Triangle applications

Contents

1.1 Using right-angled triangles 1.2 Applying bearings and direction 1.3 The sine rule 1.4 The cosine rule 1.5 Trig applications 1.6 Working in three dimensions Chapter summary Chapter review

Syllabus subject matter

Periodic functions and applications ■ Trigonometry including the definition and practical applications of the sine, cosine and tangent ratios ■ Simple practical applications of the sine and cosine rules (the ambiguous case is not essential)

Quantitative concepts and skills ■ Metric measurement including measurement of mass, length, area and volume in practical contexts ■ Calculation and estimation with and without instruments ■ Basic algebraic manipulations Syllabus

Guide Chapter 1

Triangle applications

Triangle applications

1.1

Using right-angled triangles

When working with triangles it is usual to name an angle by a single capital letter. The opposite side is usually named by the same lower-case letter.

In the triangle on the right, we could state the length of side AB as c= 44 mm or as AB= 44 mm.

We could state the size of the top angle as C= 79° or as

∠ACB= 79°. Greek letters such as θ or α are also used for angles.

Angles may be measured in degrees (of arc), shown by the symbol °. There are 360 degrees in one full turn (revolution). If greater accuracy is needed, minutes (′) and seconds (″) of arc may also be used. There are 60 minutes of arc in one degree and 60 seconds of arc in one minute.

Nowadays, it is becoming more common to express angles in degrees only, for example, as 36.5° rather than 36°30′.

Most scientific calculators have a function marked as or to change from decimal degrees to degrees, minutes and seconds of arc, and vice versa. Make sure that you know how to use your calculator to convert between decimal degrees and degrees, minutes and seconds. In a right-angled triangle, the longest side is called the hypotenuse. It is also the side opposite the right angle. You may remember that Pythagoras’s Theorem relates the lengths of the sides of a right-angled triangle.

The word ‘trigonometry’ comes from Greek and means Earth measurement. Even today, trigonometry is still used to check property boundaries and to survey for new bridges, roads and other civil engineering. However, the importance of trigonometry now extends far beyond the original purposes of the ancient Greeks. It is used in astronomy, navigation, optics, sound engineering, building and construction, and many other areas of modern life.

c b

A

a

B C

1 revolution= 360° 1° = 60′ 1′ = 60″

!

°′ ″ DMS

Pythagoras’s Theorem

In any right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

a2= _b2+ _c2

a

c

b

!

The sides other than the hypotenuse are named as opposite and adjacent (next to) to the angle we are considering.

a Find the unknown side in the triangle on the right.

b A vertical post stands 3.7 m above the ground. A taut wire joins the top of the post to an anchor point 5.6 m from the base of the post.

How long is the wire?

Solution

a The 73 cm side is the hypotenuse. 732 = ₂₇2+ _a2

Evaluate squares. 5329 = 729 + a2

Subtract to get a2 _{and reverse the equation. a}2 = ₅₃₂₉ − ₇₂₉

= 4600

Find the square root. a =

= 67.8232 …

Round and write the answer as a sentence. a is about 68 cm.

b Re-draw the diagram with the known

information.

The wire is the hypotenuse of the triangle. w2 = _3.72+ _5.62

Evaluate. = 13.69 + 31.36

= 45.05

Find the square root. w =

= 6.7119 …

Round and write the answer as a sentence. The wire is about 6.8 m long.

73 cm

27 cm _a

Wire

Post

4600

3.7 m

5.6 m w

45.05

Example

1

For angle θ For angle φ

Adjacent

Opposite Hypotenuse

θ

φ

Opposite

Adjacent Hypotenuse

θ

φ

!

Triangle applications

We use the trigonometric ratios sine, cosine and tangent for right-angled triangles. They are abbreviated to sin, cos and tan.

The ratios can be remembered using the following mnemonic. Trigonometric ratios

sin X= =

cos X= =

tan X= =

opposite side hypotenuse --- x

y

--adjacent side hypotenuse --- z

y

--opposite side adjacent side --- x

z

--Z

Y X

x

z y

!

SOH CAH TOA (sock-ah-towa)

Sin = O/H Cos = A/H Tan = O/A

!

Calculate the value of the pronumeral in each of the following triangles.

Solution

a First choose the appropriate trig ratio.

The known side is the hypotenuse and the side we want is the opposite, so use sin.

sin 52°7′ =

Multiply by 26 to obtain x. x = 26 sin 52°7′

Use your calculator to obtain sin. = 26 × 0.7892 …

= 20.5208 …

Round appropriately. ≈ 20.5

Write the answer. x is approximately 20.5 cm.

b The side we want is the hypotenuse.

The known side is the adjacent, so use cos. cos 34.6° =

Rearrange. gcos 34.6° = 16.4

Isolate g. g =

When using a calculator, don’t round off until the last step.

= 19.9237 …

≈ 19.9

Write in words. g is about 19.9 mm.

x 26 cm 52°7′

16.4 mm

a b

g 34.6°

x

26

---16.4

g

---16.4 cos 34.6°

---Example

2

Sometimes we use basic geometry to find right-angled triangles in which to apply trigonometric ratios.

Find the magnitude of the angle θ.

Solution

The cosine ratio should be used here. cos θ = = 0.3265 …

Use the inverse cos function cos−1 _{on your calculator.} θ = _{70.9416 …}°

Round off. ≈ 70.9°

Write in words. The angle θ is about 71°.

θ

49 m

16 m

16 49

---Example

3

To find the approximate length of the following cleared area we need to find the length of MN in the figure shown here.

Solution

First draw a perpendicular from

O to meet MN at Q as shown on

the right.

