4A Calculating trigonometric ratios 4B Finding an unknown side 4C Finding angles 4D Applications of right-angled triangles 4E Using the sine rule to ﬁnd side lengths 4F Using the sine rule to ﬁnd angle sizes 4G The cosine rule

(1)

4

syllabus

rref

efer

erence

ence

Topic:

• Periodic functions and

applications

In this

cha

chapter

pter

4A Calculating trigonometric

ratios

4B Finding an unknown side

4C Finding angles

4D Applications of right-angled

triangles

4E Using the sine rule to find

side lengths

4F Using the sine rule to find

angle sizes

4G The cosine rule

Triangle

(2)

Introduction

Rick is standing on the southern side of the Grand Canyon, wondering how wide it is. Through his binocu-lars, Rick sights a lookout on the northern side of the canyon, bearing N5°E. He then heads 8.1 km east along the canyon’s southern rim and notes that the bearing of the lookout is now N12°W. Using this information, Rick is able to calculate the width of the Grand Canyon. After studying this chapter you will be able to make this calculation for yourself, as well as being able to solve many other practical problems about sides and angles within a triangle.

Calculating trigonometric ratios

From previous years work you should be familiar with Pythagoras’ theorem. This enabled us to find the length of one side of a right-angled triangle given the length of the other two. To use Pythagoras’ theorem, we had to recognise the hypotenuse in a right-angled triangle. In trigonometry, we need to be able to name the two shorter sides as well. We do this with reference to a given

angle, and label them opposite and adjacent. They are the sides opposite and adjacent to the given angle. The diagram shows the sides labelled with respect to the angle, θ.

Looking at the tangent ratio

The tangent ratio is a ratio of sides in similar

right-angled triangles, such as those in the diagram. ∠BAC is common to each triangle and is equal to 30°. We are going to look at the ratio of the opposite side to the adjacent side in each triangle. You can do this either on your calculator or by completing the spreadsheet ‘Tangent’ on your Maths Quest CD-ROM.

Complete each of the following measurements and calculations.

1 a BC = mm b AB = mm c =

2 a DE = mm b AD = mm c =

3 a FG = mm b AF = mm c =

4 a HI = mm b AH = mm c =

Remember that ∠BAC is common to each triangle. In each of the above, part c is the ratio of the opposite side to the adjacent side of ∠BAC. What do you notice about each of these answers?

hypotenuse opposite

adjacent θ

A B

C E G I

D F H

E

XCEL Sprea_dshe

et

Tangent

BC AB

---DE AD

---FG AF

(3)

---Trigonometry uses the ratio of side lengths to calculate the lengths of sides and the size of angles. The ratio of the opposite side to the adjacent side is called the tangent ratio. This ratio is fixed for any particular angle.

The tangent ratio for any angle, θ, can be found using the result:

In the investigation on the previous page, we found that for a 30° angle the ratio was 0.58. We can find a more accurate value for the tangent ratio on a calculator by pressing

and entering 30.

For all calculations in trigonometry you will need to make sure that your calculator is in DEGREES MODE. For most calculators you can check this by looking for a DEG in the display.

When measuring angles: 1 degree = 60 minutes 1 minute = 60 seconds

You need to be able to enter angles using both degrees and minutes into your calcu-lator. Most calculators use a (Degrees, Minutes, Seconds) button or a

button. Check with your teacher to see how to do this.

The tangent ratio is used to solve problems involving the opposite side and the adjacent side of a right-angled triangle. The tangent ratio does not allow us to solve problems that involve the hypotenuse.

The sine ratio (abbreviated to sin) is the name given to the ratio of the opposite side and the hypotenuse.

tan θ opposite side adjacent side ---=

tan

DMS _° ’ ”

Using your calculator, find the following, correct to 3 decimal places.

a tan 60° b 15 tan 75° c d tan 49°32′

THINK WRITE/DISPLAY

a Press and enter 60. a tan 60° = 1.732

b Enter 15, press and , enter 75.

b 15 tan 75° = 55.981

c Enter 8, press and , enter 69.

c = 3.071

d Press , enter 49, press , enter 32, press .

d tan 49°32′= 1.172 8

tan 69∞

---tan

× tan

÷ tan 8

tan 69°

---tan DMS

DMS

1

(4)

In any right-angled triangle with equal angles, the ratio of the opposite side to the hypotenuse will remain the same, regardless of the size of the triangle. The formula for the sine ratio is:

The value of the sine ratio for any angle is found using the sin function on the calculator. sin 30° = 0.5

Check this on your calculator.

Looking at the sine ratio

The tangent ratio is the ratio of the opposite side and the adjacent side in a right-angled triangle. The sine ratio is the ratio of the opposite side and the hypotenuse. Look back to the right-angled triangles used in the tangent investigation on page 150.

Complete each of the following measurements and calculations by using your calculator or the spreadsheet ‘Sine’ on your Maths Quest CD-ROM.

As we saw earlier, ∠BAC is common to all of these similar triangles and so in this exercise, we look at the ratio of the side opposite ∠BAC to the hypotenuse of each triangle.

1 a BC = mm b AC = mm c =

2 a DE = mm b AE = mm c =

3 a FG = mm b AG = mm c =

4 a HI = mm b AI = mm c =

In this exercise, part c is the ratio of the opposite side to ∠BAC to the

hypotenuse. You should again notice that the answers are the same (or very close, allowing for measurement error).

E

XCEL Sprea_dshe

et

Sine

BC AC

---DE AE

---FG AG

---HI AI

---sin θ opposite side hypotenuse ---=

Find, correct to 3 decimal places:

a sin 57° b 9 sin 45° c d 9.6 sin 26°12′.

THINK WRITE/DISPLAY

a Press and enter 57. a sin 57° = 0.839

b Enter 9, press and , enter 45. b 9 sin 45° = 6.364

c Enter 18, press and , enter 44. c = 25.912

d Enter 9.6, press and , enter 26, press , enter 12, press .

d 9.6 sin 26°12′= 4.238 18

sin 44∞

---sin

× sin

÷ sin 18

sin 44°

---× sin

DMS DMS

2

(5)

A third trigonometric ratio is the cosine ratio. This ratio compares the length of the adjacent side and the hypotenuse.

The cosine ratio is found using the formula:

To calculate the cosine ratio for a given angle on your calculator, use the cos func-tion. On your calculator check the calculation:

cos 30° = 0.866

Looking at the cosine ratio

Look back to the right-angled triangles used in the tangent investigation on page 150.

Complete each of the following measurements and calculations. You may do so by using the spreadsheet ‘Cosine’ on your Maths Quest CD-ROM.

1 a AB = mm b AC = mm c =

2 a AD = mm b AE = mm c =

3 a AF = mm b AG = mm c =

4 a AH = mm b AI = mm c =

Again for part c, you should get the same answer for each triangle. In each case, this is the cosine ratio of the common angle BAC.

EXCEL Spreadshe et Cosine

AB AC

---AD AE

---AF AG

---AH AI

---cos θ adjacent side hypotenuse ---=

Find, correct to 3 decimal places:

a cos 27° b 6 cos 55° c d .

