Trigonometrymathsbooks.net/JACPlus Books/11 Standard General/Ch07 Trigonometry.pdf1 3 3 3 = 1 3 5....

Chapter 7 Trigonometry

7a Trigonometry of right-angled triangles 7b Applications of right-angled triangles 7c Non–right-angled triangles — the sine

rule 7d Non–right-angled triangles — the cosine

rule 7e Area of triangles 7F Radian measurement 7G Arcs, sectors and segments

7

297

areaS oF STudy

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Digital doc10 Quick Questions

Right-angled triangles and solutions to problems • involving right-angled triangles using sine, cosine and tangentThe relationship sin• 2 (θ) θ) θ + cos2 (θ) θ) θ = 1, cos (θ) θ) θ =sin (90° – (θ)) and sin (θ)) and sin (θ θ) = cos (90θ) = cos (90θ ° – (θ))θ))θTwo-dimensional applications including angles • of depression and elevationExact values of sine, cosine and tangent for 30• °, 45° and 60°

Solution to triangles by the sine and cosine rules• Areas of triangles, including the formula • A = s s a s b s( )s s( )s s a s( )a s( )a s( )a s b s( )b s( )a s( )a s b s( )b s( )b s( )b s c( )cb s− − −b sa s− − −a s( )− − −( )a s( )a s− − −a s( )a s( )− − −( )a s( )a s− − −a s( )a s b s( )b s− − −b s( )b s( )− − −( )b s( )b s− − −b s( )b sCircle mensuration: radian measure, arc length, • areas of sectors and segmentsApplications such as navigation and surveying • in simple contexts

Trigonometry

Trigonometry of right-angled trianglesTrigonometry, derived from the Greek words trigon (triangle) and metron (measurement), is the branch of mathematics that deals with the relationship between the sides and angles of a triangle. It involves fi nding unknown angles, side lengths and areas of triangles. The principles of trigonometry are used in many practical situations such as building, surveying, navigation and engineering. In previous years you will have studied the trigonometry of right-angled triangles. We will review this material before considering non–right-angled triangles.

sin (θ) = opposite sidehypotenuse

which is abbreviated to sin (θ) = OH

cos (θ) = adjacent sidehypotenuse

which is abbreviated to cos (θ) = AH

tan (θ) = opposite sideadjacent side

which is abbreviated to tan (θ) = OA

The symbol θ (theta) is one of the many letters of the Greek alphabet used to represent the angle. Other symbols include α (alpha), β (beta) and γ (gamma).

Writing the mnemonic SOH–CAH–TOA each time we perform trigonometric calculations will help us to remember the ratios and solve the problem.

7a

(A)

B

C A

(O)(H )

HypotenuseOpposite

Adjacent

298

Pythagoras’ theoremFor specific problems it may be necessary to determine the side lengths of a right-angled triangle before calculating the trigonometric ratios. In this situation, Pythagoras’ theorem is used. Pythagoras’ theorem states:

In any right-angled triangle, c2 = a2 + b2.

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

Think WriTe

1 Label the sides, relative to the marked angles.

x

O

4H

50

2 Write what is given. Have: angle and hypotenuse

3 Write what is needed. Need: opposite side

4 Determine which of the trigonometric ratios is required, that is, SOH–CAH–TOA.

sin (θ) = OH

5 Substitute the given values into the appropriate ratio. sin (50°) = x4

6 Transpose the equation and solve for x. 4 × sin (50°) = xx = 4 × sin (50°)

7 Round the answer to 2 decimal places. = 3.06

Find the length of the hypotenuse, correct to 2 decimal places. All lengths are in cm.

Think WriTe

1 Label the sides, relative to the marked angle.7

A

H

24º 25

2 Write what is given. Have: angle and adjacent side

3 Write what is needed. Need: hypotenuse

ca

b

Worked examPle 1

x4

50

Worked examPle 2

7A

H

24º 25

maths Quest 11 Standard General mathematics for the Casio ClassPad

299Chapter 7 Trigonometry

4 Determine which of the trigonometric ratios is required, that is, SOH–CAH–TOA.

cos (θ) = AH

5 Substitute the given values into the appropriate ratio.

cos (24°25′) = 7H

6

7 Round the answer to 2 decimal places and include the appropriate unit.

∴ H = 7.69 cm

Find the angle θ, giving the answer in degrees and minutes.

Think WriTe

1 Label the sides, relative to the marked angles.

12

A

O 18

2 Write what is given. Have: opposite and adjacent sides

3 Write what is needed. Need: angle

4 Determine which of the trigonometric ratios is required; that is, SOH–CAH–TOA.

tan (θ) = OA

5 Substitute the given values into the appropriate ratio. tan (θ °) = 1812

6 Transpose the equation and solve for θ, using the inverse tan function. Convert the decimal part of the degree to minutes.

θ ° = tan−1 1812

= 56.309 932 47°= 56° 19′

Worked examPle 3

12

18

First ensure the calculator is in degree mode.On the Main screen, complete the entry line as:

solve cos( ) ,24 257 ′ =

h

h

Then press E.Note: To enter an angle in degrees and minutes, tap:• Action• Transformation• dms

300

Find the perimeter of the following composite shape,correct to 2 decimal places. The length measurements are in metres.

Think WriTe

1 Divide the composite shape into two parts: a rectangle and a right-angled triangle. Label each of the unknown side lengths as x, y, z. 17

x

y

z

20

60

17 17

x

y

z

20 – x

602 Redraw the composite shape as two separate

parts.

3 Looking at the triangle fi rst, calculate the side length x. Write down what is given.

Have: angle and opposite side

4 Write what is needed. Need: adjacent side

5 Determine which of the trigonometric ratios is required; that is, SOH–CAH–TOA.

tan (θ) = OA


tan (60°) = 17x

7 Transpose the equation and solve for x. x × tan (60°) = 17

x = 1760tan( )


= 9.81 m

9 Calculate the side length y. What is given? Have: angle and opposite side

10 Write what is needed. Need: hypotenuse

11 Determine which of the trigonometric ratios is required (SOH–CAH–TOA).

sin (θ) = OH


sin (60°) = 17y

13 Transpose the equation and solve for y. y × sin (60°) = 17

y = 17

60sin( )ο


= 19.63 m

15 Looking at the square, calculate the side length z. From the original diagram, the base length is the sum of x and z.

x + z = 20z = 20 − x

= 20 − 9.81= 10.19 m

16 Calculate the perimeter of the composite shape by adding each of the outside lengths.

P = 17 + 20 + y + z= 17 + 20 + 19.63 + 10.19= 66.82 m

Worked examPle 4

17

2060

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Worked example 4



exact valuesMost of the trigonometric values that we will deal with in this chapter are only approximations. However, angles of 30°, 45° and 60° have exact values of sine, cosine and tangent. Consider an equilateral triangle, ABC, of side length 2 units.

If the triangle is perpendicularly bisected, then two congruent triangles, ABD and CBD, are obtained. From triangle ABD it can be seen that altitude BD creates a right-angled triangle with angles of 60° and 30° and base length (AD) of 1 unit. The altitude BD is obtained using Pythagoras’ theorem.

(AB)2 = (AD)2 + (BD)2

DA

B

60

30

1

23

22 = 12 + (BD)2

4 = 1 + (BD)2

4 − 1 = (BD)2

(BD)2 = 3 BD = 3

Using triangle ABD and the three trigonometric ratios the following exact values are obtained:

sin (B) = OH

⇒ sin (30°) = 12

sin (A) = OH

⇒ sin (60°) = 32

cos (B) = AH

⇒ cos (30°) = 32

cos (A) = AH

⇒ cos (60°) = 12

tan (B) = OA

⇒ tan (30°) = 1

3 or 3

3 tan (A) =

OA

⇒ tan (60°) = 31

or 3

Consider a right-angled isosceles triangle EFG whoseequal sides are of 1 unit. The hypotenuse EG is obtained by using Pythagoras’ theorem.

(EG)2 = (EF)2 + (FG)2

= 12 + 12

= 2 EG = 2

Using triangle EFG and the three trigonometric ratios, the following exact values are obtained:

sin (E) = OH

⇒ sin (45°) = 1

2 or 2

2

cos (E) = AH

⇒ cos (45°) = 1

2 or 2

2

tan (E) = OA

⇒ tan (45°) = 11

or 1

Trigonometric identities and other relationshipsbetween sin (θ ), cos (θ ) and tan (θ )• An identity is a relationship that holds true for all values of a pronumeral or pronumerals.• The Pythagorean Identity states that sin2 (θ) + cos2 (θ) = 1.• Sine and cosine are called complementary functions since:

sin (90° − θ) = cos (θ) and cos (90° − θ ) = sin (θ)

• The tangent function may also be written as tan (θ) = sin ( )cos ( )

θθ

.

D CA

B

60

30

2

2 2

1

F

G

E45

2

1

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InvestigationTrigonometric

identities

302

For any right-angled triangle:1.

sin (θ) = OH

cos (θ) = AH

tan (θ) = OH

To determine which trigonometric ratio to use when2. solving a right-angled triangle, follow these steps:(a) Label the diagram using the symbols θ, O, A, H.(b) Write what is given.(c) Write what is needed.(d) Determine which of the trigonometric ratios is required; that is,

SOH–CAH–TOA.(e) Substitute the given values into the rule.Pythagoras’ theorem, 3. c2 = a2 + b2, may also beused to solve right-angled triangles.Angles of 304. °, 45° and 60° have exact values of sine, cosine and tangent.