40°

N M

P

O 43 m

30 m

43 m

40°

O P

Q

M 30 m N

43 m

30 m

Exercise 1.1

Using right-angled triangles

1 Change the following angles to degrees only.

a 42°36′ b 67°21′ c 28°17′24″ d 16°24′36″ e 7°11′19″

2 Change the following angles to degrees, minutes and seconds.

a 19.2° b 49.67° c 56.97° d 87.25° e 22.16°

3 Find the unknown side or angle marked in each of the following triangles.

4 Find the value of the unknown length in each of the following.

MQOP is a rectangle, so MQ = PO and OQ = PM. MQ = 30 m and OQ = 43 m

Now find QN using QNO. tan 40° =

Rearrange to isolate QN. QNtan 40°= 43

QN =

Evaluate and round off. ≈ 51 m

Now use the diagram to help calculate MN. MN = MQ + QN = 30 + 51

= 81 m

Write in words. MN is about 81 m.

43

QN

---43 tan 40°

---Additional Exercise

1 .1

21

28°12′ m

48°28′ p

17.9

25

29

θ

42°36′ 48.1

x

a b c

d e f

35 47

φ

78 92

u

31 m

d e f

a b c

7 m

74°

d

47° 16·8 m

e

65 m a

39°

17 m

58° b

c 3·2 m 54°

8·2 m

5 Find the angle marked in each of the following triangles.

6 In PMQ, M = 90°, Q = 60° and PQ = 40 cm. Find the length of:

a MQ b PM

7 In MNO, MN = 24 mm, MO = 16 mm and M = 60°. OQ is drawn perpendicular to MN so that it meets MN at Q. Find the length of:

a MQ b QO c NO

8 In the diagram shown, BC = 14 cm. Find the length of:

a AC b AB

c AD d CD

9 In the diagram, calculate the length of:

a MO b NO

c MN d OL

Modelling and problem solving

10 Calculate the length of AB.

g h i

g

17·4 m 29°

i 41° 14 m 4·7 cm

26° h

a b c

d e f

a

k

16

18

24 48

2.3

1.8

β

5.1 cm 1860 m

3744 m 4·7 cm

9·2 m

7·6 m

2·2 m

g

x f

A B

C D

15°

30°

14 cm

60°

30°

40 mm L

M N O

49 cm

C

A 31° B

11 Find the length of MN.

12 PQRS is a rectangle with PS = 97 cm and SR = 64 cm. If PT = 35 cm, find ∠QTR.

13 A civil engineering company is using tunnel-boring machine to construct a tunnel into a hill that is inclined an angle of 15° to the horizontal tunnel. After the tunnel has been drilled 90 m into the hill, a ventilation shaft needs to be driven vertically up to the surface. What will be the depth of the shaft?

14 A chest of drawers 75 cm high is placed in an attic in which the roof slopes down towards the horizontal floor. The closest that the chest of drawers can stand to the corner formed by the roof and the floor is 28 cm. Find the angle of incline of the roof to the floor.

15 Find the perimeter of WXYZ shown on the right.

16 A builder wants to construct a roof as shown on the right with a pitch of 32°30′. How long should the beams be cut?

17 A circle of radius 30 cm has a chord drawn 20 cm from its centre, O. Calculate:

a the magnitude of ∠POM

b the length of the chord PM.

1.9 m

70 cm

116° L

M N O

Q R

S P

T

32 cm

64·2°

78 cm X

Y

Z W

32°30′ 32°30′

Beam

11.8 m

Beam

Chord

P M

18 The roof of a new home is designed so that the solar panel can collect the maximum energy from the sun. Calculate:

a the height of the roof

b the length of the beam AB.

19 This diagram shows part of a contour map. The contours are at 20 m intervals and the map is drawn to a scale of 1 : 25 000. An all-terrain vehicle can manage slopes inclined at less than 18° to the horizontal. Would it be possible for the vehicle to proceed directly from X to Y on the route shown?

20 A pendulum 1.2 m long is designed for a grandfather clock. It swings through an angle of 12°. How wide should the case of the clock be made?

1.2

Applying bearings and direction

Surveyors and navigators use directions and bearings in their work. A bearing gives the direction in the horizontal plane of an object from an observation point.

As shown in the diagram on the right, the direction of A from an observer at O can be stated as N40°E or 040°T.

The direction of B from O can be written as S20°W or 200°T. The direction of C from O can be written as S20°E or 160°T.

34°28′

A B

C Solar

panel

Roof

62°12′

24 m

400

420 460 X

Y

0 250 500 750 1000 m

Scale 1 : 25 000 Contour interval: 20 m

The bearing of the object is given as the 3-digit clockwise angle the observer turns from due north in order to face the object. So an object due west has a bearing of 270°T (T for true bearing).

A bearing can also be given as the angle from north or south towards either east or west. So a bearing of 160°T can be expressed as S20°E.

!

N

S

E W

40°

20° O

20°

C B

Exercise 1.2

Applying bearings and direction

1 The bearing from C to A is N50°E. The bearing from

C to B is N80°W. D is due north of B. What is the bearing of:

a C from A? b C from B?

c B from D? d D from B?

Two boats, A and B, are situated so that B is 26 km due east of A. Boat A sets off for an island in the direction 113°T, and B takes the course 203°T for the same island.

a After they meet, how far has A travelled?

b What is the bearing from the island to B’s starting point?

Solution

Draw a diagram showing the starting positions and directions taken by A and B. Put compass direction points NSWE at the crucial positions. Label angles and show all given

information.

a The distance travelled by A is AC.

Write down the information from ACB. ∠BAC = 23° and ∠ABC = 67°

Calculate the other angle in ACB. ∠BCA = 90°

Use ACB to write an equation with AC. sin 67°=

Isolate AC. AC = 26 sin 67°

Evaluate and round off. ≈ 23.9

Write the answer in words. A has travelled approximately 23.9 km.

b Locate the island and B’s starting point

on the diagram. We want to know the bearing from C to B.