THINK WRITE/DISPLAY

a Press and enter 27. a cos 27° = 0.891

b Enter 6, press and , enter 55. b 6 cos 55° = 3.441

c Enter 21.3, press and , enter 74.

c = 77.275

d Enter 4.5, press and , enter 82, press , enter 46, press

.

d = 35.740

21.3 cos 74∞

--- 4.5

cos 82∞46¢

---cos

× cos

÷ cos 21.3

cos 74°

---÷ cos

DMS

4.5 cos 82°46′

---3

(6)

Similarly, if we are given the sin, cos or tan of an angle, we are able to calculate the size of that angle using the calculator. We do this using the inverse functions. On most calculators these are the 2nd function of the sin, cos and tan functions and are denoted sin−1, cos−1 and tan−1. On most calculators the shift key is used as shown in the following worked examples.

So far, we have dealt only with angles that are whole degrees. You need to be able to make calculations using minutes as well. On most calculators, you will use the DMS (Degrees, Minutes, Seconds) function or the function.

Find θ, correct to the nearest degree, given that sin θ= 0.738.

THINK WRITE/DISPLAY

Press [sin–1] and enter .738. Round your answer to the nearest degree.

θ= 48°

1 SHIFT

2

4

WORKED

E

xample

° ’ ”

Given that tan θ= 1.647, calculate θ to the nearest minute.

THINK WRITE/DISPLAY

Press [tan–1] and enter 1.647. Convert your answer to degrees and minutes by pressing .

θ= 58°44′

1 SHIFT

2

DMS

5

WORKED

E

xample

remember

1. The tangent ratio is the ratio of the opposite side and the adjacent side.

tan θ=

2. The sine ratio is the ratio of the opposite side and the hypotenuse.

sin θ=

3. The cosine ratio is the ratio of the adjacent side and the hypotenuse.

cos θ=

4. The value of the trigonometric ratios can be found using the sin, cos and tan functions on your calculator.

5. The angle can be found when given the trigonometric ratio using the sin−1_, cos−1 and tan−1 functions on your calculator.

opposite side adjacent side

---opposite side hypotenuse

(7)

Calculating trigonometric

ratios

1 Calculate the value of each of the following, correct to 3 decimal places.

4 Calculate the value of each of the following, correct to 4 significant figures.

6 Find θ, correct to the nearest degree, given that sin θ= 0.167.

7 Find θ, correct to the nearest degree, given that:

8 Find θ, correct to the nearest minute, given that cos θ= 0.058.

9 Find θ, correct to the nearest minute, given that:

Finding an unknown side

We can use the trigonometric ratios to find the length of one side of a right-angled triangle if we know the length of another side and an angle. Consider the triangle at right.

In this triangle we are asked to find the length of the opposite side and have been given the length of the adjacent side.

a tan 57° b 9 tan 63° c d tan 33°19′

a sin 37° b 9.3 sin 13° c d

a cos 45° b 0.25 cos 9° c d 5.9 cos 2°3′

a sin 30° b cos 15° c tan 45°

d 48 tan 85° e 128 cos 60° f 9.35 sin 8°

g h i

a sin 24°38′ b tan 57°21′ c cos 84°40′ d 9 cos 55°30′ e 4.9 sin 35°50′ f 2.39 tan 8°59′

g h i

a sin θ= 0.698 b cos θ= 0.173 c tan θ= 1.517.

a tan θ = 0.931 b cos θ= 0.854 c sin θ= 0.277.

4A

W

WORKEDORKED

E Example

1 SkillS

HEET

4.1

8.6

tan 12°

---W WORKEDORKED

E Example

2 14.5

sin 72°

--- 48 sin 67°40′

---W WORKEDORKED

E Example

3 6

cos 24°

---SkillS

HEET

4.2

4.5

cos 32°

--- 0.5 tan 20°

--- 15 sin 72°

---19 tan 67°45′

--- 49.6 cos 47°25′

--- 0.84 sin 75°5′

---W WORKEDORKED

E Example

4

W WORKEDORKED

E Example

5

hyp

opp

adj

14 cm 30°

(8)

We know from the formula that: tan θ= . In this example, tan 30° = . From

our calculator we know that tan 30°= 0.577. We can set up an equation that will allow us to find the value of x.

tan θ=

tan 30° =

x = 14 tan 30° ≈ 8.083 cm

In the example above, we were told to use the tangent ratio. In practice, we need to be able to look at a problem and then decide if the solution is found using the sin, cos or tan ratio. To do this we need to examine the three formulas.

tan θ =

We use this formula when we are finding either the opposite or adjacent side and are given the length of the other.

sin θ =

The sin ratio is used when finding the opposite side or the hypotenuse and we are given the length of the other.

cos θ =

The cos ratio is for problems where we are finding the adjacent side or the hypot-enuse and are given the length of the other.

To make the decision we need to label the sides of the triangle and make a decision based on these labels.

opposite adjacent

--- x 14

---opp adj

---x 14

---Use the tangent ratio to find the value of x in the triangle at right,

correct to 2 decimal places.

THINK WRITE

Label the sides of the triangle opp, adj and hyp.

Write the tangent formula. tan θ=

Substitute for θ (55°) and the adjacent side (17 m).

tan 55° =

Make x the subject of the equation. x= 17 tan 55°

Calculate and round to 2 decimal places. ≈ 24.28 cm

17 m 55°

h

1

hyp

opp

adj

17 cm 55°

h

2 opp

adj

---3 x

17

---4 5

6

WORKED

E

xample

opposite side adjacent side

---opposite side hypotenuse

(9)

---To remember each of the formulas more easily, we can use this acronym: SOHCAHTOA

We may pronounce this acronym as ‘Sock ca toe her’. The initials of the acronym represent the three trigonometric formulas.

sin θ= cos θ= tan θ=

Care needs to be taken at the substitution stage. In the above examples, the unknown side was the numerator in the fraction, hence we multiplied to find the answer. If after substitution, the unknown side is in the denominator, the final step is done by division.

Find the length of the side marked x, correct to 2 decimal places.

THINK WRITE

Label the sides of the triangle.

x is the opposite side and 24 m is the hypotenuse, therefore use the sin formula.

sin θ=

Substitute for θ and the hypotenuse. sin 50° =

Make x the subject of the equation. x= 24 sin 50°

Calculate and round to 2 decimal places. ≈ 18.39 m

24 m

50°

x

1

hyp

opp

adj

24 m

50°

x

2 opp

hyp

---3 x

24

---4 5

7

WORKED

E

xample

opp hyp

--- adj

hyp

--- opp

adj

---Find the length of the side marked z in the triangle at right.

Continued over page

THINK WRITE

Label the sides opp, adj and hyp.

Choose the cosine ratio because we are finding the hypotenuse and have been given the adjacent side.