θ 30° 45° 60°

sin (θ)12

1

2

22

= 32

cos (θ) 32

1

2

2

2= 1

2

tan (θ)1

3

3

3= 1 3

Trigonometric identities and other relationships between sin (5. θ), cos (θ) and tan (θ)An identity is a relationship that holds true for all values of a pronumeral or • pronumerals.The Pythagorean Identity states that sin• 2 (θ) + cos2 (θ) = 1.Sine and cosine are called • complementary functions since:

sin (90° − θ) = cos (θ) and cos (90° − θ) = sin (θ)

The tangent function may also be written as tan • (θ) = sin ( )cos ( )

θθ

.

(A)

B

C A

(O)(H )

HypotenuseOpposite

Adjacent

ca

b

remember

Trigonometry of right-angled triangles 1 Copy and label the sides of the following right-angled triangles using the words hypotenuse,

adjacent, opposite and the symbol θ.

a b Adjacent c Opposite d

exerCiSe

7a 1

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Digital docsSkillSHEET 7.1

Labelling right-angled triangles

SkillSHEET 7.2Using

trigonometric ratios



2 We 1,2 Find the value of the pronumerals, correct to 2 decimal places.a

x40

10 b

x

32 14'

7.5 c

x

47 8'17

d

x

62 38'684

e

x

14 25'

1.03 f

x78

3.85

g x

27 47'

504

h

xy

38 48'

17

3 We 3 Find the angle θ, giving the answer in degrees and minutes.

a

7 10

b 5

12

c

2820

d

6.8

2.1

e

11.7

4.2 f

48

30

4 mC In the triangle ABC, cos (α) is given by:

a ab

b ba

c a bb

2 2+

d a bb

2 2− e b a

b

2 2−

5 An isosceles triangle has a base of 12 cm and equal angles of 30°. Find, in surd form:a the height of the triangleb the area of the trianglec the perimeter of the triangle, giving your answers in

simplest surd form.

6 We 4 Find the perimeter of the composite shape at right, correct to 2 decimal places. The length measurements are in metres.

7 A ladder 6.5 m long rests against a vertical wall and makes an angle of 50° to the horizontal ground.a How high up the wall does the ladder reach?b If the ladder needs to reach 1 m higher, to the nearest minute,

what angle should it make to the ground?

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Digital docSkillSHEET 7.3

Degrees and minutes

B a

b

C

A

30 3012 cm

26

14

60

7 eBookpluseBookplus

Digital docSkillSHEET 7.4Composite

shapes 1

304

8 A 400-m-long road goes straight up a slope. If the road rises 50 m vertically, what is the angle that the road makes with the horizontal?

9 An ice-cream cone has a diameter of 6 cm and a sloping edge of 15 cm. Find the angle at the bottom of the cone.

10 A vertical flagpole is supported by a wire attached from the top of the pole to the horizontal ground, 4 m from the base of the pole. Joanne measures the angle the wire makes with the ground and finds this is 65°. How tall is the flagpole?

11 A stepladder stands on a floor, with its feet 1.5 m apart. If the angle formed by the legs is 55°, how high above the floor is the top of the ladder?

12 The angle formed by the diagonal of a rectangle and one of its shorter sides is 60°. If the diagonal is 8 cm long, find the dimensions of the rectangle, in surd form.

13 In the figure at right, find the value of the pronumerals,correct to 2 decimal places.

14 In the figure at right, find the value of the pronumerals,correct to 2 decimal places.

15 In the figure at right, find the value of the pronumeral x,correct to 2 decimal places.

16 An advertising balloon is attached to a rope 120 m long. Therope makes an angle of 75° to level ground. How high above theground is the balloon?

17 An isosceles triangle has sides of 17 cm, 20 cm and 20 cm. Find the magnitude of the angles.

18 A garden bed at right is in the shape of a trapezium. Whatvolume of garden mulch is needed to cover it to a depth of 15 cm?

19 A gable roof has sloping sides of 8.3 m. It rises to a heightof 2.7 m at the centre. a What is the angle of slope of the two sides?b How wide is the roof at its base?

20 A ladder 10 m long rests against a vertical wall at an angle of 55° to the horizontal. It slides down the wall, so that it now makes an angle of 48° with the horizontal.a Through what vertical distance did the top of the ladder slide?b Does the foot of the ladder move through the same distance? Justify your answer.

21 mC In the diagram at right the size of angle θ is:a 8°10′b 10°8′c 18°10′d 18°11′e 19°8′

ad

50

730

b c

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Digital docSkillSHEET 7.5Composite

shapes 2

ba

14

70

48

x6

33

58

4 m

12 m

120

2.7 m8.3 m 8.3 m

2 cm

7 cm

4 cm



applications of right-angled trianglesThe principles of trigonometry have been used throughout the ages, from the construction of ancient Egyptian pyramids through to many modern-day activities like engineering projects and architecture. Trigonometry is especially useful for measuring distances and heights which are difficult or impractical to access. For example, two important applications of right-angled triangles involve:1. angles of elevation and depression, and2. bearings.

angles of elevation and depressionAngles 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.

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

1 Draw a diagram and label all the given information. Include the unknown length, x, and the angle of elevation, 12°.

12

12

50 m

x

2 Write what is given. Have: angle and opposite side


7b

Angle of elevation

Line of

sigh

t

Line of s

ight

Angle of depression

D

E

D and E are alternate angles. D = E

Worked examPle 5

306


tan (θ) = OA

5 Substitute the given values into the appropriate ratio. tan (12°) = 50x


x = 5012tan( )


8 Answer the question. The boat is 235.23 m away from the base of the cliff.

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/draW

a Draw a diagram and label all the given information.

a

40 60

1800

Helicopter

S1S2

b For survivor number 1: b Let x represent the horizontal distance from the helicopter to a survivor.




tan (θ) = OA


tan (60°) = 1800

x


x = 1800

60tan ( )

6 Round the answer to 2 decimal places. = 1039.23 m

For survivor number 2:




tan (θ) = OA

Worked examPle 6




tan (40°) = 1800

x


x = 1800

40tan( )

6 Round the answer to 2 decimal places. = 2145.16 m

7 Determine the distance between the two survivors.

Distance apart = 2145.16 − 1039.23= 1105.93

8 Answer the question. The two survivors are 1105.93 m apart.

bearingsBearings 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).

Conventional or compass bearings are measured:

first, relative to north or south, and second, 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.

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

Think WriTe/draW

1 Draw a diagram of the situation, labelling each of the compass points and the given information.

52

Nx

S

EW

40 km

Compass bearing equivalent is S30 E

150 T

N

N20 WTrue bearing equivalent

is 340 T

S70 ETrue bearing equivalent

is 110 T

2020

N

S

EW

N

S

EW

S60 W

240 T 60

N

S

EW

N

S

EW

Worked examPle 7

308

2 Write what is given for the triangle. Have: angle and hypotenuse

3 Write what is needed for the triangle. Need: opposite side


sin (θ) = OH

5 Substitute the given values into the appropriate ratio. sin (52°) = x

40

6 Transpose the equation and solve for x. 40 × sin (52°) = xx = 40 × sin (52°)


8 Answer the question. The ship is 31.52 km west of the starting point.

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

Think WriTe/draW

1 Draw a diagram of the situation, labelling each of the compass points and the given information. 10 km

4 km

N

S

2 Write what is given for the triangle. Have: adjacent and opposite sides

3 Write what is needed for the triangle. Need: angle


tan (θ) = OA


tan (θ) = 4

10

6

7 State the value of θ to the nearest degree. θ = 21°48′

Worked examPle 8eBookpluseBookplus

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Worked example 8


On the Main screen, complete the entry line as:

solve tan ( ) ,θ θ=

410

Then press E.To convert the angle to degrees and minutes, highlight it, copy it and tap:• Action• Transformation• toDMSPaste the angle and press E.


8 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°48′. To obtain the final bearing these values are added.

Bearing = 90° + 21°48′= 111°48′ T

9 Answer the question. The bearing of the ship from its starting point is 111°48′ T.

Angles of elevation and depression are each measured from the horizontal.1. The angle of elevation is equal to the angle of depression since they are alternate ‘Z’ 2. angles.True bearings are measured in a clockwise direction, starting from north (03. ° T).Conventional or compass bearings are measured first, relative to north or south, and 4. second, relative to east or west.Whenever solving problems involving either angles or elevation and depression 5. bearings, you should always draw a diagram and label all the given information.Set up a compass as the basis of your diagram for bearings questions.6.

remember

applications of right-angled triangles 1 We 5 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, Denis measures the angle of elevation to the bottom and top of the crane. These were 62° and 68° respectively. If the building is 42 m high find, to 2 decimal places:a how far Denis is from the buildingb the height of the crane.

6 A new skyscraper is proposed for the Melbourne Docklands region. It is to be 500 m tall. What would be the angle of depression, in degrees and minutes, from the top of the building to the island on Albert Park Lake, which is 4.2 km away?

7 We 6 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.

exerCiSe

7b

310

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.

a N35°W b S47°W c N58°E d S17°E

11 Express the following true bearings in conventional form.

a 246° T b 107° T c 321° T d 074° T

12 mC a A bearing of S30°E is the same as:

a 030° T b 120° T c 150° T d 210° T e 240° Tb A bearing of 280° T is the same as:

a N10°W b S10°W c S80°W d N80°W e N10°E

13 We 7 A pair of canoeists paddle 1800 m on a bearing of N20°E. How far north of their starting point are they, 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 We 8 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 campsite. 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 isb how far west of the original position it isc 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 Bracknawb 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 towerb the distance from B to the foot of the tower c the true bearing of the tower from B.