β is alternate to 67°.

β = 67°

θ = 90° − 67° = 23°

We know that β and θ form a right angle.

Now we state the bearing.

The bearing from the island to B’s starting point is 023°T.

67° N

S E

N

S E W

203° B

C

θ β

Island N

S W

113°

23° A

26 km

AC

26

---Example

5

Additional Exercise

1 .2

N

D

B

C

2 X is 45 m due west of P, and the bearing of Y from P is 120°T. Find:

a the bearing of P from X

b the bearing of P from Y

c the distance from X to Y

d the bearing of X from Y.

3 A fishing boat leaves port and travels on a bearing of N60°E for 1.6 km until it is due north of a buoy. The buoy is due east of the port.

a What is the bearing of the port from the boat?

b How far does the boat have to travel to reach the buoy?

c How far is it from the buoy to the port?

4 A motorist travels 8 km due north from a petrol station and then 6 km due west. What is the bearing of the motorist from the petrol station?

5 Two towns, X and Y, are 25 km apart, with Y being due north of X. Town Z is due east of

X and 60 km away. Find the distance and bearing of Y from Z.

Modelling and problem solving

6 Julio and Maria leave school at the same time. Julio cycles due west at 18 km/h and Maria cycles due south at 24 km/h.

a Calculate the distance that separates them, after they have cycled for 2 hours.

b How long did it take for them to be 15 km apart?

c Calculate the bearing from Maria to Julio at any time.

7 A triathlete leaves the start of a race and runs 12 km on a bearing of 030°T. She then cycles for 15 km on a bearing of 060°T. What are the bearing and distance of her starting point from her position at the end of the cycle leg?

8 A yacht travels on a bearing of 052°T until it is 3.2 km north of its starting point. It then travels on a bearing of 155°T until it is due east of its starting point. How far is the yacht from the place where it started?

9 An orienteer begins from a starting point, station 1, and runs on a bearing of 350°T until he reaches station 2, which is 3.3 km north of station 1. He then runs due east for 5.2 km to reach station 3. He then changes direction to run roughly south-west until he reaches station 4, which is 2.1 km due east of station 1. What is the bearing from station 3 to station 4?

10 An aircraft flies due south for 100 km and then due west for another 250 km. Find the distance and bearing of the aircraft from its starting point.

11 P and Q are two bushfire observation stations. Q is 10 km due west of P. A fire is sighted

at a bearing of 322°T from P and 052°T from Q. Calculate the distance of the fire from each observation station.

N

X

P

Y 120°

60 m

45 m

Buoy Boat N

60°

60°

Port

1.3

The sine rule

The area of any ABC may be written as:

ab sin C or bc sin A or ca sin B

So ab sin C = bc sin A = ca sin B

Multiplying by 2 gives ab sin C = bc sin A = ca sin B

Dividing by abc, we have = =

Inverting gives the sine rule:

A formal proof of the sine rule is given on the CD-ROM.

We use the sine rule to solve non-right-angled triangles when we know two angles and one side or two sides and an opposite angle.

Extra Material

Area of a

triangle ₁ 2 --- 1 2 --- 1 2 ---1 2 --- 1 2 --- 1 2 ---sin C c

--- sin A

a

--- sin B

b

---Sine rule In any ABC

= = c b A a B C a sin A --- b sin B --- c sin C

---!

Detailed Proof

In XYZ, X = 50°, Y = 34° and x = 12. Find Z, y and z.

Solution

Sketch a triangle to show the information.

Write the sine rule for x, X, y and Y. =

Substitute. =

Isolate y and evaluate. y = ≈ 8.8

Now find Z. Z + 50° + 34°= 180°

Z = 96°

Write the sine rule for x, X, z and Z. =

Substitute. =

Isolate z and evaluate. z = ≈ 15.6

Write the answer. Z = 96°, y ≈ 8.8 and z ≈ 15.6.

z y X 12 Y Z 50° 34° x sin X --- y sin Y ---12 sin 50°

--- y sin 34°

---12 sin 34° sin 50° ---x sin X --- z sin Z ---12 sin 50°

--- z sin 96°

---12 sin 96° sin 50°

The sine rule does not always give a unique answer for the angle in a triangle. Since

sin θ= sin (180° −θ), you must try two possible answers to check whether one or both answers are possible. Your calculator gives only the acute angle, so you must work out the other angle. For example, if sin θ= 0.6, your calculator will give sin−1 _0.6 ≈ _36.87°.

Now 180° − 36.87° = 143.13°. Checking, you find that sin 36.87° ≈ 0.6 and sin143.13° ≈ 0.6. If both answers are possible, we say it is an example of the ambiguous case.

In ABC, a = 6 cm, b = 7.5 cm and A = 40°. Calculate the remaining angles and side and sketch the possible triangles.

Solution

Write the sine rule for a, A and b, B. =

Substitute relevant values. ₌

Rearrange and evaluate. sinB =

= 0.8034 …

Use sin−1 _{on the calculator.}

Keep exact values in your calculator.

So B ≈ 53.46° or 126.54°

Find the possible totals of A and B. A + B ≈ 40° + 53.46° or 40° + 126.54°

Both are less than 180°, so both are possible. ≈ 93.46° or 166.54°

Now find C. C ≈ 180° − 93.46° or 180° − 166.54°

= 86.54° or 13.46° Use the sine rule to find the possible

values of side c. =

=

c =

Use the exact values in your calculator. = or

≈ 9.32 or 2.17 This is an ambiguous case.

The possible triangles are shown below.