Write the formula. cos θ=

Substitute for θ and the adjacent side. cos 23°15′ =

12.5 m 23°15' z

1

opp hyp

adj

12.5 m 23°15' z

2

3 adj

hyp

---4 12.5

z

---8

(10)

Trigonometry is used to solve many practical problems. In these cases, it is necessary to draw a diagram to represent the problem and then use trigonometry to solve the problem. With written problems that require you to draw the diagram, it is necessary to give the answer in words.

THINK WRITE

Make z the subject of the equation. z cos 23°15′ = 12.5

z=

Calculate and round off appropriately. ≈ 13.60 m

5

12.5 cos 23°15′

---6

A flying fox is used in an army training camp. The flying fox is supported by a cable that runs from the top of a cliff face to a point 100 m from the base of the cliff. The cable

makes a 15° angle with the horizontal. Find the length of the cable used to support the

flying fox.

THINK WRITE

Draw a diagram and show information.

Label the sides of the triangle opp, adj and hyp.

Choose the cosine ratio because we are finding the hypotenuse and have been given the adjacent side.

Write the formula. cos θ=

Substitute for θ and the adjacent side. cos 15° =

Make f the subject of the equation. f cos 15° = 100

f=

Calculate and round off appropriately. ≈ 103.5 m

Give a written answer. The cable is approximately 103.5 m long.

1

15° 100 m

f

2

3

4 adj

hyp

---5 100

f

---6

100 cos 15°

---7

8

9

(11)

Finding an unknown side

1 Label the sides of each of the following triangles, with respect to the angle marked with the pronumeral.

a b c

2 Use the tangent ratio to find the length of the side marked x (correct to 1 decimal place).

3 Use the sine ratio to find the length of the side marked a (correct to 2 decimal places).

4 Use the cosine ratio to find the length of the side marked d (correct to 3 significant figures).

remember

1. Trigonometry can be used to find a side in a right-angled triangle when we are given the length of one side and the size of an angle.

2. The trig formulas are:

sin θ= cos θ= tan θ=

3. Take care to choose the correct trigonometric ratio for each question.

4. Substitute carefully and note the change in the calculation, depending upon whether the unknown side is in the numerator or denominator.

5. Before using your calculator, check that it is in degrees mode.

6. Be sure that you know how to enter degrees and minutes into your calculator.

7. Problem questions will require you to draw a diagram and give a written answer.

opp hyp

--- adj

hyp

--- opp

adj

---remember

4B

SkillS

HEET

4.3

θ α γ

W WORKEDORKED

E Example

6

51 mm 71°

x

13 m

23°

a

35 cm 31°

(12)

5 The following questions use the tan, sin or cos ratios in their solution. Find the size of the side marked with the pronumeral, correct to 3 significant figures.

a b c

6 Find the length of the side marked with the pronumeral in each of the following (correct to 1 decimal place).

a b c

7 Find the length of the side marked with the pronumeral in each of the following (correct to 3 significant figures).

a b c

d e f

g h i

j k l

W WORKEDORKED

E Example

7

68° 13 cm

x

49° 48 m

y

41° 12.5 km

z

W WORKEDORKED

E Example

8

21°

4.8 m

t

77°

87 mm

p

36° 8.2 m

q

SkillS

HEET

4.4

23°

2.3 m

a

39°

0.85 km

b

76°

8.5 km

x

116 mm 9°

m

16.75 cm 11°

d

13° 64.75 m

x f

83° 44.3 m

x

20° 15.75 km

g

2.34 m 84°9'

m

84.6 km 60°32'

q

21.4 m 75°19' t

26.8 cm

(13)

8

Look at the diagram at right and state which of the following is correct.

E To find x you will need the length of the opposite side.

9

Study the triangle at right and state which of the following is correct.

10 A tree casts a 3.6 m shadow when the sun’s angle of elevation is 59°. Calculate the height of the tree, correct to the nearest metre.

11 A 10 m ladder just reaches to the top of a wall when it is leaning at 65° to the ground. How far from the foot of the wall is the ladder (correct to 1 decimal place)?

12 The diagram at right shows the paths of two ships, A and B, after they have left port.

If ship B sends a distress signal, how far must ship A sail to give assistance (to the nearest kilometre)?

13 A rectangle 13.5 cm wide has a diagonal that makes a 24° angle with the horizontal. a Draw a diagram of this situation.

b Calculate the length of the rectangle, correct to 1 decimal place.

14 A wooden gate has a diagonal brace built in for support. The gate stands 1.4 m high and the diagonal makes a 60° angle with the horizontal.

a Draw a diagram of the gate.

b Calculate the length that the diagonal brace needs to be.

15 The wire support for a flagpole makes a 70° angle with the ground. If the support is 3.3 m from the base of the flagpole, calculate the length of the wire support (correct to 2 decimal places).

16 A ship drops anchor vertically with an anchor line 60 m long. After one hour the anchor line makes a 15° angle with the vertical.

a Draw a diagram of this situation. b Calculate the depth of water, correct

to the nearest metre.

c Calculate the distance that the ship has drifted, correct to 1 decimal place.

A x = 9.2 sin 69° B

C x = 9.2 cos 69° D

A tan φ= B tan φ= C sin φ=

D sin φ= E cos φ=

m

multiple choiceultiple choice

69° 9.2

x

x 9.2

sin 69° ---=

x 9.2

cos 69° ---=

m

17

15

8 φ

8 15

--- 15

8

--- 15

17

---8 15

--- 8

17

---W WORKEDORKED

E Example

9

60°

23 km

Port A

(14)

Finding angles

In this chapter so far, we have concerned ourselves with finding side lengths. We are also able to use trigonometry to find the sizes of angles when we have been given side lengths. We need to reverse our previous processes.

Consider the triangle at right.

We want to find the size of the angle marked θ.

Using the formula sin θ = we know that in this triangle

sin θ =

= = 0.5

We then calculate sin−1 (0.5) to find that θ = 30°.

As with all trigonometry it is important that you have your calculator set to degrees mode for this work.

In many cases we will need to calculate the size of an angle, correct to the nearest minute. The same method for finding the solution is used; however, you will need to use your calculator to convert to degrees and minutes.

10 cm

5 cm θ

opp hyp

---5 10

---1 2

---Find the size of angle θ, correct to the nearest degree, in the

triangle at right.

THINK WRITE

Label the sides of the triangle and choose the tan ratio.

tan θ=

Substitute for the opposite and adjacent sides in the triangle and simplify.

= ≈ 0.6615

Make θ the subject of the equation. θ= tan−1(0.6615)

Calculate and round off to the nearest degree.

≈ 33°

6.5 m

4.3 m θ

1

4.3

6.5

adj hyp

opp

θ

opp adj

---2 4.3

6.5

---3

4

10

(15)

The same methods can be used to solve problems. As with finding sides, we set the question up by drawing a diagram of the situation.

Find the size of the angle θ on the right, correct to the nearest minute.

THINK WRITE

Label the sides of the triangle and choose the sin ratio.

sin θ=

Substitute for the opposite side and adjacent in the triangle and simplify.