21 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 fl ying straight and level, fi nd the horizontal distance of the bird: i from my doorway at noon ii from my doorway at 12.02 pm.b Hence, fi nd: i the distance travelled by the bird in the two minutes ii its speed of fl ight in km/h.

non–right-angled triangles — the sine ruleWhen working with non–right-angled triangles, it is usual to label the angles A, B and C, and the sides a, b and c, so that side a is the side opposite angle A, side b is the side opposite angle B and side c is the side opposite angle C.

In a non–right-angled triangle, a perpendicular line, h, can be drawn from the angle B to side b.

This divides the triangle into two right-angled triangles, ABD and CBD.

Using triangle ABD and the sine trigonometric ratio

for right-angled triangles, we obtain sin (A) = hc . Using

triangle CBD and the sine trigonometric ratio

for right-angled triangles, we obtain sin (C) = ha .

Transposing each equation to make h the subject, we obtain: h = c × sin (A) and h = a × sin (C). Since h is common to both triangles the two equations may be equated and we get c × sin (A) = a × sin (C).

Dividing both sides of the equation by sin (A) gives:

c = a C

A× sin ( )sin ( )

Dividing both sides of the equation by sin (C) gives:c

Csin ( ) =

aAsin ( )

In a similar way, if a perpendicular line is drawn from angle A to side a, the two right-angled triangles would give h = c × sin (B) and h = b × sin (C).

This would give: b

Bsin ( ) =

cCsin ( )

From this, the sine rule can be stated.

In any triangle ABC:a

Ab

Bc

Csin ( ) sin ( ) sin ( )== ==

7C

a

b

c

B

A C

D

c ah

b CA

B

h–c = sin (A) and h–a = sin (C)

ch

CA

B

h c sin (B) andh b sin (C)

b

ac

A C

B

312

Notes1. When using this rule, depending on the values given, any combination of the two equalities

may be used to solve a particular triangle.2. To solve a triangle means to find all unknown side lengths and angles.

The sine rule can be used to solve non–right-angled triangles if we are given:1. two angles and one side length, or2. two side lengths and an angle opposite one of these side lengths.

In the triangle ABC, a = 4 m, b = 7 m and B = 80°. Find A, C and c.

Think WriTe

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

b = 7

ca = 4

B

C

80

A

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

The sine rule can be used since two side lengths and an angle opposite one of these side lengths have been given.

3 Write the sine rule to find A. To find angle A:

a

Asin( ) =

bBsin( )

4 Substitute the known values into the rule. 4

sin( )A =

780sin( )

5 Transpose the equation to make sin (A) the subject.

4 × sin (80°) = 7 × sin (A)

4 80

7× sin ( )

= sin (A)

sin (A) = 4 80

7× sin( )ο

6 A = sin−1 4 80

7×

sin ( )

= sin−1 (0.562 747 287)

= 34.246 004 71°= 34°15′

Worked examPle 9


Alternatively, on the Main screen, tap:• Action• Advanced• solveComplete the entry line as:

solve4 7

80sin( ) sin ( ),

aa=

Then press E.Convert the angle to degrees and minutes.


7 Round the answer to the nearest minute. A = 34°15′8 Determine the value of angle C using the fact

that the angle sum of any triangle is 180°.C = 180° − (80° + 34°15′)

= 65°45′9 Write the sine rule to fi nd c. To fi nd side length c:

cCsin( )

= b

Bsin( )

10 Substitute the known values into the rule. c

sin ( )65 45 ′ =

780sin( )

11


c = 6.48 m

The ambiguous caseWhen using the sine rule there is one important issue to consider. If we are given two side lengths and an angle opposite one of these side lengths, then two different triangles may be drawn. For example, if a = 10, c = 6 and C = 30°, two possible triangles could be created.

a = 10c = 6

A30

C

B

a = 10c = 6

A30

C

B

In the fi rst case, angle A is an acute angle, while in the second case, angle A is an obtuse angle.When using the sine rule to fi nd an angle, we have to use the inverse sine function. If we are

fi nding an angle, given the sine value, it is important to remember that an angle between 0° and 90° has the same sine value as its supplement. For example, sin (40°) = 0.6427, and sin (140°) = 0.6427.

In the triangle ABC, a = 10 m, c = 6 m and C = 30°. Find two possible values of A, and hence two possible values of B and b.Case 1

Think WriTe/draW

1 Draw a labelled diagram of the triangle ABC and fi ll in the given information. a = 10

c = 6

A30

C

B

lengths and an angle opposite one of these side lengths, then two different triangles may be drawn. For example, if

eBookpluseBookplus

Interactivityint-0808

The ambiguous case


Tutorialint-0887

Worked example 10

On the Main screen, tap:• Action• Advanced• solveComplete the entry line as:

solvec

csin ( ) sin ( )

,60 45

780 ′

=

Then press E.

314

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

The sine rule can be used since two side lengths and an angle opposite one of these side lengths have been given.

3 Write the sine rule to find A. To find angle A:a

Asin( ) =

cCsin( )

4 Substitute the known values into the rule. 10sin( )A

= 630sin( )

10 × sin (30°) = 6 × sin (A)

10 306

× sin ( ) = sin (A)

5 Transpose the equation to make sin (A) the subject.

sin A = 10 30

6× sin ( )

6 Evaluate angle A. A = sin−1 10 30

6×

sin ( )

= sin−1(0.833 333 333)

= 56.442 690 24°7 Round the answer to degrees and minutes. A = 56°27′8 Determine the value of angle B, using the fact

that the angle sum of any triangle is 180°.B = 180° − (30° + 56°27′)

= 93°33′9 Write the sine rule to find b. To find side length b:

bBsin( )

= cCsin( )

.

10 Substitute the known values into the rule.b

sin ( )93 33 ′ =

630sin( )

11 Transpose the equation to make b the subject. b = 6 93 33

30× ′sin ( )

sin ( )

12 Evaluate. Round the answer to 2 decimal places and include the appropriate unit.

= 11.98 m

The values we have just obtained are only one set of possible answers for the given dimensions of the triangle ABC.

We are told that a = 10 m, c = 6 m and C = 30°. Since side a is larger than side c, it follows that angle A will be larger than angle C. Angle A must be larger than 30°; therefore it may be an acute angle or an obtuse angle.

Case 21 Draw a labelled diagram of the triangle ABC

and fill in the given information. a = 10c = 6

A30

C

B

2 Write the alternative value for angle A. Simply subtract the value obtained for A in Case 1 from 180°.

To find the alternative angle A:If sin (A) = 0.8333, then A could also be:A = 180° − 56°27′

= 123°33′3 Determine the alternative value of angle B,

using the fact that the angle sum of any triangle is 180°.

B = 180° − (30° + 123°33′)= 26°27′



4 Write the sine rule to find the alternative b. To find side length b:b

Bsin( ) =

cCsin( )

5 Substitute the known values into the rule.b

sin ( )26 27 ′ =

630sin( )

6 Transpose the equation to make b the subject. b = 6 26 27

30× ′sin ( )

sin ( )


= 5.34 m

Hence, for this example there were two possible solutions as shown by the diagram below.

a 10c 6

A30

C

B

a 10c 6

A30

C

B

The ambiguous case does not work for each example. Consider worked example 9, where we were required to solve the triangle ABC given a = 4 m, b = 7 m and B = 80°. For angle A, we obtained A = 34°15′. However, angle A could also have been A = 145°45′ (since there are two possible values of A between 0° and 180° whose sine is the same; that is, sin (34°15′) = 0.5628 and sin (145°45′) = 0.5628).

We will now see whether or not A = 145°45′ is a possible solution.To obtain C, subtract angles A and B from 180°:

C = 180° − (80° + 145°45′) = 180° − 225°45′ = −45°45′ (not possible)

Hence, for worked example 9 only one possible solution exists.It would be useful to know, before commencing a question, whether or not the ambiguous

case exists and, if so, to then find both sets of solutions.

The ambiguous case exists if C is an acute angle and a > c > a × sin (C), or any equivalent statement; for example, if B is an acute angle and a > b > a × sin (B), and so on.

In worked example 10 where a = 10 m, c = 6 m and C = 30°, there were two possible solutions because C was an acute angle and a > c > a × sin (C), since 10 > 6 > 10 × 0.5.

In worked example 9 where a = 4 m, b = 7 m and B = 80°, there was only one possible solution because even though B was an acute angle, the condition a > b > a × sin (B) could not be satisfied.

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

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

6052 120

A DB

C

h

x – 20x

20

Worked examPle 11

316

2 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.

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

∠ACB = 180° − (52° + 120°)= 8°

4 Write the sine rule to find b. To find side length b of triangle ABC:b

Bsin ( ) =

cCsin( )

5 Substitute the known values into the rule. bsin ( )120

= 20

8sin ( )

6 Transpose the equation to make b the subject. b = 20 120

8× sin ( )sin ( )


= 124.45 m

8 Draw a diagram of the situation, that is, triangle ADC, labelling the required information.Note: There is no need to solve the rest of the triangle in this case as the values will not assist in finding the height of the building.

52

124.45 m

A D

C

h

9 Write what is given for the triangle. Have: angle and hypotenuse

10 Write what is needed for the triangle. Need: opposite side


sin (θ) = OH

12 Substitute the given values into the appropriate ratio. sin (52°) = h

124 45.

13 Transpose the equation and solve for h. 124.45 × sin (52°) = hh = 124.45 × sin (52°)


15 Answer the question. The height of the building is 98.07 m.

The sine rule states that for any triangle ABC:1.

aA

bB

cCsin ( ) sin ( ) sin ( )

= =

When using this rule it is important to note that, depending on the values given, any 2. combination of the two equalities may be used to solve a particular triangle.The sine rule can be used to solve non–right-angled triangles if we are given:3. (a) two angles and one side length, or(b) two side lengths and an angle opposite one of these side lengths.The ambiguous case exists if 4. C is an acute angle and a > c > a × sin (C).

remember



non–right-angled triangles — the sine rule 1 We 9 In the triangle ABC, a = 10, b = 12 and B = 58°. Find A, C and c.

