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1010Trigonometric ratios and

applicationsTrigonometry comes from the Greek words ‘trigon’ (triangle) and ‘metron’(measurement). Trigonometry has many applications in mathematics; however,in this chapter we will investigate its use in triangles. In simple termstrigonometry involves finding unknown side lengths or unknown angles. Wecan apply trigonometry of triangles in many areas of real life. Surveyors usetrigonometry to calculate fence lines of a house. Ship captains usetrigonometry to get a safe passage out of a harbour or navigate around theworld. Engineers use trigonometry to determine how high a bridge can bebuilt. The applications for trigonometry go on and on. In the past you wouldhave investigated the trigonometry of right-angled triangles. Let’s revisit thistheory and also investigate trigonometry of non right-angled triangles.

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MathsWorld General Mathematics Units 1 & 2

Right-angled triangles

Trigonometric ratios of right-angled trianglesWe first need to identify names or references for the side lengths and angles of a right-angled triangle. Call one of the angles, other than the right angle, θ (theta).

The diagrams show how to name the sides of a right-angled triangle, relative to the angle θ.

. The hypotenuse (H) faces the right angle.

. The opposite side (O) faces θ.

. The adjacent side (A) is next to θ .

The trigonometric ratios linking the three sides and the angle are known as sine, cosine and tangent. They are abbreviated to sin, cos and tan, and they are defined as shown.

All the ratios involve three parts,namely two sides and the angle θ . It depends on the question(s) posedwhich one of the three ratios you willneed to use. This is the first decision that needs to be made. You are probably familiar with the mnemonicSOH-CAH-TOA, to help with remembering the rules for each ratio.

Forms for anglesAngles are usually expressed in one of two ways: in decimal degree form (e.g. 21.25°), or in degrees and minutes form (e.g. 21°15′), where there are 60 minutes (60′) in one degree. (In some applications there is a further division into seconds, where there are 60 seconds (60″) in one minute.)

It is possible to convert from one form to another.

Hypotenuse

Side opposite

Side adjacent to

θ

A H

O

θ

θ

θ

AH

O

OR

Standard trigonometric ratios

(SOH)

(CAH)

(TOA)

sin θ Opposite sideHypotenuse

-------------------------------------= sin θ OH-----=

cos θ Adjacent sideHypotenuse

---------------------------------------= cos θ AH----=

tan θ Opposite sideAdjacent Side-------------------------------------= tan θ O

A-----=

10.110.1

Trigonometric ratios and applications

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10Converting angle forms

Example 1a Convert 35.13° into degrees and minutes.

b Convert 27°14′ into decimal form, correct to two decimal places.

Solutiona Since there are 60′ in one degree,

0.13° represents 0.13 × 60 = 7.8′. To the nearest minute, this is 8′.So 35.13° = 35°8′

b 14′ represents of one degree.

= 0.2333 …

So 27°14′ = 27.23° to two decimal places.

Example 2Find the value of the following, correct to three decimal places, using a graphics calculator.a sin 33° b cos 76.87° c tan 12°55′

SolutionFirst make sure your calculator is set in degree mode. The screenshots show the results.a b

So sin 33° = 0.545 So cos 76.87° = 0.227

c As there are 60′ in one degree, 55′ represents

of a degree. You can then use your calculator in two different ways as shown in the screenshot.

So tan 12°55′ = 0.229.

tipYou can use your graphics calculator to automate this process. For example 1a, use the DMS (degrees, minutes, seconds) command from the ANGLE menu. For example 1b, just enter the angle.

If the number of seconds ( ″ ) is 30 or more, round up to give the nearest minute. So for part a, the angle is 35°8′.

GC 1.6CAS 1.6

1460------

1460------

GC 1.6CAS 1.6

tipWhen set in DEGREE mode, you do not have to type the degree symbol. Thus, SIN(33) and SIN(33°) will both give the same answer. However, if the angle is expressed in degrees and minutes, you need to change to degrees or identify the number of degrees and minutes as in example 2c.

5560------

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exercise 10.11 Use your graphics calculator to find the value of the following to 3 decimal places.

a sin 43° b cos 5° c tan 77° d cos 56°

e sin 59° f tan 12° g sin 23.55° h tan 16°15′i cos 88.95° j cos 57°3′ k sin 0.99° l tan 11°11′

2 Convert the following angles into degrees and minutes.

a 23.55° b 16.99° c 253.9° d 56.087°

e 150.5° f 1.75° g 33.555° h 12.25°

3 Convert the following angles into decimals (to two decimal places).

a 32°25′ b 54°36′ c 113°55′ d 66°59′e 45°01′ f 2°2′ g 22°30′ h 72°45′

4 For the following right angled triangles:

i name all the parts of the triangle labelled using O, A, H and θ .

ii choose which trigonometric ratio could be used to find the pronumeral (but don’t solve).

a b

c d

e f

33°

3 mx m

57°

56 mmx mm

575 mm

335 mmx °

47°

3.45 cmx cm

20 m

12 m

x°13.66 cm

°

21.80 cm25.73 cm

x


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10Finding a sideExample 3Find the value of x, correct to three decimal places.

SolutionThe required side is opposite the known angle, and we are also given the hypotenuse, so we use the sine ratio.

sin θ =

sin 35° =

x = 5 × sin 35°= 2.868

Example 4Find the value of x, correct to three decimal places.

SolutionThe required side is facing the right angle(the hypotenuse), and we are given theadjacent side, so we use the cosine ratio.

cos θ =

cos 56° =

x =

=

= 20.208

Finding an angleExample 5Find the angle θ in this right-angled triangle,giving the answer in degrees and minutes.

35°

5 cmx cm

On a graphics calculator, the calculation looks like this:

OH-----

x5---

GC 1.6CAS 1.6

56° 11.3 mm

x mm

On a graphics calculator, the calculation looks like this:

AH----

11.3x

-----------

11.3 cos 56°÷11.3

cos 56°-------------------

24 m 17 m

θ

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SolutionIdentifying the information given we have the opposite and the adjacent side lengths. So we use the tangent ratio.

tan θ =

=

To find θ , use the inverse of tan on your calculator, shown as Tan−1 on the TI-83/84.

θ = 35.31°On a graphics calculator the calculation looks like this.

So in degrees and minutes the angle is 35°19′.

Example 6Find the values of x and y in the given composite shape, correct to two decimal places.

SolutionFirst we need to identify the shapes that make up the composite shape.These are two right-angled triangles and a rectangleas shown.We can use the left triangle to find the height,h cm, of the composite shape.Since this is opposite the known angle, and we are given the hypotenuse, we use the sine ratio.

sin θ =

sin 70° =

h = 19 × sin 70°= 17.854

OA-----

1724------

GC 1.5, 1.6CAS 1.5, 1.6

WarningIn many problems involving trigonometric ratios, errors are made when the incorrect trigonometric ratio is used to find an unknown, or when the trigonometric ratio used is not correctly transposed when finding the unknown.

70° 80°

19 cm

11 cm

y cm

x cm

70°

80°

19 cm x cmh cm

tipWhen two-decimal-place accuracy is required, work to at least three decimal places and correct at the end. Alternatively, store intermediate values in your graphics calculator to work with later. For example, the answer for h could be stored in location H and retrieved for further calculation.

h cm

70°

19 cm

OH-----

h19------


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10Using this height we can determine the value of x in the right triangle.

sin θ =

sin 80° =

x = 17.854 ÷ sin 80°

=

= 18.129

To find the value of y we need to find the lengths of the adjacent sides in the two triangles, labelled as a cm and b cm below.