= ≈ 0.6479

Make θ the subject of the equation. _θ₌_sin−1_(0.6479)

Calculate and convert your answer to degrees and minutes.

≈ 43°23′

4.6 cm

7.1 cm θ

1 opp

hyp adj

4.6 cm

7.1 cm θ

opp hyp

---2 4.6

7.1

---3

4

11

WORKED

E

xample

A ladder is leant against a wall. The foot of the ladder is 4 m from the base of the wall and the ladder reaches 10 m up the wall. Calculate the angle that the ladder makes with the ground.

THINK WRITE

Draw a diagram and label the sides.

Choose the tangent ratio and write the formula.

tan θ=

Substitute for the opposite and adjacent side, then simplify.

= = 2.5

Make θ the subject of the equation. θ= tan−1(2.5)

Calculate and round to the nearest minute. = 68°12′

Give a written answer. The ladder makes an angle of 68°12′ with the ground.

1

4 m 10 m

adj opp

hyp

θ

2 opp

adj

---3 10

4

---4 5 6

12

(16)

Finding angles

1 Use the tangent ratio to find the size of the angle marked with the pronumeral in each of the following, correct to the nearest degree.

a b c

2 Use the sine ratio to find the size of the angle marked with the pronumeral in each of the following, correct to the nearest minute.

a b c

3 Use the cosine ratio to find the size of the angle marked with the pronumeral in each of the following, correct to the nearest minute.

a b c

remember

1. Make sure that the calculator is in degrees mode.

2. To find an angle given the trig ratio, press and then the appropriate ratio button.

3. Be sure to know how to get your calculator to display an answer in degrees and minutes. When rounding off minutes, check if the number of seconds is greater than 30.

4. When solving triangles remember the SOHCAHTOA rule to choose the correct formula

5. In written problems draw a diagram and give an answer in words. SHIFT

remember

4C

7 m

12 m

θ 11 m

3 m φ

25 mm

162 mm γ

24 m 13 m

θ

4.6 m

6.5 m θ

9.7 km

5.6 km α

15 cm

9 cm θ

2.6 m 4.6 m

α

27.8 cm

(17)

4 In the following triangles, you will need to use all three trig ratios. Find the size of the angle marked θ, correct to the nearest degree.

a b c

d e f

5 In each of the following find the size of the angle marked θ, correct to the nearest minute.

a b c

d e f

6

Look at the triangle drawn at right. Which of the statements below is correct?

7

The exact value of sin . The angle θ=

8 A 10 m ladder leans against a 6 m high wall. Find the angle that the ladder makes with the horizontal, correct to the nearest degree.

A ∠ABC = 30° B ∠ABC = 60°

C ∠CAB = 30° D ∠ABC = 45°

E ∠CAB = 45°

A 15° B 30° C 45° D 60° E 90°

W WORKEDORKED

E Example

10

7 cm 11 cm

θ

15 cm

8 cm

θ 14 cm

9 cm

θ

3.6 m

9.2 m

θ

196 mm θ 32 mm _{14.9 m}

26.8 m θ

W WORKEDORKED

E Example

11

30 m

19.2 m θ

10 cm 63 cm

θ

2.5 m 0.6 m

θ

3.5 m

18.5 m

θ 8.3 m 16.3 m

θ

6.3 m

18.9 m

θ

m

10 cm 5 cm

A

B

C θ

m

θ 3

2 ---=

W WORKEDORKED

E Example

(18)

9 A kite is flying on a 40 m string. The kite is flying 10 m away from the vertical as shown in the figure at right.

Find the angle the string makes with the horizontal, correct to the nearest minute.

10 A ship’s compass shows a course due east of the port from which it sails. After sailing 10 nautical miles, it is found that the ship is 1.5 nautical miles off course as shown in the figure below.

Find the error in the compass reading, cor-rect to the nearest minute.

11 The diagram below shows a footballer’s shot at goal.

By dividing the isosceles triangle in half calculate, to the nearest degree, the angle within which the footballer must kick to get the ball to go between the posts.

12 A golfer hits the ball 250 m, but 20 m off centre. Calculate the angle at which the ball deviated from a straight line, correct to the nearest minute.

40 m

10 m kite

10 nm 1.5 nm

7 m

(19)

Applications of right-angled triangles

The principles of trigonometry have been used throughout the ages, from the construction of ancient Egyptian pyramids through to modern-day architecture, as well as for measuring distances and heights which are either inaccessible or impractical. Two important applications of right-angled triangles involve:

1. angles of elevation and depression, and 2. bearings.

Angles of elevation and depression

Angles of elevation and depression are employed when dealing with directions which require us to look up and down respectively.

An angle of elevation is the angle between the horizontal and an object which is higher than the observer (for example, the top of a mountain or flagpole).

An angle of depression is the angle between the horizontal and an object which is lower than the observer (for example, a boat at sea when the observer is on a cliff).

Unless otherwise stated, the angle of elevation or depression is measured and drawn from the horizontal.

Angles of elevation and depression are each measured from the horizontal.

When solving problems involving angles of elevation and depression, it is best always to draw a diagram.

The angle of elevation is equal to the angle of depression

since they are alternate ‘Z’ angles.

Angle of elevation

Line of sight

θ

Line of sight

Angle of depression

θ

D

E

D and E are alternate angles ∴ ∠D = ∠ E

From a cliff 50 metres high, the angle of depression of a boat at sea is 12°. How far is the boat from the base of the cliff?

THINK WRITE

Draw a diagram and label all the given information. Also, label the sides of the triangle.

Choose the tangent ratio because we are finding the adjacent side and have been given the opposite side.

tan θ=

1 ₁₂_°

12°

50 m

x

opp

adj hyp

2 opp

adj

---13

(20)

THINK WRITE

Substitute for θ and the opposite side. tan 12° =

Make x the subject of the equation. x tan 12° =50

x=

Calculate and round off appropriately. ≈235.23

Give a written answer. The boat is 235.23 m away from the base of the cliff.

3 50

x

---4

50 tan 12°

---5 6

From a rescue helicopter 1800 m above the ocean, the angles of depression of two shipwreck survivors are 60° (survivor 1) and 40° (survivor 2).

a Draw a labelled diagram which represents the situation.

b Calculate how far apart the two survivors are.

THINK WRITE

a Draw a diagram and label all the given information. Also, label the sides of the triangles.

a

b For survivor number 1: b Let x represent the horizontal distance

from the helicopter to a survivor.

tan θ=

Make x the subject of the equation. x tan 60° =1800

x=

Calculate and round off appropriately. ≈1039.23 m

For survivor number 2:

tan θ=

Make x the subject of the equation. x tan 40° = 1800

x =

40° 60° 1800 Helicopter

S₁ S₂

opp

adj hyp

hyp

1 opp

adj

---2 1800

x

---3

1800 tan 60°

---4

1 opp

adj

---2 1800

x

---3

1800 tan 40°

---14

(21)

Bearings

Bearings measure the direction of one object from another. There are two systems used for describing bearings.