2 In the triangle ABC, c = 17.35, a = 26.82 and A = 101°47′. Find C, B and b.

3 In the triangle ABC, a = 5, A = 30° and B = 80°. Find C, b and c.

4 In the triangle ABC, c = 27, C = 42° and A = 105°. Find B, a and b.

5 In the triangle ABC, a = 7, c = 5 and A = 68°. Find the perimeter of the triangle.

6 Find all unknown sides and angles for the triangle ABC, given A = 57°, B = 72° and a = 48.2.

7 Find all unknown sides and angles for the triangle ABC, given a = 105, B = 105° and C = 15°.

8 Find all unknown sides and angles for the triangle ABC, given a = 32, b = 51 and A = 28°.

9 Find the perimeter of the triangle ABC if a = 7.8, b = 6.2 and A = 50°.

10 mC In a triangle ABC, A = 40°, C = 80° and c = 3. The value of b is:a 2.64 b 2.86 c 14 d 4.38 e 4.60

11 mC

MP

48 mm

N

64 mm

In the above triangle, sin (P) = 34. The value of sin (M) is:

a 34

b 43

c 916

d 38

e 316

12 We 10 In the triangle ABC, a = 10, c = 8 and C = 50°. Find two possible values of A, and hence two possible values of b.

13 In the triangle ABC, a = 20, b = 12 and B = 35°. Find two possible values for the perimeter of the triangle.

14 Find all unknown sides and angles for the triangle ABC, given A = 27°, B = 43° and c = 6.4.

15 Find all unknown sides and angles for the triangle ABC, given A = 100°, b = 2.1 and C = 42°.

16 Find all unknown sides and angles for the triangle ABC, given A = 25°, b = 17 and a = 13.

17 We 11 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?

18 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 treeb the width of the river, to the nearest metre.

19 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?

20 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?

exerCiSe

7C

318

21 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?

22 mC 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:a 10 km b 38 km c 110 km d 113 km e 114 km

23 mC 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:a 7.37 m b 8.68 m c 10.84 m d 15.04 m e 39.89 m

24 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?

non–right-angled triangles — the cosine ruleIn 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)

eBookpluseBookplus

Digital docWorkSHEET 7.1

25

50

roperock

37 m

7d

D

c

b – xxb

ah

CA

B

b

ac

A C

B



The cosine rule can be used to solve non–right-angled triangles if we are given:1. three sides of the triangle, or2. two sides of the triangle and the included angle (the angle between the given sides).

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

Think WriTe/draW

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

b

a 6c 10

A C

B

76

2 Check that one of the criteria for the cosine rule has been satisfied.

Yes, the cosine rule can be used since two side lengths and the included angle have been given.

3 Write the appropriate cosine rule to find side b. To find side b:b2 = a2 + c2 − 2ac cos (B)

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

5 Evaluate. = 36 + 100 − 120 × 0.241 921 895= 106.969 372 5

b = 106 969 372 5.


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) = b c a

bc

2 2 2

2+ −

b2 = a2 + c2 − 2ac cos (B) ⇒ cos (B) = a c b

ac

2 2 2

2+ −

c2 = a2 + b2 − 2ab cos (C) ⇒ cos (C) = a b cab

2 2 2

2+ −

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

Think WriTe/draW

1 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.

a 4c 7

b 9A C

B

Let a = 4b = 7c = 9

Worked examPle 12

Worked examPle 13

320

2 Check that one of the criteria for the cosine rule has been satisfi ed.

The cosine rule can be used since three side lengths have been given.

3 Write the appropriate cosine rule to fi nd angle A.

cos (A) = b c abc

2 2 2

2+ −

4 Substitute the given values into the rearranged rule.

cos (A) = 7 9 42 7 9

2 2 2+ −× ×

5 Evaluate. = 49 81 16126

+ −

= 114126

6 Transpose the equation to make A the subject by taking the inverse cos of both sides.

A = cos−1

114126

= 25.208 765 3°7

8 Express the answer to the nearest minute. A = 25°13′

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

Think WriTe/draW

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

B

C

A2000 m

1800 m

N

15

70

2 Check that one of the criteria for the cosine rule has been satisfi ed.

The cosine rule can be used since two side lengths and the included angle have been given.

3 Write the appropriate cosine rule to fi nd side c.

To fi nd side c:c2 = a2 + b2 − 2ab cos (C)


Tutorialint-0888

Worked example 14



solve cos ( ) ,a a= + −× ×

7 9 42 7 9

2 2 2

Then press E.Convert the angle to degrees and minutes.


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

5 Evaluate. = 40 000 000 + 3 240 000 − 7 200 000 × −0.573 576 436

= 11 369 750.342c = 11 369 750 342.

= 3371.906 04


7 Answer the question. The rowers are 3371.91 m apart.

In any triangle ABC:1. 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:2. (a) three sides of the triangle, or(b) two sides of the triangle and the included angle (that is, the angle between the two

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


bc

2 2 2

2+ −


ac

2 2 2

2+ −

c2 = a2 + b2 − 2ab cos (C) ⇒ cos (C) = a b c

ab

2 2 2

2+ −

remember

non–right-angled triangles — the cosine rule 1 We 12 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 We 13 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 We 14 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?

8 Maria cycles 12 km in a direction N68°W, then 7 km in a direction of N34°E.a How far is she from her starting point?b What is the bearing of the starting point from her finishing point?

9 A garden bed is in the shape of a triangle, with sides of length 3 m, 4.5 m and 5.2 m.a Calculate the smallest angle.b Hence, find the area of the garden. (Hint: Draw a diagram, with the longest length as the

base of the triangle.)

exerCiSe

7d

322

10 A hockey goal is 3 m wide. When Sophie is 7 m from one post and 5.2 m from the other, she shoots for goal. Within what angle, to the nearest degree, must the shot be made if it is to score a goal?

11 An advertising balloon is attached to two ropes 120 m and 100 m long. The ropes are anchored to level ground 35 m apart. How high can the balloon fly?

12 mC A giant letter M is to be constructed from fluorescent tubes and to be placed above the entrance to a club. The specifications for the manufacture are as follows: Sides AB and ED are vertical and are each 1.2 m long; BC = DC = 80 cm; A and E are 1.3 m apart. The size of angle BCD is closest to:a 148° b 138° c 118°d 108° e 98°

13 A plane flies in a direction of N70°E for 80 km, then on a bearing of S10°W for 150 km.a How far is the plane from its starting point?b What direction is the plane from its starting point?

14 Ship A is 16.2 km from port on a bearing of 053° T and ship B is 31.6 km from the same port on a bearing of 117° T. Calculate the distance between the two ships.

15 A plane takes off at 10.00 am from an airfield, and flies at 120 km/h on a bearing of N35°W. A second plane takes off at 10.05 am from the same airfield, and flies on a bearing of S80°E at a speed of 90 km/h. How far apart are the planes at 10.25 am?

16 Three circles of radii 5 cm, 6 cm and 8 cm are positioned so that they just touch one another. Their centres form the vertices of a triangle. Find the largest angle in the triangle.

17 For the given shape at near right, determine:a the length of the diagonalb the magnitude (size) of angle Bc the length of x.

18 From the top of a vertical cliff 68 m high, an observer notices a yacht at sea. The angle of depression to the yacht is 47°. The yacht sails directly away from the cliff, and after 10 minutes the angle of depression is 15°. How fast does the yacht sail?

area of trianglesThe area of any triangle is given by the rule Area =

12 bh

where b is the base length and h is the perpendicular height of the triangle.

However, often the perpendicular height is not givendirectly and needs to be calculated first. In the triangle ABC, b is the base length and h is the perpendicular height of the triangle.

Using the trigonometric ratio for sine:

sin (A) = hc

Transposing the equation to make h the subject, we obtain:h = c × sin (A)

Therefore, the area of triangle ABC becomes:

Area = 12 bc sin (A)

C

A E

B D

5 cm6 cm

8 cm

x

B

10

8

760

150

7e

b

h

c

b

ah

CA

B



Depending on how the triangle is labelled, the formula could read:

Area = 12 ab sin (C) Area = 1

2 ac sin (B) Area = 1

2 bc sin (A)

The area formula may be used on any triangle provided that two sides of the triangle and the included angle (that is, the angle between the two given sides) are known.

Find the area of the triangle shown. 7 cm 9 cm120

Think WriTe/draW

1 Draw a labelled diagram of the triangle, call it ABC and fi ll in the given information. c 7 cm

A C

Ba 9 cm120

Let a = 9 cm, c = 7 cm, B = 120°2 Check that the criterion for the area rule has been

satisfi ed.The area rule can be used since two side lengths and the included angle have been given.

3 Write the appropriate rule for the area. Area = 12 ac sin (B)

4 Substitute the known values into the rule. = 12 × 9 × 7 × sin (120°)


= 27.28 cm2

Note: If you are not given the included angle, you will need to fi nd it in order to calculate the area. This may involve using either the sine or cosine rule.

A triangle has known dimensions of a = 5 cm, b = 7 cm and B = 52°. Find A and C and hence the area.

Think WriTe/draW

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

b 7A C

B

a 552

Let a = 5, b = 7, B = 52°2 Check whether the criterion for the area rule has

been satisfi ed.The area rule cannot be used since the included angle has not been given.