Use the cosine ratio in each case.

cos θ = cos θ =

cos 70° = cos 80° =

a = 19 × cos 70° b = 18.129 × cos 80°= 6.498 = 3.148

So y = a + 11 + b= 6.498 + 11 + 3.148= 20.646

Thus, correct to two decimal places, x = 18.13 and y = 20.65.

exercise 10.15 Using trigonometric ratios find the value of x in each of the following. (Give answers to

three decimal places in each case.)

a b

17.854

80°

x cm

OH-----

17.854x

------------------

17.854sin 80°------------------

a cm70°

19 cm 18.129 cm

b cm80°

AH---- A

H----

a19------ b

18.129------------------

continued

43°

x m

12 m 33°

6.25 m

x m

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c d

e f

6 Find the value of the pronumerals in each of the following, giving answers to three decimal places.

a b c

d e

7 Find the value of θ in degrees and minutes correct to the nearest minute.

a sin θ = 0.236 b cos θ = c tan θ = 2.667

d cos θ = 0.183 e sin θ = 0.977 f tan θ =

63°4’11.5 mm

x mm

23.5 m

x m

13°23′

67.35°

115x

101.1 m

x m

42°

75°

21 x

y

13

50°10 cm

x cmy cm

40°

6 mx m

52°

68°

16

xy 40° 20°

18

y

x

b

a

115237---------

1135------


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108 Find the value of θ, giving answers in degrees to two decimal places.

a b c

d e f

9 Find the value of θ, giving answers in degrees and minutes correct to the nearest minute.

a b c

d e f

10 Find the values of x and y, giving answers in degrees correct to two decimal places.

a b

θ11 m

21 m

θ

235 m

116 m 5

3

θ

θ87.2

43.6θ 21.57 mm

73.66 mm 4.04 cm

5.05 cm

θ

θ117.9

67.3

θ23.55 cm

77.89 cm

12.06 mm17.89 mm

θ

θ

61.256.7

97.1 m

117.8 mθ

θ15.22 m

167.83 m

3.34.5

x

y

3.5

6.67

4.8

x

y

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Two-dimensional applicationsExample 7A stunt woman wants to make a double-sided motorcycle ramp so that it is set at an incline of 27°10′,as shown in the diagram. How longcorrect to the nearest cm is eachside of the ramp?

SolutionDrop a perpendicular line from Q to PR, meeting at S.The width of the ramp is 11.7 m, so PS is half that distance which is 5.85 m.In the right-angled triangle PQS, we have an angle at P,the length of the adjacent side, and we want to find PQ, the hypotenuse. So we use cosine.We know the side adjacent to the hypotenuse.Substitute in the given information.

cos θ =

cos 27°10′ =

PQ × cos 27°10′ = 5.85

PQ =

= 6.5754Since we require the answer correct to the nearest centimetre, i.e. to the nearest one hundredth of a metre, round to two decimal places.Hence each side of the motorcycle ramp has length 6.58 m.

exercise 10.111 A 7.4 metre-long ladder rests against a vertical wall and makes an angle of 35° with the

horizontal ground.

a How high up does the ladder reach on the wall?

b If the wall is 5 m high, what angle must the ladder make with the horizontal for the ladder to just reach the top of the wall?

12 A wheelchair ramp needs to be at an angle of 5°13′ with the ground. If a step 75 cm high is to have a wheelchair ramp fitted onto it for wheelchair access, how long, to the nearest cm, should the wheelchair ramp be?

27°10′ 27°10′11.7 m

Q

P R

27°10'

Q

SP5.85 m

AH----

5.85PQ-----------

5.85cos 28°10′---------------------------

continued


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1013 Michael is playing golf; he is about to take a shot using a golf club 1.05 m long. While he

is holding his club his arms add 56 cm to the length of the club and his feet are 90 cm from the ball. If Michael stands at a right angle to the ground when taking the shot, what angle does the golf club make with the ground?

14 A young girl is flying her new kite. The kite has a guide rope of length 125 m and the maximum height the kite can reach is 33.5 m above the ground. If the girl is holding the end of the guide rope 1 m above the ground, find the angle made by the guide rope with the horizontal, to the nearest minute, if the kite is at its maximum height.

15 Find the values of x and y to two decimal places.The base of the wedge is a rectangle.

16 A camping tent is 1.25 m tall when fully erected. For stability the tent needs four cables attached to the ground so they each make an angle of 33°25′ with the horizontal. What is the minimum amount of cable required?

x cm

y cm

23°

115 cm

1.25 m

3325′

Numbersense with the spence

17Seventeen is important in the Islamic tradition. The great Muslim alchemist Jabir ibn

Hayyan believed that 17 was the basis for the material world, which consisted of the series 1, 3, 5 and 8. The cycles of prayer movements in the 5 daily prayer amounts to 17 and 17 is the number of words in the call to prayer.

10.1

384


Applications of trigonometry

Angles of elevation and depressionAngles of elevation and depression involve the use of trigonometry to once again find a side length or an angle. For example, angles of elevation and depression are commonly used in navigation or construction. They are used when we either have to look up to something fromthe ground or look down at something from a mountain or top of a building.

The angle of elevation is the angle θ between the horizontal and the line of sight when an object is higher than the observer. In this diagram it is the angle between the horizontal and the line of sight from the boat to the lighthouse.

The angle of depression is the angle θ between the horizontal and the line of sight when an object is below the level of the observer. In this diagram it is the angle between the horizontal and the line of sight from the lighthouse to the boat.

It is important to remember that the angles of elevation and depression are always measured from the horizontal. The angles of elevation and depression are equal becausethey are alternate angles.

line of sight

angle of elevationθ

observationpoint

line of sight

angle of depressionθ

observation point

angle of elevation

angle of depression

θ

θ

Warning Label or perish!When you are solving problems using angles of elevation and depression it is important to draw clear diagrams labelling all sides with the relevant information. In particular, think about whether the angle at the observation point (θ ) marks the angle down from the horizontal (for depression) or marks the angle up from the horizontal (for elevation).

10.210.2


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10Example 1A boat is 240 m from the base of a cliff 150 m high. Find the angle of elevation from the boat to the top of the cliff to the nearest degree.

SolutionDraw a diagram and clearly label with the relevant information.Using SOH–CAH–TOA, the opposite and adjacent sides are given so we use tan.

tan θ =

= 0.625θ = 32.005°

The angle of elevation from the boat tothe top of the cliff is 32°, correct to the nearest degree.

Example 2The angle of depression of a cyclist from the top of a building 200 m away is 50°.Calculate the height of the building to the nearest metre.

SolutionDraw a diagram and clearly label with the relevant information.Let x metres be the height of the building.∠A = 50° (alternate angles)Using SOH–CAH–TOA, the opposite and adjacent sides are involved so we use tan.

tan 50° =

x = 200 × tan 50° = 238.35

The height of the building is 238 metres, correct to the nearest metre.

exercise 10.21 The angle of elevation of a kite, K, from a point A on the

ground is 35°. If the kite is 75 m above the ground, what is the line-of-sight distance, to the nearest metre, from point A to the kite?

150 m

Clif

f

240 m

150240----------

x m

B

C200 m

50°

Ax200----------

K

A B

75 m

35°

10.2

386


2 To find the height of the skyscraper in the diagram an engineer

used the equation tan 55° = .

a Why did the engineer use tan 55°?

b What is the height of the skyscraper to the nearest metre?

3 The angle of depression from the top of a lifesaving tower 100 m high to a buoy at sea is 23°18′. Find the horizontal distance from the lifesaving tower to the buoy correct to 2 decimal places.

4 A yacht is 2.4 km away from the base of a lighthouse. The lighthouse is 300 m in height. Calculate the angle of depression to the nearest degree, from the top of the lighthouse to the yacht.

5 The angle of elevation from a boat to the top of a cliff is 15°24′. The boat is 50 m from the base of the cliff. How high is the cliff to the nearest metre?

6 Charlotte standing on the ground observes a bird on top of a tree which is 100 m away. Find the angle of elevation to the nearest degree given Charlotte is 1.6 m tall and the tree has a height of 15 m.

7 A helicopter pilot spots two distressed fishermen whoseboat has capsized. The helicopter is 200 m above sea level. The angles of depression of the two fishermenare 37°27′ and 58°24′ respectively. The fishermen andthe helicopter lie in the same vertical plane.

Complete the diagram by indicating all angles and lengths, and hence find the distance between the two fishermen, correct to the nearest metre.

8 A wheel chair access ramp rises 30 cm for every 1 m of the ramp. Find the angle of elevation of the ramp in degrees and minutes.

9 Paradise Hotel and The Venetian are two vertical hotels 68 m and 115 m high respectively. They stand directly opposite each other with a lake separating them. The angle of elevation from the top of the Paradise Hotel to the top of The Venetian is 31°.

Draw a clearly labelled diagram representing the situation, and hence find the width of the lake? Give your answer to two decimal places.

10 Use the information provided in the diagramto calculate the height of a communications tower correct to the nearest metre. Note that the tower is located at the top of a hill.

80 m

55°x80------

H

F1 F2

x m

33°18′

35°

320 m


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10BearingsBearings are used to represent the direction of one object from another. There are two different ways bearings are expressed. The first is as a true bearing. True bearings are measured in a clockwise direction beginning from north, which is 0°T. True bearings are generally expressed using 3 digits, for example 050°T or 120°T or 340°T. In practice where there is no possibility of confusion, the T is omitted as in these diagrams.