True bearings are measured in a clockwise direction, starting from north (0° T).

Conventionalorcompass bearings are measured first, relative to north or south; then, relative to east or west.

The two systems are interchangeable. For example, a bearing of 240° T is the same as S60°W.

When solving questions involving direction, always start with a diagram showing the basic compass points: north, south, east and west.

THINK WRITE

Calculate and round off appropriately. ≈2145.16 m

Determine the distance between the two survivors.

Distance apart = 2145.16 − 1039.23 = 1105.93

Give a written answer. The two survivors are 1105.93 m apart.

4 5

6

Compass bearing equivalent is S30°E

150° T N

True bearing equivalent True bearing equivalent N20°W

is 340° T

S70°E

is 110° T 20°

20° N

S E W

N

S E W

S60°W 240° T ₆₀_° N

S E W

N

S E W

A ship sails 40 km in a direction of N52°W. How far west of the starting point is it?

THINK WRITE

Draw a diagram of the situation, labelling each of the compass points and the given information. Also, label the sides of the triangle.

Choose the sine ratio as we are finding the opposite side and have been given the hypotenuse.

sin θ=

1

52° N

x

S E W

40 km opp

adj

hyp hyp

2 opp

hyp

---15

(22)

THINK WRITE

Make x the subject of the equation. 40 sin 52° =x

x= 40 sin 52°

Calculate and round appropriately. ≈ 31.52

Give a written answer. The ship is 31.52 km west of the starting point.

3 x

40

---4

5 6

A ship sails 10 km east, then 4 km south. What is its bearing from its starting point?

THINK WRITE

Draw a diagram of the situation, labelling each of the compass points and the given information. Also, label the sides of the triangle.

Choose the tangent ratio, as we have been

given the opposite and adjacent sides. tan θ= Substitute for opposite and adjacent. tan θ =

Make θ the subject of the equation. θ = tan−1

Calculate and round to the nearest minute. = 21°49′

Express the angle in bearings form. The bearing of the ship was initially 0° T; it has since rotated through an angle of 90° and an additional angle of 21°49′. To obtain the final bearing these values are added.

Bearing = 90° + 21°49′ = 110°49′ T

Give a written answer. The bearing of the ship from its starting point is 110°49′ T.

1

10 km

4 km N

S

θ opp

adj

hyp

2 opp

adj

---3 4

10

---4 4

10 ---   

5 6

7

16

WORKED

E

xample

remember

1. Angles of elevation and depression are each measured from the horizontal. 2. The angle of elevation is equal to the angle of depression since they are

alternate ‘Z’ angles.

3. True bearings are measured in a clockwise direction, starting from north (0° T). 4. Conventional or compass bearings are measured first, relative to north or south;

then, relative to east or west.

5. Whenever solving problems involving either angles or elevation and depression bearings, it is best always to draw a diagram and to label all the given information. 6. Set up a compass as the basis of your diagram for bearings questions.

(23)

Applications of

right-angled triangles

1 From a vertical fire tower 60 m high, the angle of depression to a fire is 6°. How far away, to the nearest metre, is the fire?

2 A person stands 20 m from the base of a building, and measures the angle of elevation to the top of the building as 55°. If the person is 1.7 m tall, how high, to the nearest metre, is the building?

3 An observer on a cliff top 57 m high observes a ship at sea. The angle of depression to the ship is 15°. The ship sails towards the cliff, and the angle of depression is then 25°. How far, to the nearest metre, did the ship sail between sightings?

4 Two vertical buildings, 40 m and 62 m high, are directly opposite each other across a river. The angle of elevation of the top of the taller building from the top of the smaller building is 27°. How wide is the river? (Give the answer to 2 decimal places.) 5 To calculate the height of a crane which is on top of a building, Dennis measures the angle of elevation to the bottom and top of the crane. These were 62° and 68° respec-tively. If the building is 42 m high, find, to 2 decimal places:

a how far Dennis is from the building b the height of the crane.

6 A new skyscraper is proposed for the Brisbane city region. It is to be 200 m tall. What would be the angle of depression, in degrees and minutes, from the top of the building to the base of the Kangaroo Point cliffs, which is 4.2 km away?

7 From a rescue helicopter 2500 m above the ocean, the angles of depression of two shipwreck survivors are 48° (survivor 1) and 35° (survivor 2).

a Draw a labelled diagram which represents the situation. b Calculate how far apart the two survivors are.

8 A lookout tower has been erected on top of a mountain. At a distance of 5.8 km, the angle of elevation from the ground to the base of the tower is 15.7° and the angle of elevation to the observation deck (on the top of the tower) is 15.9°. How high, to the nearest metre, is the observation deck above the top of the mountain?

9 From a point A on level ground, the angle of elevation of the top of a building 50 m high is 45°. From a point B on the ground and in line with A and the foot of the building, the angle of elevation of the top of the building is 60°. Find, in simplest surd form, the distance from A to B.

10 Express the following conventional bearings as true bearings.

11 Express the following true bearings in conventional form.

12

a A bearing of S30°E is the same as:

b A bearing of 280° T is the same as:

a N35°W b S47°W c N58°E d S17°E

a 246° T b 107° T c 321° T d 074° T

A 030° T B 120° T C 150° T D 210° T E 240° T

A N10°W B S10°W C S80°W D N80°W E N10°E

4D

W WORKEDORKED

E Example

13

Math_ca

d

SOH CAH TOA

CabriGeom_e

try Triangle

SkillS

HEET

4.5

SkillS

HEET

4.6

W WORKEDORKED

E Example

14

SkillS

HEET

4.7

m

(24)

13 A canoeist paddles 1800 m on a bearing of N20°E. How far north of her starting point is she, to the nearest metre?

14 A yacht race consists of four legs. The first three legs are 4 km due east, then 5 km south, followed by 2 km due west.

a How long is the final leg, if the race finishes at the starting point? b On what bearing must the final leg be sailed?

15 A ship sails 20 km south, then 8 km west. What is its bearing from the starting point? 16 A cross-country competitor runs on a bearing of N60°W for 2 km, then due north for

3 km.

a How far is he from the starting point?

b What is the true bearing of the starting point from the runner?

17 Two hikers set out from the same camp site. One walks 7 km in the direction 043° T and the other walks 10 km in the direction 133° T.

a What is the distance between the two hikers?

b What is the bearing of the first hiker from the second?

18 A ship sails 30 km on a bearing of 220°, then 20 km on a bearing of 250°. Find: a how far south of the original position it is

b how far west of the original position it is

c the true bearing of the ship from its original position, to the nearest degree. 19 The town of Bracknaw is due west of Arley. Chris, in an ultralight plane, starts at a

third town, Champton, which is due north of Bracknaw, and flies directly towards Arley at a speed of 40 km/h in a direction of 110°T. She reaches Arley in 3 hours. Find: a the distance between Arley and Bracknaw

b the time to complete the journey from Champton to Bracknaw, via Arley, if she increases her speed to 45 km/h between Arley and Bracknaw.