B ≈ 53.46°, C ≈ 86.54° and c ≈ 9.32 cm or

B ≈ 126.54°, C ≈ 13.46°, and c ≈ 2.17 cm

a A

sin

--- b

B

sin

---6 40° sin

--- 7.5

B

sin

---7.5sin40° 6

---a A

sin

--- c

C

sin

---6 40° sin

--- c

C

sin

---6sinC

sin40°

---6sin86.54° 40° sin

--- 6sin13.46° 40° sin

---7.5 cm 6 cm

13.46°

40°

2.17 cm

A B

C

126.54°

7.5 cm 6 cm

86.54°

53.46° 40°

9·32 cm

A B

C

Exercise 1.3

The sine rule

1 Calculate the lengths of the unknown sides in these triangles correct to 1 decimal place. Jermaine sees that the top of a

sand dune at the end of a beach is at an angle of elevation of 16°. On walking 40 m closer, she finds that the angle of elevation increases to 22°. What is the height of the sand dune?

Solution

Make a sketch of the information. Label sides and angles as needed. The height is side TB.

Side TS is in between the known and unknown sides. Call it x and find it first.

Use the exterior angle of SFT to find

∠STF.

22° = 16° +∠STF

∠STF = 6°

Use the sine rule in SFT to find x. =

Don’t bother to calculate x. x =

Now use BST and the sine ratio to find h. sin 22° =

Isolate h. h = x sin 22°

Evaluate using the uncalculated value of x. = = 39.5129 …

Round and write the answer. The dune is about 39.5 m high.

40 m

B S F

T

h _x

22° 16°

x

16° sin

--- 40 6° sin

---40sin16° 6° sin

---h x

---40sin16° sin22° 6° sin

---Example

8

Additional Exercise

1 .3

a b c

d e f

62°

58° 14 a

b

76

115.5° 27.2°

108° 40°

8 x

y m

n c

d 43

52°

107°

32° 25

48° 18

37° q

p

h

k

2 Calculate the value of θ in each of the following triangles correct to 1 decimal place.

3 Find the length of the longest side of PQR if P =31°42′, Q =28°50′ and PR =42 mm.

4 Solve each of the following triangles. (Calculate the sizes of all unknown angles and sides, including both possibilities in ambiguous cases.)

a XYZ, where Z = 65°, y = 11 m and z = 16.2 m

b PQR, where P = 35.3°, Q = 52.8° and q = 67.5 cm

c DEF, where F = 111°, f = 12.5 km and d = 8.96 km

d ABC, where B = 124.1°, C = 18.7° and c = 94.6 cm

e XYZ, where X = 68.41°, Y = 54.23° and x = 12.75 m

f KLM, where K = 71.83°, M = 42.57° and l = 2.614 km

5 In DEF, f = 30 cm, d = 35 cm and F = 18°. Find both possible values for e and D.

6 In ABC, BC = 18 cm, AC = 15 cm and B = 38°. Find both possible values of A and make a sketch of each possible ABC.

Modelling and problem solving

7 The folding chair shown on the right has a seat which is 36 cm deep. The legs of the chair are inclined at an angle of 55° to the seat and join at an angle of 70°. The seat is 51 cm above the floor.

Find the distance from the seat, d, at which the legs are joined.

8 Three gears are arranged as shown on the right. The radii of the gears are 7.2 cm, 5.4 cm and 3.2 cm. Calculate θ.

a b c

d e f

5.8 6.29

48.3°

θ

12

16

θ θ

θ

θ 52°

8

12

23.6° 135.5°

127.6

82.9

91°

110°

θ

16

9 96 62.5

55° 55°

70° 36 cm

51 cm d

38°

3.2 cm 5.4 cm

7.2 cm

9 A helicopter is sighted at the same time by two ground observers who are 4.8 km apart, as shown in the diagram on the right.

The angles from the observers to the helicopter (measured to the horizontal) are 21.4° and 28.6°. How high is the helicopter?

28.6° 21.4°

4.8 km

10 Anne is flying from Brisbane to Dalby at a speed of 150 km/h. She thinks her altimeter, which shows a height of 15 000 feet (4600 m), is faulty. Looking ahead, she sees the top of a hill marked as 405 m on her map at an angle of depression of 13°. Two minutes later, the angle of depression has increased to 20.6°.

a Find her height above the hill and thus her altitude.

b What should the radio operator tell her about her altimeter?

1.4

The cosine rule

The sine rule can be used in triangles where an angle and its opposite side are known. However, if the known angle is included between two sides or only the sides are known, the sine rule cannot be applied. In these cases, the cosine rule is used instead.

Formal proofs of the cosine rule for acute-angled, right-angled and obtuse-angled triangles are shown on the CD-ROM.

Cosine rule In any ABC

a2= _b2+ _c2− _{2bc cos A or cos A} =

b2= _a2+ _c2− _{2ac cos B or cos B} =

c2= _a2+ _b2− _{2ab cos C or cos C} =

where the angles and sides are named in the usual way.

b2_+c2_–a2

2bc

---c b

A

a

B C

a2_+c2_–b2

2ac

---a2+b2–c2

2ab

---!

In the triangle shown at right, calculate:

a x b Y

Solution

a Write the cosine rule for XYZ. x2 = _y2+ _z2− _{2yz cos X}

Substitute values. = 2.12+ _3.92− ₂ × _2.1 × _3.9 × _{cos 50°}

Evaluate. = 9.0911 …

x = 3.0151 …

Round and write the answer. x is approximately 3.02 cm.

b You can now use the sine rule. =

Transform. sin Y =

Use the exact value of x from part a. =

= 0.5335 …

Find Y and round off. Y = 32.2°.

Write the answer in words. Y is about 32.2°.

