter)

Equivalent Sets

Two sets that have the same number of elements.

Example: {a, b, c} and {1, 2, 3} are equivalent (both have 3 elements)

Subset

Set A is a subset of set B if all elements of A are also elements of B. Denoted by A ⊆ B.

Example: If A = {1, 2} and B = {1, 2, 3}, then A ⊆ B

Universal Set

The set that contains all elements under consideration. Denoted by U.

Example: When studying numbers 1-10, U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Operations on Sets

Union of Sets

The union of sets A and B is the set of all elements that are in A or in B or in both.

Notation: A ∪ B = {x | x ∈ A or x ∈ B}

A ∪ B (shaded area)

Intersection of Sets

The intersection of sets A and B is the set of all elements that are in both A and B.

Notation: A ∩ B = {x | x ∈ A and x ∈ B}

A ∩ B (shaded area)

Complement of a Set

The complement of set A is the set of all elements in the universal set that are not in A.

Notation: A' or A^c = {x | x ∈ U and x ∉ A}

A' (shaded area)

Difference of Sets

The difference of sets A and B is the set of elements that are in A but not in B.

Notation: A - B = {x | x ∈ A and x ∉ B}

A - B (shaded area)

Example 7.3

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7}

Find:

(a) A ∪ B

(b) A ∩ B

(c) A'

(d) A - B

(e) B - A

Solution:

(a) A ∪ B = {1, 2, 3, 4, 5, 6, 7}

(b) A ∩ B = {3, 4, 5}

(c) A' = {6, 7, 8, 9, 10}

(d) A - B = {1, 2}

(e) B - A = {6, 7}

Venn Diagrams

Venn diagrams are graphical representations of sets using circles or other closed curves within a rectangle representing the universal set.

Purpose of Venn Diagrams

• Visualize relationships between sets
• Solve problems involving set operations
• Understand set concepts intuitively
• Verify set identities

Example 7.4

In a class of 40 students:

• 25 students play football
• 20 students play basketball
• 10 students play both games

Represent this information in a Venn diagram and find:

(a) How many play only football?

(b) How many play only basketball?

(c) How many play at least one game?

(d) How many play neither game?

Venn diagram for sports activities

Solution:

Let F = football players, B = basketball players

F ∩ B = 10 (both games)

Only football = 25 - 10 = 15

Only basketball = 20 - 10 = 10

(a) Only football = 15 students

(b) Only basketball = 10 students

(c) At least one game = 15 + 10 + 10 = 35 students

(d) Neither game = 40 - 35 = 5 students

Activity 7.1: Exploring Sets in Daily Life

1. Identify three different sets of objects in your classroom.
2. Classify these sets as finite, infinite, or empty.
3. Find the union and intersection of two of these sets.
4. Create a Venn diagram to show the relationship between these sets.
5. Share your findings with classmates.

Set Symbols and Notation

Symbol	Meaning	Example
{ }	Set	A = {1, 2, 3}
∈	Is an element of	2 ∈ A
∉	Is not an element of	4 ∉ A
∅ or { }	Empty set	B = ∅
⊆	Subset	{1, 2} ⊆ A
⊂	Proper subset	{1, 2} ⊂ A
∪	Union	A ∪ B
∩	Intersection	A ∩ B
A' or Ac	Complement	A'
A - B	Difference	A - B
|A| or n(A)	Number of elements	|A| = 3

Chapter Summary

1. Set: A well-defined collection of distinct objects
2. Elements: The objects that belong to a set
3. Types of Sets:
   • Empty set: ∅ or { }
   • Finite set: Countable number of elements
   • Infinite set: Unlimited number of elements
   • Singleton set: One element only
   • Equal sets: Same elements
   • Equivalent sets: Same number of elements
4. Set Operations:
   • Union: A ∪ B = {x | x ∈ A or x ∈ B}
   • Intersection: A ∩ B = {x | x ∈ A and x ∈ B}
   • Complement: A' = {x | x ∈ U and x ∉ A}
   • Difference: A - B = {x | x ∈ A and x ∉ B}
5. Venn Diagrams: Graphical representation of sets and their relationships
6. Applications: Used in probability, statistics, logic, and data analysis

Revision Exercise 7

1. Which of the following are sets? Justify your answer.
   1. The collection of all good football players
   2. The collection of all months with 31 days
   3. The collection of all difficult mathematics problems
   4. The collection of all prime numbers less than 20
2. Let A = {2, 4, 6, 8, 10}, B = {1, 2, 3, 4, 5}, C = {3, 6, 9, 12}. Find:
   1. A ∪ B
   2. A ∩ B
   3. A ∩ C
   4. B ∪ C
   5. A - B
3. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 4, 6, 8}, B = {1, 3, 5, 7, 9}, find:
   1. A'
   2. B'
   3. (A ∪ B)'
   4. A' ∩ B'
4. In a survey of 100 students:
   • 60 students like Mathematics
   • 50 students like Science
   • 30 students like both subjects

   Draw a Venn diagram and find:

   1. How many like only Mathematics?
   2. How many like only Science?
   3. How many like at least one subject?
   4. How many like neither subject?
5. Prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using Venn diagrams.
6. If n(A) = 25, n(B) = 30, and n(A ∩ B) = 10, find n(A ∪ B).
7. Write the following sets in set-builder notation:
   1. {2, 4, 6, 8, 10}
   2. {1, 3, 5, 7, 9}
   3. {a, e, i, o, u}

Chapter Eight: Trigonometry

Introduction

Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. In this chapter, you will learn about trigonometric ratios, special angles, inverse trigonometric functions, and practical applications such as angles of elevation and depression. The competencies developed will enable you to solve real-world problems in fields such as architecture, engineering, navigation, and physics.