Bearings are also expressed as compass or conventional bearings. These bearings are measured first from north or south, then in the direction of east or west. For example, N75°E and S25°W are shown below.

It is possible to convert from one type of bearing to the other. For example N75°E is the same as 075°, S25°W is the same as 205°(180° + 25°) and 335° (360° − 25°) is equivalent to N25°W.

Example 3A hiker walks 3.2 km in a direction N34°E to base camp B. How far north is she from her starting point O, correct to two decimal places?

SolutionDraw a clearly labelled diagram. Identify the right-angled triangle.

N

W E

S

120°

N

W E

S

340°

N

W E

S

75°

N75°E

N

W E

S

25°

S25°W

tipWhen solving questions involving directions and bearings it is important to always draw compass points N, S, E, W and clearly label diagrams. Remember that all bearings are in a horizontal plane. Due north is directly north, due south is directly south, etc.

B

O

N

W E

S

34° 3.2 km

BA

O

34°3.2 kmx km

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388


Using SOH–CAH–TOA, the hypotenuse and adjacent sides are involved so use cos.

cos 34° =

x = 3.2 × cos 34°= 2.65

The hiker is 2.65 km north of her starting point O.

Example 4A ferry travels 4.5 km from a point A on a bearing of 150° to a point B.a How far south has the ferry travelled, correct to one decimal place?

b From this point if the ferry turns and travels on a bearing of 100° for 7.1 km to a point C, how far south is the ferry from point A, correct to one decimal place?

Solutiona Draw a clearly labelled diagram. Identify the right-angled triangle.

Using SOH–CAH–TOA, the hypotenuse and adjacent sides are involved so we use cos.

cos 30° =

x = 4.5 × cos 30° = 3.897

The ferry has travelled 3.9 km south.

b Draw a clearly labelled diagram Identify the right-angled triangle.representing the new information.

x3.2--------

A

B

4.5 km

150°

N

W E

S

A

B

4.5 km30°

x km

x4.5--------

N

A

B

C

4.5 km

150°

100°

7.1 km

N

y kmB

C

80°7.1 km


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10Using SOH–CAH–TOA, the hypotenuse and adjacent sides are involved so we use cos.

cos 80° =

y = 7.1 × cos 80° = 1.233

To find how far south the ferry is from point A, simply add x and y.

x + y = 3.987 + 1.233= 5.130

So the ferry is 5.1 km south of point A.

exercise 10.211 Match the true and compass

bearings in the left-hand list to the true and compass bearings in the right-hand list.

12 A car travels due south for 200 m and then turns due east and travels a further 300 m. What is the bearing of the car from its starting point (to the nearest degree)?

13 X is on a bearing of 050° from Y. What is the bearing of Y from X?

14 A plane travels on a bearing of S12°E for 2 km. How far south is the plane from its starting point now?

15 A speedboat travels 1900 metres on a bearing of N18°E. How far east has the boat travelled from its starting point to the nearest metre?

16 A helicopter flies due north for 260 km and then turns due west and flies a further distance of 410 km. Calculate the bearing, to the nearest degree, of the helicopter from its starting point.

17 A hiker on an expedition walks 5 km on a bearing of N60°E to a point A. She then turns and walks due south to a marker M. Her final leg is from M, due east of her starting point O, back to O. Find the total length of the expedition to the nearest km.

18 A boat travels 3.3 km from a point A on a bearing of 080° to a point B.

a How far north of the starting point has the boat travelled, correct to one decimal place?

b From this point the boat turns and travels on a bearing of 045° for 5.2 km to a point C. How far north is the boat from point A, correct to one decimal place?

y7.1--------

continued

N23°W 080°340° 225°S12°E 357°N80°E 337°269° S80°ES45°W N20°WN3°W 168°025° N25°E100° 165°S15°E S89°W

CD10.1Sailing

TAI

10.2

390


19 From a point P at sea one yacht sailed 5 km on a bearing of 040° and a second yacht sailed 8 km on a bearing of 130°.

a Draw a diagram of the situation, and hence find how far apart the two yachts are at this time.

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

20 A cruiser starting from a point P travels on a bearing of 205° for 22 km. It then changes course and travels to a point Q on a bearing of 240° for 30 km. Calculate:

a how far south the cruiser is from its original starting point.

b how far west the cruiser is from its original starting point.

c the bearing of the cruiser from its original starting point.

CD10.2B

aywatch rescue

SAC

analysis task

A lifeguard observes a rescue boat due north of his position on the beach. The lifeguard also observes a swimmer in distress on a bearing of N52°E at a distance of 470 m. The swimmer is due east of the rescue boat.

a Draw a diagram representing the above situation.

b i Calculate the distance between the lifeguard and the rescue boat.

ii Calculate the distance between the rescue boat and the swimmer.

The rescue boat is experiencing difficulties and is unable to reach the swimmer. The lifeguard looks a further 25° to the right of the swimmer in distress and spots a shark 900 m away.

c Add the shark to the original diagram.

d What is the bearing of the shark from the lifeguard?

e How far north is the shark from the lifeguard?

f How far east is the shark from the lifeguard?

g What is the distance between the swimmer and the shark? Who is closer to the swimmer: the lifeguard or the shark?

In order for the lifeguard to rescue the distressed swimmer the shark needs to be caught. A second rescue boat on the beach due east of the lifeguard is equipped to catch sharks. The second rescue boat is in a direct line with the shark and the swimmer.

h Add the second rescue boat to the diagram.

i Calculate how far the boat must travel and on what bearing to catch the shark, given that the shark has not begun to approach the swimmer yet.

analysis task 1—baywatch rescue

SAC


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10Non-right angled triangles

A useful way to label non-right angled triangles is to labelthe three vertices with uppercase letters for example A, B and C, and then label the sides opposite these vertices withthe corresponding lowercase letters, a, b and c. Using this notation, A also represents the angle ∠BAC and a representsthe length of the side opposite angle A.

Sine ruleFor a triangle ABC, the sine rule states that the ratios of each side to the sine of its opposite angle are equal:

Note that a is opposite A, b is opposite B and c is opposite C.

The sine rule is used when we know either:

. one side and two angles, or

. two sides and an angle opposite one of the sides.

Example 1For ∆ABC find the remaining side lengths, correct to two decimal places, given that B = 33°, C = 75° and AB = 7.9 mm.

SolutionFirst draw a sketch of the triangle and label it appropriately. Since we know c, B and C, we can find b by substituting the given information into the appropriate formula:

=

b =

= 4.45The angles in a triangle sum up to 180°, so if B = 33° andC = 75°, then A = 72°.Now we can solve for a:

=

a =

= 7.78Hence AC = 4.45 mm and BC = 7.78 mm.

B

A C

ac

b

asin A------------ b

sin B------------ c

sin C------------= =

A75°

33°

a mm

B

C

7.9 mm

b mm

bsin 33°------------------ 7.9

sin 75°-----------------

7.9sin 75°----------------- sin 33°×

asin 72°----------------- 7.9

sin 75°-----------------

7.9sin 75°----------------- sin 72°×

10.310.310.3

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The screenshot at the right shows the calculations used to find b and a.

Example 2For this triangle, find the remaining angles and remaining side lengths.

SolutionIdentify the information:

c = 15.9a = 23.6A = 112°

We can use these values to solve for C:

=

=

= sin Csin C = 0.6247

C = 38.66°Now we can find B, since A + B + C = 180°.

B = 180° − 150.66°= 29.34°

We can now find b using these values:a = 23.6A = 112°B = 29.34°

Substitute into the appropriate formula to solve for b:

=

= b

b = 12.47In summary, AC = 12.47 cm, B = 29.34° and C = 38.66°.

GC 1.6CAS 1.6

B

A C

15.9 mm

23.6 mm

112°b mm

tipYou can use your calculator to find C asshown in the following screenshot.

23.6sin 112°-------------------- 15.9

sin C------------

sin 112°23.6

-------------------- sin C15.9------------

sin 112°23.6

-------------------- 15.9×

23.6sin 112°-------------------- b

sin 29.34°-------------------------

23.6sin 112°-------------------- sin 29.34°×


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10exercise 10.31 a In ∆ABC, a = 11, b = 12 and B = 48°. Find A, C and c.

b In ∆ABC, c = 13.75, a = 28.62 and A = 110°37′. Find C, B and b.

c In ∆ABC, a = 5.5, A = 25° and B = 80°. Find C, b and c.

d In ∆ABC, c = 2.73, C = 52.5° and A = 102.4°. Find B, a and b.