3.9 cm 2.1 cm

X

x

Y Z

50°

x X

sin

--- y

Y

sin

---y sinX x

---2.1×sin50° 3.0151 …

---Example

9

Calculate the largest angle in the triangle with sides 12 cm, 14 cm and 20 cm.

Solution

Make a rough sketch.

In triangles, the largest angle is opposite the longest side. So the triangle can be drawn as shown.

It’s easier to use the second form of the

cosine rule. cos A =

Substitute using the information from the

diagram. =

Keep exact number on calculator. =

A is obtuse because cos A is negative. =−0.1785 …

Most calculators will give the correct angle. A ≈ 100.29°

Write the answer in words. Largest angle is about 100.29° or 100°17′.

20 cm A

14 cm 12 cm

C B

b2_+c2_–a2

2bc

---142₊₁₂2_–202

2×14×12

---60 – 336

Exercise 1.4

The cosine rule

1 Calculate the unknown side lengths in the following triangles.

2 Find θ in each of the following triangles.

A tower in a direction N52°E is found to be at a range of 4.86 km. The range finder shows that another tower in a direction 106°T is at a range of 7.96 km. Find the distance between the towers.

Solution

Draw a sketch and name the points. Name the observation point P. Name the tower positions F and S. Draw in the bearings.

The angle between the towers must be 54°. We want to find the distance between the towers, FS. Label FS as x.

Write the cosine rule for x. FS2= _PF2+ _PS2− ₂ × _PF × _PS × _{cos P}

Substitute. = 4.862+ _7.962− ₂ × _4.86 × _7.96 × _{cos 54°}

Keep exact value on calculator. = 41.5035 …

Use the exact value for the square root. FS=

Evaluate and round off. ≈ 6.44

Write the answer in words. Distance between the towers is about 6.44 km.

N

52° 54°

106°

7.96 km

4.86 km

F

S x

P

41.5035 …

Example

11

Additional Exercise

1 .4

a b c

d e f

4 cm

6 cm 61°

121° 3 m

5 m

30 m

15 m

42.25°

30 m

100 cm 55°

90 cm

150° 27 m

100° 35 cm

21 cm

20 8

16

θ 20 24

12.6

9.6 10.4 12

18

15

θ

θ θ

θ

25.6

54.2 27

21

33.1 10

θ

a b c

d e f

3 Find the third side of each of the following triangles.

a MNO, where M = 41.1°, n = 27.8 and o = 39.2

b XYZ, where Z = 28.3°, y = 571 and x = 421

c DEF, where F = 45.6°, d = 72.3 and e = 89.4

d GHI, where G = 67.3°, h = 37.9 and i = 40.8

e STU, where S = 112.8°, t = 62.8 and u = 122

f PQR, where Q = 74.8°, p = 891.9 and r = 642.7

4 Find all unknown angles and sides of each of the following triangles.

a DAG, where D = 121°, a = 3 m and g = 5 m

b UVW, where W = 55°, u = 45 cm and v = 50 cm

c XYZ, where x = 7.2 m, y = 12.5 m and z = 8.3 m

d ABC, where a = 8.5 m, b = 20 m and c = 15.2 m

e LMN, where L = 42.3°, m = 15.4 km and n = 12.9 km

f REF, where E = 168.2°, f = 192 mm and r = 151 mm

5 In ABC, a = 12, b = 9 and c = 8. Calculate the greatest angle.

6 In XYZ, x = 8.8, y = 11 and z = 17.6. Calculate the smallest angle.

7 In PQR, p = , q = 1 and r = 2. Calculate the greatest angle.

Modelling and problem solving

8 The diagram on the right shows a crane with two arms, each 15 m long. The arms of the crane are separated by an angle of 128°.

The load is attached at point X and the counterweight is attached at point Z.

Find XZ.

9 A weight is supported by cables attached to a horizontal beam as shown in the diagram on the right.

Find the angle that each cable makes with the beam.

10 A hiker encounters a swamp at point X as shown in the diagram on the right.

To travel around the swamp, the hiker walks 3.4 km from X on a bearing of 325°T and then walks 1.8 km to point Y on the other side of the swamp.

Find the bearing of X from Y. 5

X Z

15 m 128° 15 m

Y

18 m

9 m 12 m

N

325°T

3.4 km

Swamp

1.8 km

4.2 km

Y

1.5

Trig applications

When an object is above or below eye level, an observer must look up or down to see the object. The angles up and down have special names.

The angle of depression of an object is measured downwards from the horizontal.

Angles of elevation and depression

The angle of elevation of an object is the angle an observer must look upwards from the horizontal to see the object.

Line of sight

Angle of elevation

Horizontal

Line of sight Horizontal

Angle of depression

!

A satellite receiver on top of a tower is at an angle of elevation of 38° from a point M on the ground that is 72 m from the centre of the base of the tower. There is a warning sign on the tower, with its top positioned one-third of the vertical distance up the tower.

a Calculate the height of the tower.

b Calculate the angle of elevation from the point M to the top of the warning sign.

Solution

Draw a diagram showing all the relevant information. Then show just the triangles and name the vertices.

72 m

M

72 M

R

P

38°

θ

Q

Alternative Method

a From the diagram, we want to find QP.

Form an equation involving QP. tan 38° =

Isolate QP. QP = 72 tan 38°

Retain the exact answer in your calculator. = 56.2526 …

Round off for the answer. ≈ 56.3

State the answer in words. The tower is about 56.3 m high.

b First form an equation to find RP. Use the

exact answer in your calculator from part a. RP =

Retain the exact answer in your calculator. = 18.7508 …

Now we can use tan to find θ. tan θ =

Use the exact figure on your calculator. = = 0.2604 …

Use tan−1 _{to find} θ_. θ = _{14.5972 …°}

Now round off. ≈ 14.6°

Write the answer in words. The angle of elevation is about 14.6°.