20 From a point, A, on the ground, the angle of elevation of the top of a vertical tower due north of A is 46°. From a point B, due east of A, the angle of elevation of the top of the tower is 32°. If the tower is 85 m high, find:

a the distance from A to the foot of the tower b the distance from B to the foot of the tower c the true bearing of the tower from B.

Fly like a bird

A bird flying at 50 m above the ground was observed at noon from my front door at an angle of elevation of 5°. Two minutes later its angle of elevation was 4°. a If the bird was flying straight and level, find the horizontal distance of the bird:

i from my doorway at noon ii from my doorway

at 12.02 pm. b Hence, find:

i the distance travelled by the bird in the two minutes

ii its speed of flight in km/h.

W WORKEDORKED

E Example

15

W WORKEDORKED

E Example

16

Work

(25)

The sine rule

Finding side lengths

The trigonometry we have studied so far has been applicable to only right-angled triangles. The sine rule allows us to calculate the lengths of sides and the size of angles in non-right-angled triangles. Consider the triangle drawn on the right.

Derivation of the sine rule

A, B and C represent the three angles in the triangle ABC and a, b and c represent the three sides, remembering that each side is named with the lower-case letter of the opposite vertex.

Construct a line from C to a point, D, perpendicular to AB. CD is the perpendicular height of the triangle, h.

Now consider ∆ACD and ∆BCD separately.

Using the formula for the sine ratio:

sin θ= sin θ=

sin A = sin B =

h = b sin A h = a sin B We are now able to equate these two expressions for h.

a sin B = b sin A Dividing both sides by sin A sin B we get: =

=

Similarly, we are able to show that each of these is also equal to . Try it! C

A A B

b c C

a

B

C

A A _D B

b a

h

c

B

C

A D

b h

C

D B

a h

opp hyp

--- opp

hyp

---h b

--- h

a

---a sin B sin A sin B

--- b sin A sin A sin B

---a sin A --- b

sin B

(26)

---The sine rule states that in any triangle, ABC, the ratio of each side to the sine of its opposite angle will be equal.

This formula allows us to calculate the length of a side in any triangle if we are given the length of one other side and two angles. When using the formula we need to use only two parts of it.

To use the sine rule we need to know the angle opposite the side we are finding and the angle opposite the side we are given. In some cases these are not the angles we are given. In such cases we need to use the fact that the angles in a triangle add to 180° to calculate the required angle.

a

sin A

--- b

sin B

--- c

sin C

---= =

Calculate the length of the side marked x in the

triangle on the right, correct to 1 decimal place.

THINK WRITE

Write the formula. =

Substitute a = x, b = 16, A = 80° and B = 40°.

=

Make x the subject of the equation by multiplying by sin 80°.

x=

Calculate and round to 1 decimal place. x≈ 24.5 cm

A

B 40° x C

80°

16 cm

1 a

sin A --- b

sin B

---2 x

sin 80°

--- 16 sin 40°

---3 16 sin 80°

sin 40°

---4

17

WORKED

E

xample

Calculate the length of the side labelled

m in the figure on the right, correct to

4 significant figures.

THINK WRITE

Calculate the size of angle C. C= 180° − 65° − 75° = 40°

1

2 a

sin A --- c

sin C

---18

WORKED

E

xample

A

B C

m

75° 65°

(27)

As mentioned in the previous investigation, we need to apply the sine rule to obtuse-angled triangles. In such examples the method used is exactly the same with the sub-stitution of an obtuse angle.

Using the sine rule allows us to solve a number of more complex problems. As with our earlier trigonometry problems, we begin each with a diagram and give a written answer to each.

The sine rule can be used to calculate the height of objects that it would otherwise be difficult to measure. Problems such as this can be solved by combining the use of the sine rule with the trigonometry of right-angled triangles covered earlier in this chapter.

THINK WRITE

Substitute a = 16, c = m, A = 65° and C = 40°. =

Make m the subject of the equation. m=

Calculate and round to 4 significant figures. = 11.35 m

3 16

sin 65°

--- m sin 40°

---4 16 sin 40°

sin 65°

---5

George looks south and observes an aeroplane at an angle of elevation of 60°. Henrietta is 20 km south of where Georg is and she faces north to see the aeroplane at an angle of elevation of 75°.

Calculate the distance of the aeroplane from Henrietta’s observation point, to the nearest metre.

THINK WRITE

Calculate the size of ∠GAH. A= 180° − 60° − 75° = 45°

Substitute g = x, a = 20, G = 60° and H = 75°. =

Make x the subject. x=

Calculate and round to 3 decimal places (nearest metre).

x= 24.495 km

Give a written answer. The distance of the aeroplane from Henrietta’s observation point is 24.495 km.

1

2 g

sin G --- a

sin A

---3 x

sin 60°

--- 20 sin 45°

---4 20 sin 60°

sin 45°

---5

6

19

WORKED

E

xample

A

G H

x

60° 75°

(28)

To calculate the height of a building, Kevin measures the angle of elevation to the top as 52°. He then walks 20 m closer to the building and measures the angle of elevation as 60°. How high is the building?

THINK WRITE

Draw a labelled diagram of the situation and fill in the given information.

Check that one of the criteria for the sine rule has been satisfied for triangle ABC.

The sine rule can be used for triangle ABC since two angles and one side length have been given.

Determine the value of angle ACB, using the fact that the angle sum of any triangle is 180°.

∠ACB = 180° − (52° + 120°) = 8°

Write down the sine rule to find b.

Substitute the known values into the rule.

Make b the subject of the equation. Calculate and round off the answer to 2 decimal places and include the appropriate unit.

To find side length b of triangle ABC:

=

b=

= 124.45 m

Draw a diagram of the situation, that is, triangle ADC, labelling the required information. Also label the sides of the triangle.

Choose the sine ratio as we are finding the opposite side and have been given the hypotenuse.

sin θ =

Make h the subject of the equation. 124.45 sin 52° =h

h= 124.45 sin 52°

Calculate and round off appropriately. = 98.07

Give a written answer. The height of the building is 98.07 m.

1

60° 52° 120°

A B D

C

h

x – 20

x

20

2

3

4

5

6

7

b sin B --- c

sin C

---b sin 120°

--- 20 sin 8°

---20×sin 120° sin 8°

---8

52° 124.45 m

A D

C

h

9 opp

hyp

---10 h

124.45

---11

12

13

20

(29)

Using the sine rule to find

side lengths

1 Write down the sine rule formula as it applies to each of the triangles below.

a b c

2 Use the sine rule to calculate the length of the side marked with the pronumeral in each of the following, correct to 3 significant figures.

a b c

3 In each of the following, use the sine rule to calculate the length of the side marked with the pronumeral, correct to 1 decimal place, by first finding the size of the third angle.

a b c

4 ABC is a triangle in which BC = 9 cm, ∠BAC = 54° and ∠ACB = 62°. Calculate the length of side AB, correct to 1 decimal place.