3 Write the sine rule to fi nd A. To fi nd angle A:a

Asin( ) =

bBsin( )

4 Substitute the known values into the rule. 5sin( )A

= 752sin( )

Worked examPle 15


Tutorialint-0889

Worked example 16

324

5 Transpose the equation to make sin (A) the subject. 5 × sin (52°) = 7 × sin (A)5 52

7× sin ( ) = sin (A)

sin (A) = 5 52

7× sin ( )

6 Evaluate. A = sin−1 5 52

7×

sin ( )

= 34.254 151 87°

7 Express the answer in degrees and minutes. = 34°15′

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

C = 180° − (52° + 34°15′)= 93°45′

9 Write the appropriate rule for the area. Area = 12 ab sin (C)

10 Substitute the known values into the rule. = 12 × 5 × 7 × sin (93°45′)


= 17.46 cm2.

heron’s formulaIf we know the lengths of all the sides of the triangle but none of the angles, we could use the cosine rule to find an angle, then use 1

2 bc sin (A) to find the area. Alternatively, we could use

Heron’s formula to find the area.Heron’s formula states that the area of a triangle is:

Area = s s a s b s c( )( )( )− − −where s is the semi-perimeter of the triangle; that is,

s = 12 (a + b + c)

Note: The proof of this formula is beyond the scope of this course.

Find the area of the triangle with sides of 4 cm, 6 cm and 8 cm.

Think WriTe/draW

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

8 cm

4 cm

A

C

B

6 cm

Let a = 4, b = 6, c = 8

2 Determine which area rule will be used. Since three side lengths have been given, use Heron’s formula.

3 Write the rule for Heron’s formula. Area = s s a s b s c( )( )( )− − −

Worked examPle 17



4 Write the rule for s, the semi-perimeter of the triangle.

s = 12 (a + b + c)

5 Substitute the given values into the rule for the semi-perimeter.

= 12 (4 + 6 + 8)

= 12 (18)

= 9

6 Substitute all of the known values into Heron’s formula.

Area = 9 9 4 9 6 9 8( )( )( )− − −

7 Evaluate. = 9 5 3 1× × ×

= 135 = 11.618 950 04


= 11.62 cm2

If two sides of any triangle and the included angle (that is, the angle between the two 1. given sides) are known, the following rules may be used to determine the area of that triangle.

Area = 12 ab sin (C) Area = 1

2 ac sin (B) Area = 1

2 bc sin (A)

Alternatively, if the lengths of three sides of a triangle are known, Heron’s formula 2. may be used to find the area of the triangle:

Area = s s a s b s c( )( )( )− − −

where s is the semi-perimeter of the triangle; that is,

s = 12 (a + b + c)

remember

area of triangles 1 We 15 Find the area of the triangle ABC with a = 7, b = 4 and C = 68°.

2 Find the area of the triangle ABC with a = 7.3, c = 10.8 and B = 104°40′. 3 Find the area of the triangle ABC with b = 23.1, c = 18.6 and A = 82°17′. 4 A triangle has a = 10 cm, c = 14 cm and C = 48°. Find A and B and hence the area.

5 We 16 A triangle has a = 17 m, c = 22 m and C = 56°. Find A and B and hence the area.

6 A triangle has b = 32 mm, c = 15 mm and B = 38°. Find A and C and hence the area.

7 mC In a triangle, a = 15 m, b = 20 m and B = 50°. The area of the triangle is:a 86.2 m2 b 114.9 m2 c 149.4 m2 d 172.4 m2 d 181.7 m2

8 Find the area of the triangle with sides of 5 cm, 6 cm and 8 cm.

9 We 17 Find the area of the triangle with sides of 40 mm, 30 mm and 5.7 cm.

10 Find the area of the triangle with sides of 16 mm, 3 cm and 2.7 cm.

11 mC A triangle has sides of length 10 cm, 14 cm and 20 cm. The area of the triangle is:a 41 cm2 b 65 cm2 c 106 cm2 d 137 cm2 e 1038 cm2

exerCiSe

7e

326

12 A piece of metal is in the shape of a triangle with sides of length 114 mm, 72 mm and 87 mm. Find its area using Heron’s formula.

13 A triangle has the largest angle of 115°. The longest side is 62 cm and another side is 35 cm. Find the area of the triangle.

14 A triangle has two sides of 25 cm and 30 cm. The angle between the two sides is 30°. Find:a its area b the length of its third side c its area using Heron’s formula.

15 The surface of a fish pond has the shape shown in the diagram at right. How many goldfish can the pond support if each fish requires 0.3 m2 surface area of water?

16 Find the area of this quadrilateral.

17 A parallelogram has diagonals of length 10 cm and 17 cm. An angle between them is 125°. Find:a the area of the parallelogramb the dimensions of the parallelogram.

18 A lawn is to be made in the shape of a triangle, with sides of length 11 m, 15 m and 17.2 m. How much grass seed, to the nearest kilogram, is needed if it is sown at the rate of 1 kg per 5 m2?

19 A bushfire burns out an area of level grassland shown in the diagram. What is the area, in hectares, of the land that is burnt?

200 m400 m

2 km

1.8 km

RoadR

iver

20 An earth embankment is 27 m long, and has a cross-sectionshown in the diagram. Find the volume of earth needed tobuild the embankment.

21 mC A parallelogram has sides of 14 cm and 18 cm, and an angle between them of 72°. The area of the parallelogram is:a 86.2 cm2 b 118.4 cm2 c 172.4 cm2

d 239.7 cm2 e 252 cm2

22 mC An advertising hoarding is in the shape of an isosceles triangle,with sides of length 15 m, 15 m and 18 m. It is to be painted with twocoats of purple paint. If the paint covers 12 m2 per litre, the amountof paint needed, to the nearest litre, would be:a 9 L b 18 L c 24 L d 36 L e 42 L

2 m

4 m

5 m

1 m

4 m

5 m

8 m

60

3.5 m

2 m

5 m80

100

13050

eBookpluseBookplus

Digital docWorkSHEET 7.2



radian measurementIn all of the trigonometry tasks covered so far, the unit for measuring angles has been the degree. There is another commonly used measurement for angles, the radian. This is used in situations involving length and areas associated with circles.

Consider the unit circle, a circle with a radius of 1 unit.OP is the radius.

If OP is rotated θ ° anticlockwise, the point P traces a path along the circumference of the circle to a new point, P1.

The arc length PP1 is a radian measurement, symbolised by θ c.

Note: 1c is equivalent to the angle in degrees formed when the length of PP1 is 1 unit; in other words, when the arc is the same length as the radius.

If the length OP is rotated 180°, the point P traces out half the circumference. Since the circle has a radius of 1 unit, and C = 2πr, the arc PP1 has a length of π.

The relationship between degrees and radians is thus established. 180° = π c

This relationship will be used to convert from one system to another. Rearranging the basic conversion factor gives:

180° = π

1° = π

180

To convert an angle in degrees to radian measure, multiply by π

180.

Also, since π = 180°, it follows that 1c = 180π

.

To convert an angle in radian measure to degrees, multiply by 180π

.

Where possible, it is common to have radian values with π in them. It is usual to write radians without any symbol, but degrees must always have a symbol. For example, an angle of 25° must have the degree symbol written, but an angle of 1.5 is understood to be 1.5 radians.

Convert 135° to radian measure, expressing the answer in terms of π.

Think WriTe

1 To convert an angle in degrees to radian measure,

multiply the angle by 180π

.

135° = 135 × π

180

= 135180

π

2 Simplify, leaving the answer in terms of π. = 34π

7F

P

OP 1 unit

O

P

P1

OP 1 unit

O

c

PP1 O

circumference1 –2

180

Worked examPle 18

328

Convert the radian measurement 45π to degrees.

Think WriTe

1 To convert radian measure to an angle in degrees,

multiply the angle by 180

π.

45π

= 45

180ππ

×

= 720

5

2 Simplify.Note: The π cancels out.

= 144°

If the calculation does not simplify easily, write the answers in degrees and minutes, or radians to 4 decimal places. If angles are given in degrees and minutes, convert to degrees only before converting to radians.

1801. ° = π c

To convert an angle in degrees to radian measure, multiply by 2. π

180.

To convert an angle in radian measure to degrees, multiply by 3. 180π

.

remember

radian measurement 1 We 18 Convert the following angles to radian measure,

expressing answers in terms of π.a 30° b 60° c 120° d 150°e 225° f 270° g 315° h 480°i 72° j 200°

2 We 19 Convert the following radian measurements into degrees.

a π4

b 32π c

76π

d 53π

e 712π

f 176π

g π12

h 1310

π

i 118π j 8π

3 Convert the following angles in degrees to radians, giving answers to 4 decimal places.a 27° b 109° c 243° d 351°e 7° f 63°42′ g 138°21′ h 274°8′i 326°53′ j 47°2′

4 Convert the following radian measurements into degrees and minutes.a 2.345 b 0.6103 c 1 d 1.61e 3.592 f 7.25 g 0.182 h 5.8402i 4.073 j 6.167

Worked examPle 19

exerCiSe

7F eBookpluseBookplus

Digital docSkillSHEET 7.6

Changing degrees to radians



arcs, sectors and segmentsarc lengthAn arc is a section of the circumference of a circle. The length of the arc is proportional to the angle subtended at the centre. For example, an angle of 90° will create an arc which is 1

4 the

circumference.We have already defined an arc length as equivalent to θ radians if the circle

has a radius of 1 unit.

Therefore, a simple dilation of the unit circle will enable us to calculate the arc length for any sized circle, as long as the angle is expressed in radians.

If the radius is dilated by a factor of r, the arc length is also dilatedby a factor of r.

Therefore, l = rθ, where l represents the arc length, r represents the radius and θ represents an angle measured in radians.