QP

72

---QP

3

---RP

72

---18.7508… 72

To an observer on top of a cliff, 120 m above sea level, a boat appears to be at an angle of depression of 17°. After the boat sails directly towards the cliff, the angle of depression increases to 37°. How far did the boat sail?

Solution

Draw the triangles, name the vertices and put all the information on the diagram.

There are two right-angled triangles, WYZ and XYZ. The length we want is WX, the

difference of the bases of these triangles.

120 m 17°

120 m 17°

W

X Y

Z 37°

73°

53°

Example

13

First find WY using the tan ratio in WYZ. In WYZ, ∠WZY = 73°. tan 73° =

WY = 120 tan 73°

Keep the exact figure in your calculator memory. = 392.5023 …

Now find XY using the tan ratio in XYZ. In XYZ, ∠XZY = 53°.

tan 53° =

Use the exact figures in your calculator in the last step of the problem.

XY = 120 tan 53°

= 159.2453 … The length we want is the difference between

WY and XY.

Now WX = WY − XY

= 233.2569 …

Round and write the answer in words. The boat sailed approximately 233 m.

WY

120

---XY

120

---This diagram shows a cinema. The screen is 2.5 m high and is mounted so that the bottom of the screen is 3 m from the floor.

Eye level for the audience is approximately 120 cm.

x

1.2 m

Screen 2.5 m

3 m

θ

Work in groups to determine the distance x (to the nearest 10 cm) that will give the viewer the maximum viewing angle θ. Use a table of values of x such as that below, or a spreadsheet, to find the best distance.

Modern cinemas have a sloping floor to make it easier for viewers to see the screen. Keep the same dimensions for the screen, but change the seating so that the floor slopes up towards the back. Make another table or spreadsheet to do your investigations. Look at the changes of angle with flat and sloping floors of different angles.

x (m) 1 2 3 4 5 6 7 8

θ

Spreadsheet

Spreadsheet

Investigation

Viewing distances and angles

When solving trigonometric problems, you must decide which rules to use. The following provides some ideas to guide you in choosing the correct (or easiest) method.

When solving triangles:

■ Look for solutions using right-angled triangles first.

■ Use the sine rule where you know one angle and the opposite side. Be careful of ambiguous cases.

■ Use the cosine rule where the known angle is enclosed by two known sides or the three sides are known. After using the cosine rule once, the sine rule may be useful for other angles and sides.

■ If the known side is in a different triangle from an unknown side, the side common to the triangles should usually be found as an intermediate step.

!

Find the marked sides and angles below.

Solution

a The angles in a triangle add up to 180°.

So the third angle must be 90°. The hypotenuse is 28 cm.

a is opposite 58°.

Use sin = . sin 58° =

Rearrange to find a. a = 28 sin 58°

= 23.7453 …

Round and write answer. a is about 23.7 cm.

b You want the third side. Use the cosine

rule.

b2 = _a2+ _c2− _{2ac cos B}

Substitute. = 162+ ₁₉2− ₂ × ₁₆ × ₁₉ × _{cos 95°}

= 256 + 361 + 52.9906 … Keep the exact number on your

calculator.

= 669.9906 …

Use . b = 25.8841 …

Round and write answer. b is about 25.9 m.

a 58°

32°

28 cm

95°

16 m

19 m

b

c

32 cm

36 cm

75° 71° 69°

a b c

a 58°

32°

28 cm

opposite hypotenuse

--- a

28

c The first triangle has three things known.

c is in a second triangle.

The common side can be found.

Label this side as x.

Use the sine rule for x. =

Rearrange to find x. x =

Keep the exact answer on your calculator. = 31.5959 …

Now use the cosine rule for c. c2 = _x2+ ₃₆2− ₂ × _x × ₃₆ × _{cos 75°}

Keep exact figures on the calculator. = 1705.5151 …

c = 41.2978 …

Round and write answer. c is about 41.3 cm.

c

32 cm

36 cm

75° 71° 69°

x

x

69° sin

--- 32 71° sin

---32 sin69° 71° sin

---Two points, P and Q, are 100 m apart on level ground. A vertical tower stands somewhere between P and Q such that the angles of elevation to the top of the tower from P and Q are 60° and 35° respectively. How high is the tower?

Solution

Draw a sketch.

The angles add up to 180°.

So ∠PRQ = 85°.

You want to find RS.

It is in the right-angled triangle QRS.

You need to find RQ first.

It is in PQR as well as QRS.

Use the sine rule to find RQ in PQR. =

Rearrange to get RQ.

Don’t bother to calculate. RQ =

Now use QRS. sin 35° =

Rearrange to get RS. RS = RQ sin 35°

Put in the expression for RQ. =

Use your calculator. = 49.8629 …

Round and write answer. The tower is approximately 50 m high.

100 m S

P Q

R

85°

35° 60°

RQ

60° sin

--- 100 85° sin

---100 sin60° 85° sin

---RS RQ

---100 sin60° sin35° 85° sin

Exercise 1.5

Trig applications

1 Find the marked angles and sides in the following triangles.

2 Find the marked sides in the following.

Modelling and problem solving

3 An electricity pole is stabilised by a wire. One end of the wire is connected to the pole 11.2 m above the ground. The other end is fixed to the ground by a spike. From the point where it is attached to the pole, the angle of depression of the wire is 65°.

How far is the spike from the base of the pole?

4 Some workers are doing repairs to an electricity pole. The worker who is at ground level is standing so that his eye is 1.8 m above the ground. His feet are at the same level as the base of the electricity pole, 13.2 m away from the pole.