5 XYZ is a triangle in which y = 19.2 m, ∠XYZ = 42° and ∠XZY = 28°. Calculate x, correct to 3 significant figures.

remember

1. The sine rule formula is .

2. The sine rule is used to find a side in any triangle when we are given the length of one other side and two angles.

3. We need to use only two parts of the sine rule formula.

4. For written problems, begin by drawing a diagram and finish by giving a written answer.

a sin A --- b

sin B --- c

sin C

---= =

remember

4E

X

Z Y

P

R Q

W WORKEDORKED

E Example

17

A

B

x

C

50° 45°

16 cm

L

M q N

59° 63° 1.9 km

R

T

t

S 84°

52°

89 mm

W WORKEDORKED

E Example

18

G x H

I 74° 74°

18.2 mm

M

N

P

m

80° 62°

35.3 cm A

C B

19.4 km

y

85°

27° A

B C

b c

(30)

6 X and Y are two trees, 30 m apart on one side of a river. Z is a tree on the opposite side of the river, as shown in the diagram below.

It is found that ∠XYZ = 72° and ∠YXZ = 59°. Calculate the distance XZ, correct to 1 decimal place.

7 From a point, M, the angle of elevation to the top of a building, B, is 34°. From a point, N, 20 m closer to the building, the angle of elevation is 49°.

a Draw a diagram of this situation.

b Calculate the distance NB, correct to 1 decimal place.

c Calculate the height of the building, correct to the nearest metre.

8 To calculate the height of a building, Kevin measures the angle of elevation to the top as 48°. He then walks 18 m closer to the building and measures the angle of elevation as 64°. How high is the building?

9 A river has parallel banks which run directly east–west. Kylie takes a bearing to a tree on the opposite side. The bearing is 047° T. She then walks 10 m due east, and takes a second bearing to the tree. This is 305° T. Find:

a her distance from the second measuring point to the tree

b the width of the river, to the nearest metre.

10 A cross-country runner runs at 8 km/h on a bearing of 150° T for 45 mins, then changes direction to a bearing of 053° T and runs for 80 mins until he is due east of the starting point.

a How far was the second part of the run?

b What was his speed for this section?

c How far does he need to run to get back to the starting point?

11 From a fire tower, A, a fire is spotted on a bearing of N42°E. From a second tower, B, the fire is on a bearing of N12°W. The two fire towers are 23 km apart, and A is N63°W of B. How far is the fire from each tower?

W

WORKEDORKED

E Example

19

X 30 m Y

Z

59° 72°

W

WORKEDORKED

E Example

(31)

12

A boat sails on a bearing of N15°E for 10 km, then on a bearing of S85°E until it is due east of the starting point. The distance from the starting point to the nearest kilometre is, then:

13

A hill slopes at an angle of 30° to the horizontal. A tree which is 8 m tall is growing at an angle of 10° to the vertical and is part-way up the slope. The vertical height of the top of the tree above the slope is:

14 A cliff is 37 m high. The rock slopes outward at an angle

of 50° to the horizontal, then cuts back at an angle of 25° to the vertical, meeting the ground directly below the top of the cliff.

Carol wishes to abseil from the top of the cliff to the ground as shown in the diagram. Her climbing rope is 45 m long, and she needs 2 m to secure it to a tree at the top of the cliff. Will the rope be long enough to allow her to reach the ground?

A 10 km B 38 km C 110 km D 113 km E 114 km

A 7.37 m B 8.68 m C 10.84 m D 15.04 m E 39.89 m

Bearing east and west

At the beginning of this chapter we looked at Rick, who wanted to calculate the width of the Grand Canyon.

From a point on the southern side of the canyon Rick sighted a lookout on a

bearing of N5°E. Rick then

headed east along the southern rim for 8.1 km, where the bearing of the

lookout was N12°W.

1 Draw a diagram of this situation and label the lookout on the northern rim A,

the easternmost point B and the westernmost point C.

2 Use the sine rule to calculate the distance AB.

3 Draw a perpendicular from A to the side AB. This will represent the width of

the canyon.

4 Use right-angled triangle trigonometry to find the width of the canyon in

kilometres, correct to 1 decimal place. m

multiple choiceultiple choice

m

25° 50°

rope

rock

(32)

Finding angle sizes

Using the sine rule result we are able to calculate angle sizes as well. To do this, we need to be given the length of two sides and the angle opposite one of them. For simplicity, in solving the triangle we invert the sine rule formula when we are using it to find an angle. The formula is written:

As with finding side lengths, we use only two parts of the formula.

As with finding side lengths, some questions will be problems that require you to draw a diagram to extract the required information and then give the answer in written form.

sin A a

--- sin B b

--- sin C c

---= =

Find the size of the angle, θ, in the figure on the right,

correct to the nearest degree.

THINK WRITE

Substitute A = 110°, C =θ, a = 20 and c = 6.

=

Make sin θ the subject of the equation. sin θ=

Calculate a value for sin θ. sin θ= 0.2819

Calculate sin−1(0.2819) to find θ. θ= 16°

B

A

C 110°

20 cm

6 cm _θ

1 sin A

a

--- sin C c

---2 sin 110°

20

--- sin θ 6

---3 6 sin 110°

20

---4 5

21

WORKED

E

xample

From a point, P, a ship (S) is sighted 12.4 km from P on a bearing of 137°. A point, Q, is due south of P and is a distance of 31.2 km from the ship. Calculate the bearing of the ship from Q, correct to the nearest degree.

THINK WRITE

Draw a diagram.

1 P

Q S 12.4 km

31.2 km 43°

137°

22

(33)

Using the sine rule to find

angle sizes

1 Find the size of the angle marked with a pronumeral in each of the following, correct to the nearest degree.

a b c

d e f

THINK WRITE

Substitute for p, q and P. =

Make sin Q the subject. sin Q=

Calculate a value for sin Q. sin Q= 0.271

Calculate sin−1(0.271) to find Q. Q= 16°

Give a written answer. The bearing of the ship from Q is 016°.

2 sin Q

q

--- sin P p

---3 sin Q

12.4

--- sin 43° 31.2

---4 12.4 sin 43°

31.2

---5 6 7

remember

1. The sine rule formula for finding an angle is .

2. We can use this formula when we are given two sides and the angle opposite one of them.

3. Problem questions should begin with a diagram and finish with a written answer.

sin A a

--- sin B b

--- sin C c

---= =

remember

4F

W WORKEDORKED

E Example

21

B

A

C 100°

46 cm 32 cm

θ

Q

P

R 60°

18.9 m 29.5 m φ

M L

N 117°

153 mm 79 mm

α

U V

W 75°

23.6 km 23.6 km

θ _Y

X

Z 86°

27.6 cm 16.5 cm

β

170°

156 mm 27 mm

(34)

2

Which of the statements below give the correct value for sin θ?

A B

C D

3

In which of the triangles below is the information insufficient to use the sine rule?