2 In ∆ABC, a = 7, b = 5 and A = 72°. Find the perimeter of the triangle.

3 Find the shortest side of ∆ABC, where A = 33°56′, B = 35° and AB = 80 mm.

4 Find the value of the pronumeral in the following, correct to two decimal places.

a b

c d

5 Find all the unknown sides and angles of a triangle, where the smallest side is 21 cm in length, the smallest angle is 29° in size, and another angle is 50° in size.

6 Find the value of x correct to three decimal places.

Sine rule: the ambiguous caseIf we are given two sides of a triangle and an angle opposite one of these sides, it is sometimes possible to create two different triangles with the same information.

For example, consider ∆ABC with A = 40°, BC = 5 cm and AC = 7 cm. If we try to construct a triangle from this information, two triangles are possible.

Since A is the only defined angle, the ambiguous case arises at angle B which can be acute (as in the first triangle) or obtuse (as in the second triangle).

7.5 cm

47°

50°

x cm

23 mm42°

33°x mm

65 mm

140 mm

103°

x°3.67 cm

5.23 cm

58° x°

x

61°

52°

7.6

9.2

C

AB′ B

40°

7 cm5 cm

5 cm

C

A B 40°

7 cm 5 cm

C

A B′40°

7 cm5 cm

Two possible ∆s

10.3

394


Example 3In ∆ABC, A = 30°, BC = 25 cm and AC = 35 cm. Find the two possible values for angles B and C.

SolutionDrawing the two possible triangles from this information we find the following.

Using the sine rule:

=

=

=

= sin B0.7 = sin B

B = 44.43°From the diagram, we can see that thesecond solution corresponds to:

B = 180° − 44.43°= 135.57°

So we find an acute answer and an obtuse answer for B depending on the orientation of the triangle. The two angles are supplementary.Now if angle B has two answers, so will angle C:A = 30°, B = 44.43°: then C = 180° − (30° + 44.43°) = 105.57°orA = 30°, B = 135.57°: then C = 180° − (30° + 135.57°) = 14.43°

C

AB′ B

30°

35 cm25 cm

25 cm

C

A B30°

35 cm 25 cm

C

A B′30°

35 cm

25 cm

asin A------------ b

sin B------------

25sin 30°----------------- 35

sin B------------

sin 30°25

----------------- sin B35

------------

sin 30°25

----------------- 35×C

AB′

B30° 135.57° 44.43° 44.43°

35 cm25 cm

25 cm

C

A B44.43°

105.57°

30°

35 cm 25 cm

B′

C

A135.57°

14.43°

30°

35 cm

25 cm

B′ B


395

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10exercise 10.37 In ∆ABC, a = 12, b = 8 and B = 40°. Find two possible sets of values for A, C and c.

8 In ∆ABC, c = 16, a = 26 and C = 33°.

a Draw two possible triangles that suit this information.

b For each triangle, find all unknown values.

9 In this diagram, find the values of x and y.

10 In ∆ABC, AC = 16, AB = 22 and B = 36°. Find two possible values for the perimeter of the triangle.

11 In ∆PQR, P = 25°, PQ = 27 and QR = 19.

Show that the information given for ∆PQR leads to two possible triangles, and hence, for each triangle, find all unknown values.

Cosine ruleFor a triangle ABC, the cosine rule is given by:

The cosine rule is used when we know either:

. all three sides, or

. two sides and the included angle.

Example 4Find the value of x.

SolutionWe need to label the triangle and identify the information given.Using the cosine rule:a2 = b2 + c2 − 2bc cos A

continued

y cm

x cm

55° 70°

7 cm

C

A B c

b aa2 = b2 + c2 − 2bc cos A

tipIf the labels are given, we can write the formula in two other ways:b2 = a2 + c2 − 2ac cos Bc2 = a2 + b2 − 2ab cos C

x m

67°83 m

47 m

B

C

Ac

a

b

x m

67°83 m

47 m

10.3

396


Substitute the given information to solve for a.a2 = b2 + c2 − 2bc cos Ax2 = 472 + 832 − 2 × 47 × 83 × cos 67°x = = 77.78

The calculation is shown in the TI-83/84 screenshot.

Example 5Stephen the farmer has to use two horses to pull his plough. He has two ropes that he attaches to each horse and the plough. The ropes are 5.2 m and 5.3 m long. The horses are kept at a distance of 1.8 m apart. At what angle do the two ropes meet with each other?

SolutionFirstly, we need to draw a labelled diagramthat correctly describes Stephen’s situation.So we now have the following information:

a = 1.8b = 5.2c = 5.3

Substitute the information into the formula that involves A:

a2 = b2 + c2 − 2bc cos A1.82 = 5.22 + 5.32 − 2 × 5.2 × 5.3 × cos A

1.82 − (5.22 + 5.32) = −2 × 5.2 × 5.3 cos A−51.89 = −55.12 cos A

cos A =

A = 19.72°

Example 6A snack food company is designing a new triangular corn chip. If the corn chip has two sides of length 25 mm and 31 mm with an included angle of 62°, what is the length of the third side and the size of the two other angles? Give answers correct to two decimal places.

SolutionConstruct a triangle from the given information and label the sides appropriately. With these labels we now have the following information:

b = 31c = 25A = 62°

Substitute the information into the cosine rule to find a:a2 = b2 + c2 − 2bc cos A

= 252 + 312 − 2 × 25 × 31 × cos 62°a =

= 29.297

GC 1.6CAS 1.6

472 832 2 47 83 cos 67°×××–+

c = 5.3 b = 5.2

a = 1.8

A

B C

51.89–55.12–

------------------

a

A

BC

25 mm 31 mm

62°

252 312 2 25 31 cos 62°×××–+


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10To find B, use the cosine rule:

b2 = a2 + c2 − 2ac cos B312 = 29.2972 + 252 − 2 × 29.297 × 25 × cos B

Transpose and solve for B.−522.314 = −1464.85 cos B

cos B =

= 0.3566B = 69.11°

To find C we use the angle sum of a triangle:A + B + C = 180°, so:

C = 180° − (62° + 69.1°)= 48.89°

Hence correct to two decimal places, the third side has length 29.30 mm and the remaining angles are 69.11° and 48.89°.

exercise 10.312 Find the third side of ∆ABC, given, b = 8.6, c = 5.2, A = 47°.

13 Find all the angles in ∆ABC, whose side lengths are 7 cm, 9 cm and 12 cm.

14 Find the largest angle of the triangle shown.

15 In ∆ABC, a = 16, b = 15, C = 107°. Find c.

16 Find all the missing side lengths and angles in this triangle.

17 In ∆ABC, b = 5.7, c = 6.6, A = 44°28′. Find a.

18 Find the value of x in the following.

a b

522.3141464.25---------------------

continued

216 mm

113 mm

186 mm

58°

10

23

14.1 mm

9.6 mm

7.2 mm

x mm

32°

43°

5 m

8 m

6 m

9 m

x°

10.3

398


19 Johnny Black is playing in his under 17s football grand final. The siren has gone and Johnny is lining up to kick the winning goal. He is 37 m from one of the goal posts and 63 m from the other goal post. The goal posts are 28.5 m apart. Within what angle must Johnny kick the winning goal?

20 A plane flies in a direction of N77°E for 67 km. Then it flies on a bearing of S25°W for 193 km. How far is the starting point of the plane from its current position?

21 The back of a picture frame is shown. The painting is hung and the wires holding it up are shown. What is the distance between the two nails?

22 A yacht race comprises three legs. The first leg is in a direction of 019° for 39 km, the second leg is a distance of 55 km at a bearing of 159°. If the race is completed when a competing yacht returns to the start, what is the bearing and distance of the third leg?

23 Mary is bushwalking through a wildlife sanctuary. As she walks she can see a falcon hovering in the air in front of her. The angle of elevation Mary sees the falcon at is 50°. Mary walks 20 m and can still see the falcon, which has not moved position, hovering in the air in front of her. The angle of elevation this time is 64°. How far above Mary is the falcon hovering?

24 A bunny rabbit hops out of her hole at 8 km/h and travels at this speed for 45 min on a bearing of 145°. Then she suddenly changes direction and hops at a new constant speed for 80 min on a bearing of 053° until she is due east of her starting point.

a How far did the bunny rabbit hop after she changed direction?

b How fast did the bunny rabbit hop after she changed direction?

c How far does the bunny rabbit need to go to get back to her hole?

25 The 18th hole at a golf course is illustrated. (The 18th hole has a par 3; that means the hole should be completed in three strokes.)