When he looks up, he sees the top of the helmet of his workmate in the cherry-picker at an angle of elevation of 40°.

What is the vertical distance from the base of the pole to the helmet of the worker in the cherry-picker?

Additional Exercise

1. 5

a _{2.8 m} b c d

36°

a b

22 mm 49°

14 m

c

69°

17 m

48 cm

54° 109°

d

18 mm

a b c d

22° 31° 105° a

36°

42° b

24 m

247 m 42° c

d

75°

32 cm 24°

45 m

69°

71°

65°

11.2 m

13.2 m 1.8 m

5 A building in the central business district is 85 m tall. The angle of depression from the top of this building to a fire alarm on the face of a building on the opposite side of the street is 58°20′. The street is 15 m wide. Calculate:

a the height of the fire alarm above the ground

b the angle of elevation from the base of the first building to the fire alarm.

6 A boat leaves a jetty at the base of a lighthouse that is 80 m high. It sails directly away from the lighthouse and after 5 minutes is at an angle of depression of 7.67° from an observer at the top of the lighthouse. At what speed did the boat sail?

7 An observer sees that the angle of elevation to the top of a tower, 80 m high and due south of her position, is 37°. The top of another tower, 60 m high and due west of her position, is at an angle of elevation of 28°. Calculate the horizontal distance between the towers.

8 From point P, level with the base of a vertical flagpole and 50 m from it, the angle of elevation to the top of the pole is 48°.

a Calculate the height of the pole.

b Calculate the angle of elevation from P to a point halfway up the pole.

9 From the cockpit of an aircraft cruising 1520 m above the ground, the angles of depression to the tops of two identical (same height) structures are 50° and 70° respectively. How far apart are the 20 m high structures?

10 An observer in a lighthouse, 120 m above sea level, is watching a fishing trawler move toward the lighthouse. The angle of depression of the trawler from the observer is 18°.

a How far is the trawler from the lighthouse?

b Some time later the angle of depression is measured as 28°. How far has the trawler travelled in this time?

11 While Ann is walking on a straight road running east–west from A to B, which are 10 km apart, she sees a church in the distance. The bearing from A to the church is N48°E and from

B the church is at a bearing of N67°W. Find the distance from A to the church.

12 The top of a lighthouse is at an angle of elevation of 35° from an observer some distance from a point on its base, as shown on the right. The top of the lighthouse is 38 m above the point on the base. This point appears to be at an angle of elevation of 8° from the observer. How far is the observer from the base of the lighthouse?

38 m

13 An aircraft flies on a course of 322°T for 250 km and then changes course to 244°T to an airstrip where it lands. If the airstrip is 480 km from the starting position, calculate:

a the distance travelled from the time that the course was changed until the plane landed

b the bearing from the airstrip to the starting position.

14 Two cyclists leave from the same shopping centre. The first travels along a straight road due west at 18 km/h. Thirty minutes later, the second leaves and proceeds along a straight road in a south-westerly direction at 24 km/h. How far apart are the two cyclists 30 minutes after the second departs from the shops?

15 A girl is flying a kite and has extended 250 m of string when an updraft makes the kite rise suddenly and she has to let out another 60 m of string. If the angle of elevation to the kite was 38° before the updraft, how far has the kite been driven vertically?

16 Two points, A and B, on the same bank of a river are 120 m apart. Point C is on the opposite bank. ∠CAB and ∠CBA are

measured to be 55° and 33° respectively. Find the width of the river.

17 Sophie and Will set out from point A at the same time. Sophie travels at 30 km/h along a straight road in the direction 042°T. Will travels at 24 km/h along another straight road in the direction 142°T. Find their distance apart after 4 hours.

18 From a distance, the angle of elevation to the top of a tree is 25.13°. Ten metres closer, the angle of elevation is 31.6°. How high is the tree?

38°

310 m

250 m

120 m B

C

A

55° 33°

19 A tower is erected at the base of a road that is inclined at a constant angle of 8° to the horizontal. Angles of elevation to the top of the tower are measured at two points that are 80 m apart, some distance up the slope away from the tower. The angles are found to be 25° and 8° respectively. What is the height of the tower?

20 A low-impact mobile phone tower is on top of a building. The angles of elevation of the top and bottom of the tower are 16° and 10° respectively. Thirty metres closer to the building, the angle of elevation of the bottom of the tower has increased to 17°. Find the height of the tower above the building and the building’s height above the ground.

You can use a small program on a graphics calculator to solve triangles. The program TRISOLVE can be used to find the unknown sides or angles in any triangle, given any set of three inputs (apart from Angle-Angle-Angle). The program is given in full on the CD-ROM. Enter the program (or load it from the CD-ROM) and try it with various triangles.

A right-angled triangle is entered by giving a right angle as one of the angles. The calculator must be set in degree mode.

1.6

Working in three dimensions

Many trigonometric problems can be solved using triangles in only two dimensions, such as the height and distance applications earlier in this chapter. Applications involving triangles in three

dimensions (3D) must be simplified before they can be solved with trigonometry.

P 8°

80 m

Q R

Technology

Solution of 3D problems

To solve three-dimensional problems:

A line box drawing may help to visualise the problem.

Make a 3D drawing including triangles and all relevant information. Label points on the drawing.

‘Unfold’ the diagram and draw the triangles flat. Label the unknowns and intermediate sides and angles.

Use the methods of right-angled triangles, sine and cosine rules to find the unknowns. Write the solution to the problem in an appropriate form.

!

TI Calculator

Program

From a certain point, A, a mountain peak due north has an angle of elevation of 20°. From another point, B, 2 km east of A and on the same level as A, the bearing of the peak is N40°W. Find the height of the peak above the level of A and B.