Find the length of the arc which subtends an angle of 75° at the centre of a circle with radius 8 cm:

Think WriTe/draW

1 Draw a diagram representing the situation and label with the given values.

r 8

l r75

2 Convert the angle from 75° to radian measure by

multiplying the angle by π

180.

Note: In order to use the formula for the length of the arc, the angle must be in radian measure.

75° = 75 × π

180

= 75180

π

3 Evaluate to 4 decimal places. = 1.3090

4 Write the rule for the length of the arc. l = rθ5 Substitute the values into the formula. = 8 × 1.3090

6 Evaluate to 2 decimal places and include the appropriate unit.

= 10.4720= 10.47 cm

Find the angle subtended by a 17 cm arc in a circle of radius 14 cm:a in radians b in degrees

Think WriTe

a 1 Write the rule for the length of the arc. a l = rθ2 Substitute the values into the formula. 17 = 14θ

7G

c

r 1

Dilation by factor of r

r

cr

Worked examPle 20

Worked examPle 21

330

3 Transpose the equation to make θ the subject.

θ = 1714

4 Evaluate to 4 decimal places and include the appropriate unit.

= 1.214 285 714= 1.2143c

b 1 To convert radian measure to an angle in

degrees, multiply the angle by 180π

.

b 1.2143c = 1.2143 × 180π

= 1 2143 180. ×

π2 Evaluate. = 218 5714.

π= 69.573 446 55°

3 Convert the angle to degrees and minutes. = 69°34′

If l > r, θ > 1 radian (or θ > 57°18′ )If l < r, θ < 1 radian (or θ < 57°18′ )

area of a sectorIn the diagram at right, the shaded area is the minor sector AOB,

A

BO

Major sector

Minorsector

and the unshaded area is the major sector AOB.The area of the sector is proportional to the arc length.

For example, an area of 14 of the circle contains an arc which

is 14 of the circumference.

Thus, in any circle: area of sectorarea of circle

= arc length

circumference of circle

A

rπ 2 =

rr

θπ2

where θ is measured in radians.

A = r r

rθ π

π× 2

2

= 12 r2θ

The area of a sector is: A = 12 r2θ

A sector has an area of 157 cm2, and subtends an angle of 107°. What is the radius of the circle?

Think WriTe

1 Convert the angle from 107° to radian

measure by multiplying the angle by π180

.

Note: In order to use the formula for the area of a sector, the angle must be in radian measure.

107° = 107 × π180

= 107180

π


3 Write the rule for the area of a sector. A = 12 r 2θ

Worked examPle 22



4 Substitute the values into the formula. 157 = 12 × r 2 × 1.8675

5

6 Round to 2 decimal places and include the appropriate unit.

r = 12.97 cm

area of a segmentA segment is that part of a sector bounded by the arc and the chord.

As can be seen from the diagram at right:Area of segment = area of sector − area of triangle

A = 12 r 2θ − 1

2 r 2 sin (θ °)

= 12 r 2 (θ − sin (θ °))

Note: θ is in radians and θ ° is in degrees.

The area of a segment: A = 12 r 2 (θ − sin (θ °))

Find the area of the segment in a circle of radius 5 cm, subtended by an angle of 40°.

Think WriTe

1 Convert the angle from 40° to radian measure

by multiplying the angle by π

180.

40° = 40 × π

180

= 40180

π

Note: For the area of a sector formula, the angle must be in radians.


3 Write the rule for the area of a segment. A = 1

2r 2 (θ − sin (θ °))

4 Identify each of the variables. r = 5, θ = 0.6981, θ ° = 40°

5 Substitute the values into the formula. A = 12 × 52 (0.6981 − sin (40°))

rSegment


Tutorialint-0890

Worked example 23


Solve 15712

1 86752= × ×

r r. ,

Then press E.

332

6 Evaluate. = 12 × 25 × 0.0553

= 0.691 25

7 Round to 2 decimal places and include the appropriate unit.

= 0.69 cm2

Arc length: 1. l = rθArea of a sector: 2. A = 1

2r 2θ

Area of a segment: 3. A = 12r 2 (θ − sin (θ °))

where r = radius, θ = angle (measured in radians) and θ ° = angle (measured in degrees).

remember

arcs, sectors and segments 1 We 20 Find the length of the arc which subtends an angle of 65° at the centre of a circle of

radius 14 cm.

2 Find the length of the arc which subtends an angle of 153° at the centre of a circle of radius 75 mm.

3 An arc of a circle is 3.5 cm long, and subtends an angle of 41° at the centre of the circle. What is the radius of the circle?

4 An arc of a circle is 27.8 cm long, and subtends an angle of 205° at the centre of the circle. What is the radius of the circle?

5 We 21 Find the angle subtended by a 20 cm arc in a circle of radius 75 cm:a in radians b in degrees.

6 Find the angle subtended by an 8 cm arc in a circle of radius 5 cm:a in radians b in degrees.

7 An arc of length 8 cm is marked out on the circumference of a circle of radius 13 cm. What angle does the arc subtend at the centre of the circle?

8 An arc of length 245 mm is marked out on the circumference of a circle of radius 18 cm. Find the angle that the arc subtends at the centre of the circle.

9 The minute hand of a clock is 35 cm long. How far does the tip of the hand travel in 20 minutes?

10 A child’s swing is suspended by a rope 3 m long. What is the length of the arc it travels if it swings through an angle of 42°?

11 Find the area of the sector of a circle of radius 17 cm with an angle of 56°.

12 Find the area of the sector of a circle of radius 6.2 cm with an angle of 256°.

13 We 22 A sector has an area of 825 cm2, and subtends an angle of 70°. What is the radius of the circle?

14 A sector with an area of 309 cm2 is part of a circle of radius 18.2 cm. Find the angle in the sector.

exerCiSe

7G



15 Find the area of a sector of a circle of radius 30 cm if the sector has an arc length of 18 cm.

16 A garden bed is in the form of a sector of a circle of radius 4 m. The arc of the sector is 5 m long. Find:a the area of the garden bedb the volume of mulch needed to cover the bed to a depth of 10 cm.

17 The minute hand on a clock is 62 cm long. What area does the hand sweep through in 40 minutes?

18 A sector whose angle is 150° is cut from a circular piece of cardboard whose radius is 12 cm. The two straight edges of the sector are joined so as to form a cone.a What is the surface area of the cone?b What is the radius of the cone?

19 We 23 Find the area of the segment in a circle of radius 25 cm subtended by an angle of 100°.

20 Find the area of the segment of a circle of radius 4.7 m that subtends an angle of 85°20′ at the centre.

21 A segment of a circle subtends an angle of 75° at the centre. The area of the segment is 100 cm2. Find the radius of the circle.

22 In a circle of radius 15 cm, a sector has an area of 100 cm2. Find the angle subtended by the sector.

23 Two circles of radii 3 cm and 4 cm have their centres 5 cm apart. Find the area of the intersection of the two circles.

24 mC The angle subtended by a 28 cm arc in a circle of radius 20 cm in radians is:a 0.71 b 40.93 c 80.21 d 1.4 e 0.4

25 mC The area of the segment in a circle of radius 12 cm, subtended by an angle of 60° is:a 6.52 cm2 b 30.31 cm2 c 26.08 cm2 d 15.24 cm2 e 13.04 cm2

26 Two irrigation sprinklers spread water in circular paths with radii of 7 m and 4 m. If the sprinklers are 10 m apart, find the area of crop that receives water from both sprinklers.

27 mC The length of the arc which subtends an angle of 50° at the centre of a circle with radius 10 cm is:a 8.73 cm b 0.87 cm c 10.43 cm d 6.25 cm e 0.63 cm

334

Summary

Trigonometry of right-angled triangles

For any right-angled triangle:•

sin (θ) = OH

cos (θ) = AH

tan (θ) = OA

Pythagoras’ theorem, • c2 = a2 + b2 may also be used to solve right-angled triangles.

ca

b

Angles of 30• °, 45° and 60° have exact values of sine, cosine and tangent.

30° 45° 60°

sin 12

1

2

2

2= 3

2

cos 32

1

2

2

2= 1

2

tan1

3

33

= 1 3

Trigonometric identities and other relationships between sin (θ ), cos (θ ) and tan (θ )

An identity is a relationship that holds true for all values of a pronumeral or pronumerals.• The Pythagorean Identity states that sin• 2 (θ) + cos2 (θ) = 1.Sine and cosine are called • complementary functions since:

sin (90° − θ) = cos (θ) and cos (90° − θ ) = sin (θ)

The tangent function may also be written as tan (• θ) = sin ( )cos ( )

θθ

.

applications of right-angled triangles

Angles of elevation and depression are each measured from the horizontal.• The angle of elevation is equal to the angle of depression since they are alternate ‘Z’ angles.• True bearings are measured in a clockwise direction, starting from north (0• ° T).

Non–right-angled triangles — the sine rule

The sine rule states that for any triangle ABC:•

aA

bB

cCsin( ) sin( ) sin( )

= =

When using this rule, it is important to note that, depending on the values given, any combination of the two equalities may be used to solve a particular triangle.The sine rule may be used to solve non–right-angled triangles if we are given:•

(a) two angles and one side length, or (b) two side lengths and an angle opposite one of these side lengths.The ambiguous case exists if • C is an acute angle and a > c > a × sin (C ).