Two golfers, Norman and Greg, are having a playoff to see who will be the next champion. Norman knows he can make the 18th hole in two strokes. If Norman tees up at a bearing of 032° and hits at a distance of 130 m he will make it to point B. Then at a bearing of 100° hitting a distance 100 m, he can get to the hole C, shooting one under par (i.e. one less than par 3).

Greg needs to beat Norman, so he has to get a hole in one. What bearing and distance must Greg take his shot to beat Norman?

15.4 cm 14.2 cm

150°

hook

nailnail

130 m 100 m

B

A CA


399

chap

ter

1026 A plane takes off from an airport and flies at a bearing of S25°W for 145 km, then changes

direction and flies for 215 km and lands at another airport. The bearing from the second airport to the original airport is N60°W. How far are the two airports from each other?

27 A construction company is constructing a five-storey building. They are currently workingon the 3rd floor. To bring the required materialup to the work site the workmen use a rope topull their equipment up the side of the building.To stop any damage to the building as the materials and equipment are pulled up they have attached a ramp that keeps the rope andmaterials off the side of the building (illustrated).

The 3rd floor is 25 m above the ground. The ramp slopes outward at an angle of depression of 50°, then cuts back at an angle of 25° to the vertical meeting the ground directly below the ramp. A winch is attached to the floor 2 m from the edge so as to pull up the equipment and materials. How much rope will need to be fitted to the winch to pull up the required materials?

50°

25°

25 m

winch

3rd floor

A race car is testing a new track that has been constructed. One of the corners in this race track is in the shape of a quarter circle of a circular annulus, so that as a race car drives through this corner the width of the track does not change. While the race car is in the corner, it can leave the track as long as part of the racing car is on the track (this is called cutting the corner). The race car being used is able to take very sharp turns.

On the race car’s first run of the trackthe driver decides to stay in the outsidelane as he approaches the turn; the on-board computer in the race car has calculated for the driver that there aretwo possible driving lines through the corner. They are illustrated in figure 1.

15 m

w m

Figure 1

analysis task 2—which way should I go?

SAC

CD10.3W

hich way should I go?

SAC

analysis task

10.3

400


a The length of the longest driving line in the corner is 33.6 m and the shortest driving line in the corner is 28.5 m, with a turning angle between these lines of 10°8′. If this driving line places the racing car in the centre of the outside or inside lanes when leaving the corner, as shown in figure 1, what is the width, w m, of the track?

On the second series of testing, the driver approaches the corner while in the centre of the track, as illustrated in figure 2.

b What is the length of each driving line, and the angle of the turn between these two driving lines?

c Will the racing car cut the corner during this second series of testing?

In the final series of testing the driver wants to push the racing car to the extreme through the corner. He approaches the corner centred in the inside lane, as illustrated in figure 3.

d What is the length of each driving line, and the angle of the turn between these two driving lines?

15 m

w m

Figure 2

15 m

w m

Figure 3


401

chap

ter

10Area of a triangle

Recall that the area of a triangle is given by , where the height, h, is the

perpendicular height. If h is unknown, but c and A are known, then

sin A =

h = c × sin A

Area =

That is,

Example 1Find the area of the triangle correct to one decimal place.

SolutionSubstitute the given information into the area formula.

Area = bc sin A

=

= 20.673So the area of the triangle is 20.7 cm2 correct to one decimal place.

Example 2A triangle has dimensions as shown in the diagram. Calculate the area to the nearest mm2.

SolutionIt is evident that the area rule cannot be applied because we do not have the anglebetween the two side lengths given.In this case we can use the sine rule to find B and then find angle C.

12--- b× h×

A C

B

ah

c

b

hc---

12--- b× c sin A×( )×

tipIf labels are given, we can write this formula in two other ways:

Area = ac sin B

Area = ab sin C

12---

12---

Area = 12---bc sin A

A

CB

11 cm 4 cm110°

12---

12--- 4× 11× sin 110°×

70.8 mm

40.3 mm

68°

A

B C

10.410.410.4

402


Substitute the given values into the sine rule:

=

=

=

sin B =

= 0.5278B = 31.85°

If B = 31.85° and A = 68° thenC = 180 − (31.85° + 68°).

= 80.15°Write down the appropriate rule for the area and substitute:

Area = ab sin C

= × 70.8 × 40.3 × sin 80.15°

= 1405.6So the area of the triangle is 1406 mm2, correct to the nearest mm2.

Heron’s formulaHeron, a Greek mathematician from Alexandria, is credited with a formula for the area of a triangle given its three side lengths. This formula is called Heron’s formula.

Heron’s formula states that the area of a triangle is given by:

Notice that the perimeter of the triangle is a + b + c, so s is the semi-perimeter of the triangle, that is, half the perimeter.

asin A------------ b

sin B------------

70.8sin 68°----------------- 40.3

sin B------------

sin 68°70.8

----------------- sin B40.3------------

40.3 sin 68°×70.8

-----------------------------------

12---

12---

a

b

c

A

B

C

Area of ∆ = where s = s s a–( ) s b–( ) s c–( ) 12--- a b c+ +( )


403

chap

ter

10Example 3Find the area of the triangle with side lengths 6.8 cm, 5.2 cm and 9 cm.

SolutionWe have three given side lengths so we can use Heron’s formula.Calculate the value for s by substituting values fora, b and c.

s =

=

= 10.5Write down Heron’s formula, substitute values for a, b, c and s and solve.

Area = = = 17.57

So the area of the triangle is 17.57 cm2.

exercise 10.41 Calculate the area of each of the following triangles to the nearest squared unit.

b c

2 Find the area of the triangle ABC with a = 4, b = 6 and C = 76°.

3 Find the area of ∆ABC, given ∠A = 48°, b = 5.6 and c = 7.2.

4 Calculate the area of the triangle with side lengths 3.4 mm, 6.5 mm and 7.3 mm.

5 Find the area of the triangle shown to the nearestsquare centimetre.

6 A triangle ABC has side lengths BC = 57 mm, AB = 79 mm and angle C = 37°25′.a Find the size of each of angles A and B.

b Find the area of the triangle to the nearest square millimetre.

tipThe TI-83/84 program MENSURAT can be used to check youranswer.

CD10.4M

ENSU

RAT

GC

program

12--- a b c+ +( )

12--- 6.8 5.2 9+ +( )

s s a–( ) s b–( ) s c–( )10.5 10.5 6.8–( ) 10.5 5.2–( ) 10.5 9–( )

15 cm

8 cm

A

B C

a

67 mm

89 mmA

B

C 33°

D

E

12.3 cm

5.4 cm

15°23′ F

19 cm

12 cm

112°

10.4

404


7 The area of triangle DEF is 64.4 cm2. If DF = 12 cm and DE = 21.4 cm, calculate angle D to the nearest degree, assuming it is acute.

8 In the triangle pictured:

a find the length of the hypotenuse.

b use Heron’s formula to find the area of the triangle.

c check that your answer to part b is correct by findingthe area of the triangle in a simpler way.

9 Find the area of a triangle with sides 35 mm, 46 mm and 53 mm.

10 In triangle ABD, the point C lies on BD so that AC = AB. AK is perpendicular to BD.

a Find the value of θ.

b Calculate AK to the nearest cm.

c Find the area of:

i triangle ABC.

ii triangle ACD.

11 A triangle has an area of 60 cm2. Two of its side lengths are 16 cm and 14 cm. Calculate the magnitude of the angle between the two sides, assuming it is acute.

12 Find the magnitude of the smallest angle in ∆ABC if the side lengths are 16 cm, 11 cm and 8 cm, and hence find the area of the triangle. Check your answer by using Heron’s formula.

13 A triangular picture frame has two side lengths, 8 cm and 10 cm and a base length 12 cm. Find, correct to one decimal place:

a the area enclosed by the picture frame.

b the height of the picture frame when it is standing on its 12 cm base.

14 Calculate the area of the regular hexagon shown in the diagram.

15 Find the area of this block of land to the nearest m2.

16 A parallelogram has sides with lengths of 30 cm and 45 cm and an angle of 81° between them. Calculate the area of the parallelogram.

7

24

A

B C DK

18°

31 cm

35 cm

5 cm

30 m

35 m

15 m

25 m


405

chap

ter

10Circle mensuration

Radian measure of an angleThroughout the chapter, we have measured angles in degrees. This unit of angle measurement is commonly used in navigation and surveying. However, there are other units of angle measurement and one of these is the radian. Radian measure is often used when measuring lengths and areas associated with circles.

This diagram shows a unit circle (it has radius 1 unit).