A B

C D

4 In ∆PQR, q = 12 cm, r = 16 cm and ∠PRQ = 56°. Find the size of ∠PQR, correct to the nearest degree.

5 In ∆KLM, LM = 4.2 m, KL = 5.6 m and ∠KML = 27°. Find the size of ∠LKM, correct to the nearest degree.

6 A, B and C are three towns marked on a map. Judy calculates that the distance between A and B is 45 km and the distance between B and C is 32 km. ∠CAB is 45°. Calculate ∠ACB, correct to the nearest degree.

7 A surveyor marks three points X, Y and Z in the ground. The surveyor measures XY to be 13.7 m and XZ to be 14.2 m. ∠XYZ is 60°.

a Calculate ∠XZY to the nearest degree. b Calculate ∠YXZ to the nearest degree.

8 Two wires support a flagpole. The first wire is 8 m long and makes a 65° angle with the ground. The second wire is 9 m long. Find the angle that the second wire makes with the ground.

9 A ship sails on a bearing of S20°W for 14 km, then changes direction and sails for 20 km and drops anchor. Its bearing from the starting point is now N65°W.

a How far is it from the starting point? b On what bearing did it sail the 20 km leg?

m

multiple choiceultiple choice

θ 36°

7 13

sin θ 13 sin 36° 7

---= sin θ 7 sin 36°

13 ---=

sin θ 36 sin 13° 7

---= sin θ 7 sin 13°

36 ---=

m

θ

57°

12.6 m 14.8 m

θ

45° 16.2 m

12.7 m

θ

115°

12.9 m

6.2 m

θ 9°

8.7 m 12.7 m

W WORKEDORKED

E Example

(35)

The cosine rule

The previous two sections have looked at finding sides and angles using the sine rule. The sine rule, however, will not allow us to solve all triangles. Depending on the infor-mation provided about the triangle we may need to use the cosine rule. In this section we will use the cosine rule to find both sides and angles of non-right-angled triangles.

In any non-right-angled triangle, ABC, a perpendicular line can be drawn from angle B to side b. Let D be the point where the perpendicular line meets side b, and the length of the perpendicular line be h. Let the length AD = x units. The perpendicular line creates two right-angled triangles, ADB and CDB.

Using triangle ADB and Pythagoras’ theorem, we obtain:

c2= h2+ x2 [1]

Using triangle CDB and Pythagoras’ theorem, we obtain:

a2= h2+ (b − x)2 [2]

Expanding the brackets in equation [2]: a2= h2+ b2− 2bx + x2

Rearranging equation [2] and using c2= h2+ x2 from equation [1]: a2=h2+x2+ b2− 2bx

=c2+ b2− 2bx = b2+ c2− 2bx

From triangle ABD, x = c cos A, therefore a2= b2+ c2− 2bx becomes a2= b2+ c2− 2bc cos A

This is called the cosine rule and is a generalisation of Pythagoras’ theorem.

In a similar way, if the perpendicular line was drawn from angle A to side a or from angle C to side c, the two right-angled triangles would give c2= a2+ b2− 2ab cos C and b2= a2+ c2− 2ac cos B respectively. From this, the cosine rule can be stated:

In any triangle ABC

a2=b2+c2− 2bc cos A

b2=a2+c2− 2ac cos B

c2=a2+b2− 2ab cos C

The cosine rule can be used to solve non-right-angled triangles if we are given: 1. three sides of the triangle

2. two sides of the triangle and the included angle (the angle between the given sides).

D

c

b – x x

b a h

C A

B

b a c

A C

B

Find the third side of triangle ABC given a= 6, c= 10 and B= 76°.

THINK WRITE

Draw a labelled diagram of the triangle ABC and fill in the given information.

Write down the appropriate cosine rule to find side b.

b2=a2+ c2− 2ac cos B

Substitute the given values into the rule. = 62+ 102− 2 × 6 × 10 × cos 76°

Evaluate. ≈ 106.969 372 5

b=

Round off the answer to 2 decimal places. ≈ 10.34

1

b

a = b

c = 10

A C

B 76°

2

3 4

106.969 372 5

5

23

(36)

Note: Once the third side has been found, the sine rule could be used to find other angles if necessary.

If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject.

a2=b2+c2− 2bc cos A ⇒ cos A=

b2=a2+c2− 2ac cos B ⇒ cos B=

c2=a2+b2− 2ab cos C ⇒ cos C=

b2₊_c2_–_a2

2bc

---a2₊_c2_–_b2

2ac

---a2+_b2_–_c2

2ab

---Find the smallest angle in the triangle with sides 4 cm, 7 cm and 9 cm.

THINK WRITE

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

Note: The smallest angle will correspond to the smallest side.

Let a= 4

b= 7

c= 9

Write down the appropriate cosine rule to find angle A. cos A=

Substitute the given values into the rearranged rule. =

Evaluate. =

=

Make A the subject of the equation. A= cos−1

Calculate and round off to the nearest minute. = 25°13′

1

a = 4

c = 7

b = 9

A C

B

2 b

2+_c2_–_a2

2bc

---3 7

2+₉2_–₄2

2×7×9

---4 49+81–16

126

---114 126

---5 114

126

--- 

 

6

24

WORKED

E

xample

Two rowers set out from the same point. One rows N70°E for 2000 m and the other rows S15°W for 1800 m. How far apart are the two rowers?

THINK WRITE

Draw a labelled diagram of the triangle, call it ABC and fill in the given information.

1

B C

A 2 000 m

1800 m N

15° 70°

25

(37)

The cosine rule

1 Find the third side of triangle ABC given a = 3.4, b = 7.8 and C = 80°.

2 In triangle ABC, b = 64.5, c = 38.1 and A = 58°34′. Find a.

3 In triangle ABC, a = 17, c = 10 and B = 115°. Find b, and hence find A and C.

4 Find the smallest angle in the triangle with sides 6 cm, 4 cm and 8 cm.

5 In triangle ABC, a = 356, b = 207 and c = 296. Find the largest angle.

6 In triangle ABC, a = 23.6, b = 17.3 and c = 26.4. Find the size of all the angles.

7 Two rowers set out from the same point. One rows N30°E for 1500 m and the other rows S40°E for 1200 m. How far apart are the two rowers?

THINK WRITE

Write down the appropriate cosine rule to find side c. _c2₌_a2₊_b2₋₂_ab_cos_C

Substitute the given values into the rule. =20002+ 18002− 2 × 2000 × 1800 cos 125°

Evaluate. ≈11 369 750.342

c=

Round off the answer to 2 decimal places. ≈3371.91

Give a written answer. The rowers are 3371.91 m apart.

2

3

4

11 369 750.342

5 6

remember

1. In any triangle ABC:

a2=b2+c2− 2bc cos A b2=a2+c2− 2ac cos B c2=a2+b2− 2ab cos C

2. The cosine rule can be used to solve non-right-angled triangles if we are given: (a) three sides of the triangle

(b) two sides of the triangle and the included angle (that is, the angle between the two given sides).

3. If three sides of a triangle are known, an angle could be found by transposing the cosine rule to make cos A, cos B or cos C the subject.

a2=b2+c2− 2bc cos A ⇒ cos A=

b2=a2