(A)

B

C A

(O)(H )

HypotenuseOpposite

Adjacent


Chapter 7 Trigonometry 335

Non–right-angled triangles — the cosine rule

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:• (a) three sides of the triangle, or(b) two sides of the triangle and the included angle (that is, the angle between the two given sides).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.


bc

2 2 2

2+ −


ac

2 2 2

2+ −

c2 = a2 + b2 − 2ab cos (C ) ⇒ cos (C ) = a b c

ab

2 2 2

2+ −

area of triangles

If two sides of any triangle and the included angle (that is, the angle between the two given sides) are known, • the following rules may be used to determine the area of that triangle.

Area = 12 ab sin (C )

Area = 12 ac sin (B)

Area = 12 bc sin (A)

Alternatively, if three side lengths of a triangle are known, Heron’s formula may be used to find the area of a • triangle:

Area = s s a s b s c( )( )( )− − −

where s is the semi-perimeter of the triangle; that is,

s = 12(a + b + c)

Radian measurement

180• ° = π cTo convert an angle in degrees to radian measure, multiply by •

π180

.

To convert an angle in radian measure to degrees, multiply by • 180π

.

arcs, sectors and segments

Arc length: • l = rθ

Area of a sector: • A = 12r 2θ

Area of a segment: • A = 12r 2 (θ − sin (θ °))

where r = radius, θ = angle (measured in radians) and θ ° = angle (measured in degrees).

336

ChaPTer revieW

mulTiPle ChoiCe

1 In the triangle, the value of θ, to the nearest degree, is:a 37° b 39° c 51°d 52° e 53°

2 A ladder 4.5 m long rests against a vertical wall, with the foot of the ladder 2 m from the base of the wall. The angle the ladder makes with the wall, to the nearest degree, is:a 24° b 26° c 35° d 64° e 66°

3 A person stands 18 m from the base of a building, and measures the angle of elevation to the top of the building as 62°. If the person is 1.8 m tall, how high is the building, to the nearest metre?a 11 m b 18 m c 36 md 22 m e 34 m

4 A bearing of 310° T is the same as:a N40°W b N50°W c S50°Wd S50°E e N50°E

5 Points M and P are the same distance from a third point O. The bearing of M from O is 038° and the bearing of P from O is 152°.The bearing of P from M is:a between 000° and 090°b between 090° and 180°c exactly 180°d between 180° and 270°e between 270° and 360°

[©Vcaa 2006]

6 In triangle ABC, a = 10, b = 7 and B = 40°. A possible value for C, to the nearest degree, is:a 37° b 52° c 68°d 73° e 113°

7 Two boats start from the same point. One sails due north for 10 km and the other sails south east for 15 km. Their distance apart is:a 10.62 km b 14.83 km c 17.35 kmd 21.38 km e 23.18 km

8 A triangle has sides measuring 5 cm, 8 cm and 10 cm. The largest angle in the triangle, to the nearest degree, is:a 52° b 82° c 98°d 128° e 140°

9

20 35

H

G

4 cm

12 cm

9 cmF

T

In the diagram above, the length of FH is equal to:

a 4 tan (55°) b ( )12 92 2−

c ( cos ( ))12 9 216 352 2+ −

d 4 20

15sin( )

sin ( )

e 12 sin (55°)

[©Vcaa 2004]

10 The points M, N and P form the vertices of a triangular course for a yacht race.MN = MP = 4 km. The bearing of N from M is 070°. The bearing of P from M is 180°. Three people perform different calculations to determine the length of NP in kilometres.

Graeme NP = ( cos ( )16 16 2 4 4 110+ − × × ×

Shelley NP = 2 × 4 × cos (35°)

Tran NP = 4 110

35× sin ( )sin ( )

The correct length of NP would be found by:a Graeme onlyb Tran onlyc Graeme and Shelley onlyd Graeme and Tran onlye Graeme, Shelley and Tran

[©Vcaa 2007]

4

5

exam TiP Correctly answering this question, a bearings problem, depended critically on being able to construct an appropriate diagram from the information given. This was clearly beyond most students, as only 31% gave the correct response.

[Assessment report 2006]

exam TiP To answer this question, students were required to draw a diagram and then carefully assess all of the alternatives offered before arriving at a solution. The correct option was chosen by only 37% of students. The relatively even distribution of student responses over the other incorrect options suggests that most students either failed to systematically test all of the alternative methods, or failed to get started and just guessed.




11 The area of the triangle with a = 10 m, b = 8 m and C = 72° is:a 12.36 m2 b 76.08 m2 c 10.15 m2

d 38.04 m2 e 123.10 m2

12 A garden bed is in the shape of a triangle, with sides of length 4 m, 5.2 m and 7 m. The volume of topsoil needed to cover the garden to a depth of 250 mm is:a 2.32 m3 b 2.57 m3 c 2.81 m3

d 3.17 m3 e 3.76 m3

13 When 75° is converted to radian measure, the value of the angle, expressed in terms of π, is:

a 125π b π

12c 5

24π

d 512π e 7

12π

14 When 5.321 is converted to degrees and minutes, the value of the angle is:a 305°27′ b 304°52′ c 5°19′d 1°42′ e 152°26′

15 An arc in a circle of radius 5 cm is 3.5 cm long. The angle, to the nearest degree, subtended at the centre by the arc is:a 35° b 40° c 50° d 68° e 82°

16 A sector has an area of 40 cm2, and an angle of 30°. The arc length of the sector, to 2 decimal places, is:a 1.64 cm b 2.66 cm c 4.83 cmd 6.47 cm e 12.36 cm

17 The area of the shaded region in the figure at right to the nearest cm2 is:a 800 cm2 b 846 cm2

c 898 cm2 d 952 cm2

e 983 cm2

18 A clock has a minute hand 75 cm long. The area that it sweeps when passing through 48 minutes, to 2 decimal places, is:a 0.90 m2 b 1.35 m2 c 1.41 m2

d 1.88 m2 e 2.01 m2

ShorT anSWer

1 A stepladder stands on a floor with its feet 2.18 m apart. If the angle formed by the legs is 50°, how high above the floor is the top of the ladder?

2 Two buildings, 15 m and 27 m high, are directly opposite each other across a river. The angle of depression of the top of the smaller building from the top of the taller one is 52°. How wide is the river?

3 An allotment of land contains a communications tower, PQ. Points S, Q and T are situated on level ground. From S, the angle of elevation of P is 20°. Distance SQ is 125 metres. Distance TQ is 98 metres.

20

ST

P

Q

Tower

125 m98 m

a Determine the height, PQ, of the communications tower in metres, correct to 1 decimal place.

b Determine the angle of depression of T from P in degrees, correct to 1 decimal place.

[©Vcaa 2006]

4 A four-wheel-drive vehicle leaves a camp site and travels across a flat sandy plain in a direction of S65°E, for a distance of 8.2 km. It then heads due south for 6.7 km to reach a waterhole.

a How far is the waterhole from the camp site?b What is the bearing of the waterhole from the

camp site?

5 A plane is on a course that takes it over points A and B, two locations on level ground. At a certain time, from point A, the angle of elevation to the

12040 cm

exam TiP Finding the size of an incorrect angle was a common error.


338

plane is 72°. From point B, the angle of elevation is 47°. If A and B are 3500 m apart, find the height of the plane off the ground.

6 Find all unknown sides and angles of triangle ABC, given a = 25 m, A = 120° and B = 50°.

7 A triangle has sides of length 12 m, 15 m and 20 m. Find the magnitude (size) of the largest angle.

8 The land near the camping ground is flat and suitable for orienteering. Checkpoint X is situated 2.1 km from camping ground G on a bearing of 140°. Checkpoint Y is situated 1.7 km from checkpoint X on a bearing of 075°.

140

75 1.7 km2.1 km

X

Y

North

G

a i How far south of G is checkpoint X in kilometres, correct to 1 decimal place?

ii How far south of G is checkpoint Y in kilometres, correct to 1 decimal place?

b Determine the size of angle GXY.

c Calculate the distance GY in kilometres, correct to 1 decimal place.

d Determine the bearing of checkpoint Y from camping ground G, correct to the nearest degree.

[©Vcaa 2005]

9 A triangle has two sides of 18 cm and 25 cm. The angle between the two sides is 45°. Find:a its area b the length of its third sidec its area using Heron’s formula.

10 a Convert the following angles to radian measure, expressing answers in terms of π :i 80° ii 125° iii 640°

b Convert the following radian measurement into degrees.

i π20

ii 158π

iii 7πc Convert the following angles in degrees to

radians, giving answers to 4 decimal places. i 56° ii 129°15′ iii 212° 17′

d Convert the following radian measurements into degrees and minutes.

i 1.17 ii 3.57 iii 0.2539

11 A paddock is in the shape of a sector with radius of 75 m and an angle of 60°. Find:a the amount of fencing needed to enclose the

paddockb the area enclosed by the paddock.

12 A decorative glass panel is in the shape illustrated at right. Find the area of glass in the whole panel.

exam TiP A common incorrect answer was 2.1 km (the length of GX).


exam TiP Exact numbers should be retained in the calculator and used in the second step unless the first step was the answer to a specific question.

Several students applied Pythagoras’ theorem to the non–right-angled triangle GXY to obtain 2.7 km. Others assumed that X was twice as far south of G as Y.


exam TiP Common errors included answers of 75°, 150° and 215°.


exam TiP Many students who used the cosine rule forgot to take the square root at the end. Pythagoras’ theorem was also incorrectly used here by some students.


exam TiP This question was quite poorly done, and, as working out was often not clear; students were sometimes unable to even earn 1 mark for method. Bearings continued to be poorly understood by many students.


40

1.2 m

exTended reSPonSe

1 Three circles of radii 2 cm, 3 cm and 4 cm are placed so that they just touch each other. A triangle is formed by joining their three centres. Find:a the three angles of the triangleb the area of the triangle, correct to 3 decimal placesc the shaded area correct to 3 decimal places.