Imagine a piece of string of length 1 unit. We wrap the stringaround the circumference of the circle from Q to P so that the arcPQ has length 1 as shown. Then the angle subtended at the centreof the circle, that is ∠POQ, is called ‘one radian’ and is written as1c (or often just 1 when it is clear that we mean radian measure).

Converting between degrees and radiansThe circumference of a circle is given by C = 2πr, so for a unit circle, C = 2π. This means that there are 2π radians in a full circle. We also know that there are 360° in a full circle. It follows that 2πc = 360°, so πc = 180°.

As mentioned earlier, it is common to omit the symbol for radians so that πc is just denoted by π. If there is no symbol present, assume that the angle measure is radians.

To convert between degrees and radians, we use the fact that 180° = π. First we consider some special angles.

30° = , so

45° = , so

60° = , so

For these special angles and multiples of them, it is common to express radians in terms of π as shown. For more general angles, it is usual to give the radian measure as a decimal.

Example 1Convert the following angles to radian measure, expressing answers in terms of π.a 90° b 210° c 135°

Solution

a , so

x

y

P

O Q1c

1

1

180°6

------------ 30° π6---=

180°4

------------ 45° π4---=

180°3

------------ 60° π3---=

90° 180°2

------------= 90° π2---=

10.510.510.5

406


b Use the fact that 210° is a multiple of 30°, which is one of the special angles.

210° = 7 × 30°

=

=

c Use the fact that 135° is a multiple of 45°, which is one of the special angles.

135° = 3 × 45°

=

=

For other conversions, note that as 180° = π, dividing by 180 gives .

Example 2Convert the following angles to radian measure, giving answers correct to three decimal places.a 70° b 56°23′

Solution

a Using , multiply

both sides by 70.

70° =

= 1.222

To convert radians into degrees, note that as πc = 180°, dividing by π gives 1c = or,

dropping the radian symbol, .

7 π6---×

7π6

------

3 π4---×

3π4

------

1° π180---------=

b As 23′ = ° = 0.3833°, 56°23′ = 56.3833°.

Using , multiply both sides by 56.3833.

56°23′ =

= 0.984

2360------⎝ ⎠

⎛ ⎞

1° π180----------=

56.3833 π180----------×

1° π180----------=

70 π180----------×

GC 1.5CAS 1.5

tipConversion from degrees to radians in decimal form is easy to do with a graphics calculator. On the TI-83/84, make sure that you are in Radian mode. Enter the angle, with a degree symbol at the end (found in the ANGLE menu) and press ENTER. The calculator automatically does the conversion, so you do not need to explicitly multiply by π and divide by 180 (see the first screenshot).However, when the angle is expressed in degrees and minutes, you must add the degree symbol at the end; if you do not, the calculator simply converts the minutes part to a decimal as shown in the second screenshot.

180°π

------------

1 180°π

------------=


407

chap

ter

10Example 3Convert the following angles in radian measure to degree measure.

a b 1.9

Solution

a Using , multiply both sides by .

=

=

= 135°

b Using , multiply both sides by 1.9.

1.9 =

= 108.86° (correct to 2 dp)

exercise 10.51 Convert the following angles to radian measure, expressing answers in terms of π.

a 30° b 90° c 45°

d 120° e 300° f 225°

g 330° h 180° i 450°

2 Convert the following angles to radian measure, giving answers correct to three decimal places.

a 14° b 78° c 124°

d 256° e 302° f 359°

g 32°26′ h 130°14′ i 207°46′

3 Convert the following angles in radian measure to degree measure.

a b c

d e 4π f

g h i

4 Convert the following angles in radian measure to degree measure, giving answers correct to two decimal places.

a 4 b 1.5 c 3.75

d 2.49 e 3.26 f 5.32

g 0.89 h 6.87 i 4.55

3π4

------

tipThe calculation in part b is shown in the following screenshot. If the answer is required in degrees and minutes, the DMS command in the ANGLE menu can be used.

So the answer in degrees and minutes, correct to the nearest minute, is 108°52′.

GC 1.5CAS 1.5

1 180°π

------------=3π4

------

3π4

------ 3π4

------ 180°π

------------×

3 180°×4

----------------------

1 180°π

------------=

1.9 180°π

------------×

π3--- 3π

2------ π

4---

π6--- 7π

6------

3π4

------ 9π4

------ 11π6

---------

10.5

408


Arc lengthConsider a circle with centre O. Points P and Q onthe circumference of the circle divide it into two arcs as shown, a major arc and a minor arc. The length of an arc is therefore a part of the circumference of a circle and is proportional to the angle subtended at the centre. For example, an angle of 270° gives an arc length which is three-quarters of the circumference of the circle.

The circumference of a circle is given by C = 2πr and we have seen that there are 2π radians in a full circle.

Let arc PQ have length l units and let ∠POQ = θ radians. As the length of the arc is proportional to the angle it subtends at the centre, it follows that

=

=

l =

= rθ

So the length of the arc is given by l = rθ.

Example 4A circle has radius 8 cm. An arc of the circle subtends an angle of 60° at the centre of the circle. Calculate the length of the arc correct to two decimal places.

SolutionTo use l = rθ, we need to express the angle in radians. We should recognise 60° as a specialangle; in radians, it is . (Alternatively, use the conversion method of example 2.)

l = rθ=

=

= 8.38 (correct to 2 dp)The length of the arc is 8.38 cm.

Example 5Find the angle subtended at the centre of a circle of radius 7 cm by an arc of length 18 cm, expressing the answer in degrees correct to two decimal places.

SolutionSubstitute the given values into l = rθ.

l = rθ18 = 7θ

θ =

minor arc

major arc P

O

Q

r

WarningIn this formula for arc length, θ is the angle measure at the centre in radians. Always convert any angle given in degree format to radians before using mensuration formulae.

θ

P

O

Q

r

l

lC--- θ

2π------

l2πr--------- θ

2π------

θ2π------ 2πr×

π3---

8 π3---×

8π3

------

187

------


409

chap

ter

10This is the angle in radians; we need to convert it to degrees.

Using , multiply both sides by .

=

= 147.33° (correct to 2 dp)So the angle subtended by the arc is 147.33°.

exercise 10.55 Find the length of an arc that subtends an angle of 70° at the centre of a circle of radius

10 cm, giving the answer correct to two decimal places.

6 A goat is tethered to a post by a rope 18 m long. If the goat moves so that it always keeps the rope taut, calculate correct to the nearest cm how far it travels when the rope sweeps through an angle of 80°.

7 An arc of a circle of radius 98 mm subtends an angle of 55° at the centre of the circle. Find:

a the length of this minor arc, correct to two decimal places.

b the length of the corresponding major arc, correct to two decimal places.

8 The length of an arc of a circle is 8.5 cm and it subtends an angle of 65° at the circle’s centre. Calculate the circle’s radius, correct to two decimal places.

9 The minute hand of a clock is 30 cm long. How far does its tip travel in 25 minutes?

10 Find the angle subtended at the centre of a circle of radius 6 cm by an arc of length 14 cm, expressing the answer in degrees correct to two decimal places.

Area of a circle sectorConsider a circle with centre O and points P and Q on its circumference. Two sectors are formed: the shaded area is the minor sector POQ andthe unshaded area is the major sector POQ. The area of either sector is proportional to the angle it subtends at the centre of the circle.

Let A be the area of the minor sector POQ and let ∠POQ = θ radians.Recall that the area of a circle is πr2. As the area of the sector is proportional to the angle it subtends at the centre, it follows that

=

A =

=

So the area of the sector is given by .

1 180°π

------------=187

------

187

------ 187

------ 180°π

------------×

continued

major sector

minor sector

θ

P

O

Q

r

Aπr2-------- θ

2π------

θ2π------ πr2×

12---r2θ

A12---r2θ=

10.5

410


Example 6Find the area of a sector which subtends an angle of 45° at the centre of a circle of radius 10 cm. Give the answer correct to two decimal places.

Solution

To use , we need to express the angle in radians. We should recognise 45° as a

special angle; in radians, it is . (Alternatively, use the conversion method of example 2.)

A =

=

= 39.27 (correct to 2 dp)

So the area of the sector is 39.27 cm2.

Example 7Using the information on the diagram, calculate the radius of thecircle, correct to one decimal place.

Solution

To use , we need to express the angle in radians.

Use the method of example 2, or just use a graphics calculator in Radian mode as shown in the earlier tip (see screenshot).

A =

120 =

r2 =

= 125.00897r =

= 11.2 (correct to 1 dp)So the radius is 11.2 cm correct to one decimal place.