Think

How can we measure the height of a tall building without physically climbing it? Trigonometry provides the mathematical tools to solve such problems!

Trigonometric Ratios

In a right-angled triangle, there are six trigonometric ratios that relate the angles to the ratios of the sides.

Right-angled triangle with angle θ

For a right-angled triangle with angle θ:

• Opposite (opp): Side opposite to angle θ
• Adjacent (adj): Side adjacent to angle θ (excluding hypotenuse)
• Hypotenuse (hyp): Longest side, opposite the right angle

Basic Trigonometric Ratios

Ratio	Definition	Abbreviation
Sine	opposite/hypotenuse	sin θ = opp/hyp
Cosine	adjacent/hypotenuse	cos θ = adj/hyp
Tangent	opposite/adjacent	tan θ = opp/adj

Reciprocal Trigonometric Ratios

Ratio	Definition	Abbreviation
Cosecant	1/sine = hypotenuse/opposite	cosec θ = hyp/opp
Secant	1/cosine = hypotenuse/adjacent	sec θ = hyp/adj
Cotangent	1/tangent = adjacent/opposite	cot θ = adj/opp

Example 8.1

In a right-angled triangle, if the opposite side is 3 cm and the hypotenuse is 5 cm, find all trigonometric ratios for angle θ.

Solution:

Given: opp = 3 cm, hyp = 5 cm

First, find the adjacent side using Pythagoras theorem:

adj = √(hyp² - opp²) = √(25 - 9) = √16 = 4 cm

Now calculate the ratios:

sin θ = opp/hyp = 3/5 = 0.6

cos θ = adj/hyp = 4/5 = 0.8

tan θ = opp/adj = 3/4 = 0.75

cosec θ = 1/sin θ = 5/3 ≈ 1.667

sec θ = 1/cos θ = 5/4 = 1.25

cot θ = 1/tan θ = 4/3 ≈ 1.333

Trigonometric Ratios of Special Angles

Certain angles (0°, 30°, 45°, 60°, 90°) have exact trigonometric values that are important to memorize.

Trigonometric Values for Special Angles

Angle	sin	cos	tan
0°	0	1	0
30°	1/2	√3/2	1/√3
45°	1/√2	1/√2	1
60°	√3/2	1/2	√3
90°	1	0	∞

Example 8.2

Find the exact value of:

(a) sin 30° + cos 60°

(b) tan 45° × cos 30°

(c) sin² 45° + cos² 45°

Solution:

(a) sin 30° + cos 60° = 1/2 + 1/2 = 1

(b) tan 45° × cos 30° = 1 × (√3/2) = √3/2

(c) sin² 45° + cos² 45° = (1/√2)² + (1/√2)² = 1/2 + 1/2 = 1

Trigonometric Ratios of Any Angle

Trigonometric ratios can be extended to angles greater than 90° using the concept of reference angles and the unit circle.

ASTC Rule (All Students Take Calculus)

This rule helps determine the sign of trigonometric ratios in different quadrants:

• Quadrant I: All ratios are positive
• Quadrant II: Only Sine and its reciprocal are positive
• Quadrant III: Only Tangent and its reciprocal are positive
• Quadrant IV: Only Cosine and its reciprocal are positive

ASTC Rule - Signs of trigonometric ratios in different quadrants

Relationship Between Trigonometric Ratios

Fundamental Trigonometric Identities

Pythagorean Identities:

• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

Reciprocal Identities:

• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• cot θ = 1/tan θ

Quotient Identities:

• tan θ = sin θ/cos θ
• cot θ = cos θ/sin θ

Example 8.3

If sin θ = 3/5 and θ is acute, find cos θ and tan θ.

Solution:

Using sin²θ + cos²θ = 1

(3/5)² + cos²θ = 1

9/25 + cos²θ = 1

cos²θ = 1 - 9/25 = 16/25

cos θ = 4/5 (positive since θ is acute)

tan θ = sin θ/cos θ = (3/5)/(4/5) = 3/4

Calculating Trigonometric Ratios with a Calculator

Using a Scientific Calculator

Most scientific calculators have dedicated buttons for trigonometric functions:

• sin, cos, tan: For calculating trigonometric ratios
• sin⁻¹, cos⁻¹, tan⁻¹: For finding angles (inverse functions)

Important: Ensure your calculator is in the correct angle mode (degrees or radians).