2 A farmer owns a large triangular area of flat land, bounded on one side by an embankment to a river flowing NE, on a second side by a road which meets the river at a bridge where the angle between river and road is 105°, and on the third side by a long fence. Find:a the length of the river frontage, correct to 3 decimal placesb the area of the land correct to 3 decimal places.The farmer decides to divide the land into two sections of equal area, by running a fence from the bridge to a point on the opposite side.c On what bearing must the fence be built?d What is the length of the fence, correct to 3 decimal places?

3 Christopher lives on a farm. He has decided that this year he will plant a variety of crops in his large but unusually shaped vegetable garden. He has divided the vegetable garden into six triangular regions, which he will fence off as shown in the diagram at right. Christopher needs to calculate the perimeter and area of each region so he can purchase the correct amount of fencing material and seedlings.a Separate each of the regions into single triangles and

label each with the information provided.b Use the appropriate rules to determine all unknown

lengths and relevant angles.c How much fencing material is required to section off the

six regions?d If fencing material is $4.50 per metre (and only

sold by the metre) what will the cost be?e Calculate the area of each region and hence

determine the total area available for planting.

4 A farmer owns an allotment of land in the shape of triangle ABC shown at right. Boundary AB is 251 metres. Boundary AC is 142 metres. Angle BAC is 45°. A straight track, XY, runs perpendicular to the boundary AC. Point Y is 55 m from A along the boundary AC.a Determine the size of angle AXY.b Determine the length of AX in metres, correct

to 1 decimal place.

Road

N

3.2 km

River

Fence105

42

4533

A

B

DE

F

C95º 64º

80º58º

38º

85 m

43 m68 m

52 m

56 m124 m

1 2

3

4

5

6

4555 m

142 m

North

251 m

B

CY

X

A

340

c The bearing of C from A is 078°. Determine the bearing of B from A.

d Determine the shortest distance from X to C in metres, correct to 1 decimal place.

e Determine the area of triangle ABC correct to the nearest square metre.

The length of the boundary BC is 181 metres (correct to the nearest metre).f i Use the cosine rule to show how this length can be found. ii Determine the size of angle ABC in degrees, correct to

1 decimal place.

A farmer plans to build a fence, MN, perpendicular to the boundary AC. The land enclosed by triangle AMN will have an area of 3200 m2.g Determine the length of the fence MN.

[©Vcaa 2006]

exam TiP Bearings proved to be a problem for many students.


exam TiP This question was often poorly done. Some students incorrectly assumed that angle AXC was 90° or that triangle YXC was isosceles.


exam TiP Students who used Heron’s formula to get an answer of 12602 were awarded full marks.


exam TiP Students were required to substitute into the formula. Listing the inputs for a cosine rule calculator application was not acceptable.


45

B

C

N

MArea3200 m2

A

exam TiP This question was very poorly answered. Most students began with ½ bc = 3200 or bc = 6400 but then failed to recognize the equality of b and c and did not take the square root. Some students seemed to confuse the line MN with the line XY in the previous diagram, and some tried to use an area ratio but were often unsuccessful.


eBookpluseBookplus

Digital docTest Yourself Chapter 7



eBookpluseBookplus aCTiviTieS

chapter openerDigital doc

10 Quick Questions: Warm up with ten quick • questions on trigonometry. (page 297)

7a Trigonometry of right-angled trianglesTutorial

We4 • int-0885: Watch how to fi nd the perimeter of a composite shape. (page 300)

Digital docs

Investigation: Investigate trigonometric identities. • (page 301)SkillSHEET 7.1: Practise labelling right-angled • triangles. (page 302)SkillSHEET 7.2: Practise using trigonometric ratios. • (page 302)SkillSHEET 7.3: Practise using degrees and • minutes. (page 303)SkillSHEET 7.4: Practise fi nding measurements of • composite shapes 1. (page 303)SkillSHEET 7.5: Practise fi nding measurements of • composite shapes 2. (page 304)

7b applications of right-angled trianglesTutorial

We8 • int-0886: Watch how to determine the bearing of a ship from its starting point after travelling fi xed distances south and east. (page 308)

7c Non–right-angled triangles — the sine ruleInteractivity

The sine rule • int-0808: Consolidate your understanding of using the sine rule with non–right angled triangles. (page 313)

Tutorial

We 10 • int-0887: Watch how to apply the ambiguous case of the sine rule to determine two sets of values of a side length and its opposite angle. (page 313)

Digital doc

WorkSHEET 7.1: Trigonometric ratios and their • applications. (page 318)

7d Non–right-angled triangles — the cosine rule

Tutorials

We 14 • int-0888: Watch how to calculate the distance between two rowers after they have rowed in different directions from the same position. (page 320)

7e area of trianglesTutorials

We 16 • int-0889: Watch how to calculate the area of non–right-angled triangle. (page 323)

Digital doc

WorkSHEET 7.2: Calculate length and area of • 2-decimal shapes. (page 326)

7F Radian measurementDigital doc

SkillSHEET 7.6: Practise changing degrees to • radians. (page 328)

7G arcs, sectors and segmentsTutorial

We23 • int-0890: Watch how to fi nd the area of a segment in a circle. (page 331)

chapter reviewDigital doc

Test Yourself: Take the end-of-chapter test to test • your progress. (page 340)

To access eBookPLUS activities, log on to

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342

exam PraCTiCe 2 ChaPTerS 5 To 7

mulTiPle ChoiCe 15 minutes

Each question is worth 1 mark.

1

( p 4)

2p

p

The value of p in the triangle above is:a 30 b 44 c 58d 60 e 120

2

t

2 m54

The value, in metres, of t in the triangle above is closest to:a 1.18 b 1.45 c 1.62d 2.47 e 2.75

3

Level 4 Level 5 Level 6

To develop hand and eye coordination in children, a game is set up in which the child is required to roll a ball at a stack of cans. The aim is for the ball to make contact with at least one of the cans. The number of cans decreases as the child advances through the different levels of difficulty. Some of the different levels are shown in the diagram above. Assuming that this pattern continues, the number of cans in level 2 would be:a 5 b 6 c 10 d 15 e 21

The following information relates to questions 4 and 5.

Sarah begins a new fitness routine, each day she walks further than the day before. On day one she walks 1.5 km, on day two she walks 1.65 km and day three 1.8 km. Sarah continues to follow this pattern each day for 14 days.

4 Sarah’s routine would best be described by which one of the following:a arithmetic sequence with a common difference

of 0.15b arithmetic series with a common difference

of 0.15c geometric sequence with a common ratio of 0.1d geometric sequence with a common ratio of 1.1e Fibonacci sequence

5 The total distance, in kilometres, Sarah walks in 14 days, will be closest to:a 5 b 21 c 23d 35 e 42

6 A ladder is placed 2.5 metres from the base of a brick wall. The angle the ladder makes with the brick wall is 37°. The length of the ladder, to the nearest metre, can be found from which one of the following expressions?a 2.5 sin (37°) b 2.5 cos (37°)

c cos ( )

.53

2 5

d

sin ( ).53

2 5

e 2 5

53.

cos ( )

This information relates to questions 7 and 8.

40 cm

96.6 cmx

The local council is constructing larger stop signs to be placed in the area. The new stop sign will still be in the shape of a regular octagon, but the side length will be increased to 40 cm, and the height of the sign will be 96.6 cm as shown in the diagram above.

7 The value of x° in diagram above is:a 45 b 60 c 120d 135 e 180

8 The minimum amount of metal, in cm2, required to construct the new sign is closest to:a 4525 b 5543 c 5913d 7728 e 9329


343exam practice 2

exTended reSPonSe 30 minutes

1

T

R

Rapid Rise Village Mt Blizzard

S

Summit

Figure 1

The Mt Blizzard Corporation is constructing a new chair lift to connect the recently established ski village, Rapid Rise (point R) to the Mt Blizzard Summit, (point S) as shown in Figure 1.

Stephanie who works for the corporation has been asked to determine the distance between Rapid Rise and the summit. She measured an angle of elevation of 18° from R to point T on the mountain. She then measured the distance between R and T which she determined to be 200 metres. At point T, she measured an angle of elevation of 55° from T to a point S on the summit. The angle of depression from S to R was measured at 47°.a On Figure 2, clearly label: i the distance of 200 metres between points R and T ii the angle of elevation of 55° from T to S. 1 + 1 = 2 marks

T

R 18

47

Mt Blizzard

S

Figure 2b Show that the size of angle ∠RTS is 143°.c Determine the size of angle ∠RST.d Using the sine rule, determine the distance, in metres, between points R and S, to the nearest metre.e Determine the vertical height, in metres, of the summit from Rapid Rise, correct to the nearest metre.

2 + 1 + 2 + 3 = 8 marks

2 Due to increased fuel costs and wages, the price of chairlift tickets has risen at a constant amount each season. In 2005 the cost of an adult all day ticket was $66, in 2006 the cost was $69.30 and in 2007 the cost had risen to $72.75.a If the cost of the ticket continued to follow this pattern, determine the cost of the ticket in 2008.b Show that this pattern does not follow an arithmetic sequence.c Using the common ratio of 1.05, fi nd an equation for the cost of an adult all day ticket, C, in any season, n.d Using your answer to part c, determine the difference in ticket price between 2007 and 2012. In 2005, the number of adult all day tickets that were sold on the opening day was 1245. In 2006, the number sold on the opening day had decreased to 1183, and in 2007, the number of tickets sold was 1124. e If this decline continues to follow this pattern, determine

i the common ratio, r ii the percentage decrease.1 + 2 + 2 + 3 + (1 + 1) = 10 marks

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