Area of a circle segmentConsider a circle with centre O and points P and Q on its circumference. The first diagram shows that two segments are formed: the shaded area is the minor segment cut off by the chord PQ and theunshaded area is the major segment cut off by thechord PQ. From the second diagram, we see that the area of the minor segment can be found by subtracting the area of triangle POQ from the area of the sector POQ.

A 12---r2θ=

π4---

12---r2θ

12--- 102× π

4---×

Area of sector = 120 cm2

A

O

B110°

A 12---r2θ=

GC 1.5CAS 1.5 1

2---r2θ

12--- r2× 1.91986×

120 2×1.91986---------------------

125.00897

major segment

P Q

minor segment

O

θr r

P Q

O


411

chap

ter

10Let A be the area of the minor segment and let ∠POQ = θ radians. Recall that the area of a

triangle is . In this case, a = b = r, so the area of triangle POQ is .

A = area of minor sector POQ − area of ∆POQ

=

=

So the area of the segment is given by

Example 8A chord subtends an angle of 50° at the centre of a circle of radius 8 cm. Calculate the area of the minor segment cut off by the chord.

Solution

To use , we need to express the angle

in radians. Use the method of example 2, or just use a graphics calculator in Radian mode as shown in the earlier tip (see screenshot at right).

A =

=

= 3.41 (correct to 2 dp)So the area of the sector is 3.41 cm2.

Example 9An arc AB of length 22 cm is drawn on a circle of radius 12 cm. Find:a the angle subtended at the centre of the circle by the arc AB, giving the answer in:

i radians. ii degrees.

b the area of the minor segment cut off by the chord AB, correct to 2 decimal places.

12---ab sin θ 1

2---r2 sin θ

12---r2θ 1

2---r2 sin θ–

12---r2 θ sin θ–( )

A12---r2 θ sin θ–( )=

GC 1.5, 1.6CAS 1.5, 1.6

A 12---r2 θ sin θ–( )=

12---r2 θ sin θ–( )

12--- 82 0.87266 sin 0.87266–( )×

tipYou can express the angle in degrees for the sine part of the calculation if you wish. This screenshot shows the calculation of A in example 8 in two ways.Although the calculator is in Radian mode, adding the degree sign overrides the mode and the calculator correctly finds sin 50°. The answers are slightly different since .87266 has rounding error, whereas 50° is exact.

10.5

412


Solutiona Substitute the given values into l = rθ where θ is in radians.

l = rθ22 = 12θ

θ =

This is the angle in radians.

To convert it to degrees, use and multiply both sides by .

=

= 105.04° (correct to 2 dp)

So the angle subtended by the arc is:

i radians (or ); or

ii 105.04°

b Use the formula for area of a segment:

A =

=

= 62.47 (correct to 2 dp)So the area of the minor segment is 62.47 cm2.

exercise 10.511 Find the area of a sector which subtends an angle of 36° at the centre of a circle of radius

11 cm. Give the answer correct to two decimal places.

12 An arc of length 125 mm is drawn on a circle of radius 80 mm. Find the area of the corresponding sector.

13 A sector of a circle has an area of 205 cm2 and makes an angle of 64° at the centre of a circle. Calculate the radius of the circle correct to two decimal places.

14 The hour hand of a clock is 6 cm long. Find the total area it sweeps out in 3 hours, correct to two decimal places.

15 A slice of pizza is in the shape of a sector of a circle, whose radius is 8 cm.

If there are 8 equal slices cut from a whole pizza, find:

a the angle in degrees made at the centre of the circle.

b the length of the curved edge of the pizza slice.

c the area of this slice of pizza.

16 A sector of a circle has an area of 185 cm2 and the circle has a radius of 9 cm. Find the angle made at the centre of the circle, in degrees and minutes.

116

------

1 180°π

------------=116

------116

------ 116

------ 180°π

------------×

CD10.4M

ENSU

RAT

GC

program

116

------ 11c

6--------

tipThe MENSURAT program can be used to check these answers.

12---r2 θ sin θ–( )

12--- 122 11

6------ sin 11

6------–⎝ ⎠

⎛ ⎞×

continued


413

chap

ter

1017 In a circle of radius 305 mm, calculate the area of the minor segment whose arc subtends

an angle of 88° at the centre of the circle. Give your answer correct to one decimal place.

18 Calculate the area of the segment of a circle of radius 4 cm that has an arc length of 16 cm.

Numbersense with the spence

18If there are 18 or more people at a party, then there must be either a group of 4 people who all know each other or a group of 4 complete strangers.

CD10.5Location, location, location

SAC

analysis task

This analysis task involves the use of measurements collected by a geographical surveyor. A surveyor’s job uses many of the methods of geometry and trigonometry.

Surveying land can be a very difficult task. An error in calculations can cost someone their property. With this in mind the techniques and methods a surveyor uses to measure land need to be as accurate as possible.

There are two main techniques a surveyor uses to measure a section of land. They are:

. traverse surveying

. radial surveying.

Traverse surveyingThe technique of traverse surveying involves ameasured line that stretches from one side ofthe block of land being measured to the other.This is called the traverse line. Then the distance of all corners or major features of the block of land at a 90° angle to the traverse lineare measured. It is assumed that the boundariesof the land are all straight edges.

An example of a surveyor’s sketch for a traverse survey is shown.

A surveyor’s sketch can be used to create ascale diagram as shown.

As we can see in the diagram the traverseline is AB.

The length of the traverse line in the surveyor’ssketch is 70 m.

In the scale map the total of the section lengths of thetraverse line is 70 m. The block of land comprises 7 shapes:3 to the left and 4 to the right. There are 4 triangles and 3 trapeziums.

A

B

1

2

7

6 3

5

4

10 m

12 m

7 m5 m

5 m

15 m

20 m

15 m

15 m

13 m

10 m

A

5 7

15 12

10 30

15 50

65 13

70

B

analysis task 3—location, location, location

SAC

10.5

414


Part 1A surveyor has taken measurements of a block of land for a farmer in country Victoria.

He has used the technique of traverse survey. His sketch is shown(all measurements are in metres).

a Draw a scale diagram on A4 paper for the surveyor’s sketch using the scale 1 : 500 (1 cm = 5 m).

b Look at the shape of the block of land that has been drawn.How many shapes make up this block of land? Number and name them, as in the example above.

c Find the area of each shape, and hence the total area of the farmer’s block of land to the nearest m2.

d Find the perimeter of the farmer’s block using the shapes identified in question b, correct to the nearest m.

e Complete a table of your results from question d using the following headings.

f Using your scale diagram, measure the lengths of each boundary edge with a ruler and convert the measured lengths in cm to the actual lengths in m.

Add these measurements to the previous table in a new column and use them to give an estimate for the total perimeter of the block

g Compare and comment on any differences or similarities between the calculated and measured values that lead to the perimeter. Give reasons for your observations.

Radial surveyingThe technique of radial surveying involves a chosen central point in the block of land. A surveyor then measures the distance and the bearing of each corner or feature of the section of land.

Shape number Length of boundary edges (calculated)

12..Total perimeter of block

Shape number Length of boundary edges (calculated)

Length of boundary edges (measured)


A

7 16

29 21

17 38

37 62

70 19

95 11

8 110 4

B


415

chap

ter

10An example of the surveyor’s sketch for a radial survey is illustrated below left. From this a scale diagram can be drawn (below right). Using the gathered information it would look like this. Note that this method divides the block into triangular shapes.

Part 2The farmer in country Victoria needs the surveyor’s services once more. The surveyor uses the technique of radial surveying on another block of land. His sketch is shown opposite.

h Draw a scale diagram on A4 paper of the surveyor’s sketch using the scale 1 : 1000 (1 cm = 10 m).

i Look at the shape of the block of land that has been drawn. How many triangles make up this block of land? Number them.

j Find the area of each triangle, and hence the total area of the farmer’s block of land to the nearest m2.

k Find the perimeter of the farmer’s block using the triangles identified in question i, correct to the nearest m.

l Complete a table of your results from question k using the following headings.

Triangle number Length of boundary edges (calculated)


E

D

C

B

A

O

45 m303°

37 m047°

42 m122°

52 m202°

50 m255°

N

E

C

B

A

D

47°57°48°

53°80°

75°

N

O

104°

A

B

C

D

E

F80 m346°

93 m293°

40 m195° 84 m

168°

112 m082°

76 m035°

N

O

10.5

416


CD10.6D

am it

SAC

modelling task

m Using your scale diagram, measure the lengths of each boundary edge with a ruler and convert the measured lengths in cm to the actual lengths in m.