Example 8.4

Use a calculator to find:

(a) sin 25°

(b) cos 1.2 rad

(c) tan⁻¹(0.75)

Solution:

(a) sin 25° ≈ 0.4226

(b) cos 1.2 rad ≈ 0.3624

(c) tan⁻¹(0.75) ≈ 36.87°

Inverse of Trigonometric Ratios

Inverse trigonometric functions are used to find angles when the trigonometric ratio is known.

Inverse Trigonometric Functions:

• If sin θ = x, then θ = sin⁻¹x
• If cos θ = x, then θ = cos⁻¹x
• If tan θ = x, then θ = tan⁻¹x

These are also called arcsin, arccos, and arctan respectively.

Example 8.5

Find the angle θ for which:

(a) sin θ = 0.5

(b) cos θ = 0.8660

(c) tan θ = 1

Solution:

(a) θ = sin⁻¹(0.5) = 30°

(b) θ = cos⁻¹(0.8660) = 30°

(c) θ = tan⁻¹(1) = 45°

The Angle of Elevation and Depression

Angle of Elevation: The angle between the horizontal line and the line of sight when looking upward at an object.

Angle of Depression: The angle between the horizontal line and the line of sight when looking downward at an object.

Angle of elevation and angle of depression

Real-World Applications

• Measuring heights of buildings, trees, and mountains
• Navigation and surveying
• Aviation and space science
• Architecture and construction
• Sports and recreation

Example 8.6

A person standing 50 meters away from a tree measures the angle of elevation to the top of the tree as 30°. Find the height of the tree.

Finding the height of a tree using trigonometry

Solution:

Let h be the height of the tree

tan 30° = opposite/adjacent = h/50

1/√3 = h/50

h = 50/√3 ≈ 50/1.732 ≈ 28.87 meters

Therefore, the tree is approximately 28.87 meters tall.

Example 8.7

From the top of a 100-meter cliff, a boat is spotted at an angle of depression of 15°. How far is the boat from the base of the cliff?

Solution:

Angle of depression = 15°, so angle at the boat = 15° (alternate angles)

Let d be the distance from the boat to the cliff base

tan 15° = opposite/adjacent = 100/d

d = 100/tan 15° ≈ 100/0.2679 ≈ 373.2 meters

Therefore, the boat is approximately 373.2 meters from the base of the cliff.

Activity 8.1: Measuring Heights Using Trigonometry

1. Choose a tall object (tree, building, flagpole) whose height you want to measure.
2. Measure the distance from your observation point to the base of the object.
3. Using a clinometer or protractor, measure the angle of elevation to the top of the object.
4. Calculate the height using trigonometric ratios.
5. Compare your calculated height with actual measurements if possible.
6. Discuss any sources of error in your measurement.

Chapter Summary

1. Trigonometric Ratios:
   • sin θ = opposite/hypotenuse
   • cos θ = adjacent/hypotenuse
   • tan θ = opposite/adjacent
   • cosec θ = 1/sin θ
   • sec θ = 1/cos θ
   • cot θ = 1/tan θ
2. Special Angles: 0°, 30°, 45°, 60°, 90° have exact trigonometric values
3. ASTC Rule: Determines sign of ratios in different quadrants
4. Fundamental Identities:
   • sin²θ + cos²θ = 1
   • 1 + tan²θ = sec²θ
   • 1 + cot²θ = cosec²θ
5. Inverse Trigonometric Functions: Used to find angles from ratios
6. Angle of Elevation: Looking upward at an object
7. Angle of Depression: Looking downward at an object
8. Applications: Height and distance measurements, navigation, engineering

Revision Exercise 8

1. In a right-angled triangle, if the adjacent side is 12 cm and the hypotenuse is 13 cm, find:
   1. sin θ
   2. cos θ
   3. tan θ
   4. cosec θ
2. Find the exact values of:
   1. sin 60° + cos 30°
   2. tan 45° - sin 30°
   3. 2 cos 45° × sin 45°
   4. sin² 60° + cos² 60°
3. If sin θ = 4/5 and θ is acute, find cos θ and tan θ.
4. Use a calculator to find:
   1. sin 35°
   2. cos 1.05 rad
   3. tan⁻¹(0.5)
   4. sin⁻¹(0.7071)
5. A ladder leans against a wall making an angle of 60° with the ground. If the foot of the ladder is 2 meters from the wall, find the length of the ladder.
6. From the top of a 50-meter building, the angle of depression to a car on the road is 30°. How far is the car from the base of the building?
7. A flagpole casts a shadow of 15 meters when the sun's altitude is 45°. Find the height of the flagpole.
8. Two buildings are 100 meters apart. From the top of the taller building (120 meters), the angle of depression to the top of the shorter building is 20°. Find the height of the shorter building.
9. Prove the identity: (1 - cos²θ)(1 + tan²θ) = tan²θ
10. A ship is spotted from a lighthouse at an angle of depression of 15°. If the lighthouse is 80 meters tall, how far is the ship from the base of the lighthouse?