Add these measurements to the previous table in a new column and use them to give an estimate for the total perimeter of the block.

n Compare and comment on any differences or similarities between the calculated and measured values that lead to the perimeter. Give reasons for your observations.

Triangle number Length of boundary edges (calculated)

Length of boundary edges (measured)


In a dry country such as Australia, farm dams are part of the rural lifeblood of rural. Knowing the volume of water in a dam at any given time is important for such things as planning the numbers and distribution of livestock, and estimating when the supply is likely to run out if there are drought conditions. In general, dams do not have depth markers so a method is needed that can use physically observable signs. One readily available piece of data is the distance the water level has receded from its position when the dam is full—a distance obtained simply by measuring the steps taken to walk from the top of the dam directly to the current water level. The challenge then is to estimate the volume of water in a dam from this information, and the known overall dimensions of a dam.

In addition to feeding stock, or providing water for irrigation, dams lose water through evaporation, and although the evaporation rate varies throughout the year with changes in climatic conditions, rough daily rates can be calculated by dividing annual rates by 365. Published annual evaporation rates (cm/year) vary from less than 80 cm in western Tasmania to more than 340 cm in the desert regions of Western Australia and Northern Territory. Values in Victoria vary from about 140 cm in the south to 180 cm in the north.

This problem investigates ways of estimating the volume of water remaining in a dam of which the design enables the use of basic mensuration formulae. By specifying a geographical location for the dam (e.g. northern Victoria) the evaporation rate can be included to modify the calculations. Implications then follow if the dam is to serve a specified herd, for example under drought conditions.

modelling task—dam it

SAC

Chapter reviewTrigonometric ratios and applications 10ch

apte

r

417

SummaryTrigonometric ratios of right-angled triangles

. For any right-angled triangle , ,

Forms for angles. Angles are usually expressed in decimal degree form or in degrees and minutes, where

there are 60′ in one degree.

Applications of trigonometryAngles of elevation and depression. Angles of elevation and depression are measured from

the horizontal.

. The angle of elevation and depression are equal because they are alternate angles.

Bearings. Bearings are a measure used to represent the direction of one object from another.

. Bearings can be expressed as true bearings measured in a clockwise direction from north, or as compass/conventional bearings measured first from north or south, then in the direction of east or west.

. When answering questions involving bearings it is very important to always draw the compass points N, S, E, W together with clearly labelled diagrams.

Non right-angled triangles. With non right-angled triangles, it is useful to label the three

vertices with uppercase letters, A, B and C and then label the sides opposite these with the corresponding lowercase letters, a, b and c.

. For a triangle ABC, the sine rule is given by

It is used to find unknown lengths and angles when given:

a two angles and one side length

a two side lengths and an angle opposite one of the sides.

. For a triangle ABC, the cosine rule is given by a2 = b2 + c2 − 2bc cos A. It is used to find unknown lengths and angles when given:

a all three side lengths

a two side lengths and the included angle.

. Alternative forms for the cosine rule are:

b2 = a2 + c2 − 2ac cos B, or c2 = a2 + b2 − 2ab cos C

θ

OH

A

sin θ OH----= cos θ A

H----= tan θ O

A----=

angle of elevation

angle of depression

θ

θ

A C

B

ac

b

asin A------------- b

sin B------------ c

sin C------------= =

418


Area of a triangle. If two sides of a triangle and the included angle are given then we can use the following

rule to find the area.

=

. If all three side lengths of the triangle are known then Heron’s formula can be used to find the area of the triangle.

Area of ∆ = where s =

Circle mensurationRadian measure of an angle. A radian is the angle subtended at the centre of a unit circle by an arc of length one unit.

. 360° = 2π c; 180° = π c. The symbol c is used to indicate radian measure, but is often omitted when the meaning is clear.

. To convert between radians and degrees:

1° =

1c =

Arc length. Two radii divide the circumference of a circle into two arcs

the minor and major arc.

. Arc length is given by l = rθ , where r is the radius of the circle and θ is the angle measure at the centre in radians.

Area of a sector. Two radii divide the circle into two sectors, the minor sector and

the major sector.

. The area of the minor sector is given by , where r is the

radius of the circle and θ is the angle measure at the centre in radians.

Area of a segment. A chord divides the area of a circle into two parts, the minor and

major segment.

. The area of a segment is given by , where r is

the radius of the circle and θ is the angle measure at the centre in radians.

Area12---bc sin A=

12---ac sin B

12---ab sin C=

s s a–( ) s b–( ) s c–( ) 12--- a b c+ +( )

π180---------

180°π

------------

minor arc

major arc P

O

Q

r

θ

major sector

minor sector

θ

P

O

Q

rA12---r2θ=

major segment

P Q

minor segment

OA

12---r2 θ sin θ–( )=

Trigonometric ratios and applications 10chap

ter

419

Revision questionsShort answer1 Lim walks 4 km due north, then 5 km due east. Find:

a his bearing, to the nearest degree, from his starting point.

b the length of the last leg of the journey if Lim decides to head straight back to his starting point.

2 A ladder 5 m long rests against a vertical wall, with the foot of the ladder 1.8 m from the base of the wall. Calculate the angle the ladder makes with the wall correct to two decimal places.

3 In the triangle, find the value of the pronumeralcorrect to one decimal place.

4 In a triangle ABC, a = 8, A = 110° and B = 35°. Find the unknown angle and lengths in this triangle.

5 From the top of a lighthouse 85 m high, Frank observes a boat at an angle of depression of 33°22′. If the foot of the lighthouse is at sea level, how far is the boat from the base of the lighthouse, correct to the nearest metre?

6 In ∆ABC, b = 5, c = 7 and A = 58°42′. Find the value of a.

7 Calculate the angle subtended by an arc 9 cm long on a circle of radius 3 cm in:

a radians. b degrees.

8 A triangular flowerbed has side lengths of 4 m, 5 m and 6 m. Find the area of this flowerbed.

9 Three towns, Diamond, Ruby and Emerald, are famous for each of their stones. Ruby is 20 km from Diamond on a bearing of 050°. Emerald is 28 km from Ruby on a bearing of 260°. A jeweller is making his monthly trip to buy stones to make his jewellery. Calculate the total distance covered by the jeweller if he visits all three towns, starting and ending at Diamond.

10 Two vertical goal posts of heights 5 m and 7 m stand apart on horizontal ground. If the angle of elevation from the top of the shorter goal post to the top of the taller goal post is 26°, find the distance between the two goal posts correct to the nearest metre.

11 In triangle ABC, B = 55°, b = 15, c = 18. Calculate the two possible values for a.

12 Calculate the area of the shaded region shown in the diagram to the nearest square centimetre.

7.5 cm

37° 52°x cm

P Q

O

120°35 cm

420


13 The arc of a circle with radius 5.2 cm subtends an angle of 82° at the centre. Find:

a the length of the arc.

b the area of the sector formed by the arc and two radii.

14 The area of a sector of a circle is 50 cm2 and its bounding arc subtends an angle of 65° at the centre. Calculate the perimeter of the sector correct to two decimal places.

Extended response1 A yacht travels 3 km from a point A on a bearing of 110° to a point B and then travels for

4 km on a bearing of 168° to a point C.

a Draw a clearly labelled diagram representing the situation.

b How far south is the yacht from point A?

c How far east is the yacht from point A?

d What is the direct distance between A and C?

e Calculate the bearing of A from C.

f If the yacht travels back to point A from point C, find the area that is enclosed by the yacht’s course.

2 Examine the diagram. Using the information provided find:

a the circumference of the circle.

b the angle θ in:

i degrees

ii radians.

c the length of the minor arc PQ and of the major arc PQ.

d the area of the minor sector OPQ.

e the area of the shaded section.

3 A spy plane is flying above the ocean at an altitude of 3000 m. The sensors of the spy plane detect an enemy submarine travelling on the water’s surface directly in front of the spy plane at an angle of depression of 27°.

a Draw a diagram to represent the situation.

b If the plane is travelling at 500 m/min, how long before the plane is directly above the enemy submarine?

c Three minutes after the spy plane passes over the enemy submarine, the enemy submarine detects the spy plane. At what angle of elevation is the spy plane from the enemy submarine at that instant?

The spy plane has signalled a friendly submarine to intercept the enemy submarine it originally detected. The friendly submarine is behind the enemy submarine at a depth of 1750 m and is a direct distance of 3850 m from the enemy submarine.

d Draw a diagram to represent the situation.

e At what angle of elevation should the second submarine rise to come up just behind the enemy submarine?

θ

P Q

O

15 cm

20 cm

CD10.7–10.9A

trio of testsC

hapter tests

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