Solution

Make a sketch inside a box to make a 3D drawing.

Call the foot of the mountain M and the peak P. Show the directions of north and east.

∠MBN = 40° and ∠ABN = 90°, so ∠ABM = 50°.

Call the height of the mountain h. Call the distance from

A to the foot of the mountain x. Redraw the diagram with

the sides and angles marked, ignoring the parts of the box you don’t need.

Then unfold the diagram to make flat triangles. In this case they are joined along AM.

Use tan in MAB to find x. tan 50° =

Rearrange, but don’t calculate. x = 2 tan 50°

Now use tan in PAM to find h.

tan 20° =

Rearrange. h = x tan 20°

Substitute x. = 2 tan 50° tan 20°

Calculate the answer. = 0.8675 … km

Round and write the answer in appropriate form. The peak is about 868 m high.

2 km

A B

P

M

20°

E 40°

N

2 km

A B

P

M

20°

50° h

x

2 km A

B P

M

20°

50° h

x

h

A

2 km B

50° 20°

M P

x Unfold

x

2

---h x

In the next example, one side has to be duplicated to unfold the diagram to make it flat. In addition, the required side cannot be found directly. After applying trigonometry you must solve a simple equation to find the height.

Hassan observes that the top of a transmission tower at a bearing of 038°T is at an angle of elevation of 12°. Fatima is 575 m due east of Hassan, and she says the bearing of the tower is 295°T. Find the height of the tower.

Solution

Make a sketch inside a box. Call Hassan’s position H. Call Fatima’s position F. Show north and east.

Call the bottom of the tower W. Call the top of the tower T.

∠WHN = 38° and ∠FHN = 90°,

so ∠FHW = 52°.

The other bearing shows that ∠WFH = 25°.

∠TWH = 90°.

Unfold and draw the triangles flat. Call the height h.

Call the common side x.

The angles in WHF add up to 180°.

Thus ∠HWF = 103°.

Use the sine rule in WHF to find x. =

Rearrange but don’t calculate. x=

Now use tan in THW to find h. tan 12°=

Rearrange. h = x tan 12°

Substitute x and calculate. =

= 53.0110 …

Round and write the answer. The transmission tower is about 53 m high.

H F

T

W

575 m E

12° 38°

295° N

575 m

H F

12°

52° 25°

W T _h

x

x

25° sin

--- 575 103° sin

---575sin25° 103° sin

---h x

---575sin25° tan12° 103° sin

Maria and Jack are standing on a level road 150 m apart. A church spire is due north of Maria and on a bearing of 070°T from Jack. The top of the spire appears at an angle of elevation of 27° to Jack and 18° to Maria. Find the height of the spire above the road.

Solution

Make a sketch inside a box. Call Jack’s position J. Call Maria’s position M. Show north and east.

Call the bottom of the spire S. Call the top of the spire T. Call the height of the spire h. Call the distance of the spire from Jack x and its distance from Maria y. ∠NJS and ∠JSM are alternate, so

∠JSM = 70°.

Unfold the diagram to make it flat, showing the side TS on two of the flat triangles.

By the angle sum of a triangle, ∠JTS = 63° and ∠MTS = 72°.

Use tan in TSJ to find h. tan 63° =

Use tan in TSM to find h. tan 72° =

Now use the cosine rule in JMS. 1502 = _x2+ _y2− _{2xy cos 70°}

To proceed further you need x and y. Rearrange the previous expressions.

x = h tan 63° and y = h tan 72°

Now substitute x and y. 1502= _{(h tan63°)}2+ _(htan72°)2− _{2(htan63°)(htan72°) cos 70°}

Note: tan2θ= _(tan θ₎2. = _h2 _tan2 _63° + _h2 _tan2 _72° − _2h2 _{tan 63° tan 72° cos 70°}

Factorise. = h2_(tan2 _63° + _tan2 _72° − _{2 tan 63° tan 72° cos 70°)}

Calculate. 22 500 = h2× _{9.1921 …}

Rearrange and calculate. h2 = _{22 500 9.1921 …}

= 2447.7350 …

h = 49.4745 …

Round and write the answer. The height of the spire is about 49.5 m.

27° 70°

18° N

E J

M

S T

150 m

h

27°

18° J

M

S T

150 m

x 70°

h

T

T

M J

S h

h

y x

150 m

27° _70°

18° 72° 63°

y

x h

---y h

Exercise 1.6

Working in three dimensions

Modelling and problem solving

1 A rectangular-based pyramid with a base measuring 16 cm by 12 cm and a height of 10 cm is to be constructed using perspex sheeting.

Calculate the angle that the triangular faces AYZ and

AZW make with the horizontal base.

2 A flagpole 12 m high is temporarily erected for an Australia Day ceremony. It is held in place by two wires: one 15 m long fastened due west of the pole and the other 17 m long fastened due south of the pole. Both supporting wires are attached to the top of the flagpole. Calculate:

a the distance that each wire is from the pole at the point where it meets the ground

b the angle of inclination that each wire makes with the horizontal

c the distance between the points where the wires meet the ground

d the angle between the wires.

3 From a lighthouse 150 m above sea level, the lighthouse keeper observes a boat due east at an angle of depression of 25° and another boat due south at an angle of depression of 32°.

a Find the distance that separates the boats.

b Find the bearing of the second boat from the first.

4 An aircraft flying due east at 385 km/h at a constant height of 1500 m passes over a village at a particular instant. One minute later, a person who is 1.6 km south of the village sees the plane. In what direction and at what angle of elevation does the person see the plane?

5 A sculptor has removed a section from a copper block with dimensions as shown in the diagram on the right. Calculate:

a QR b x c y

d the angle that the triangular face PQR makes with the horizontal base.

6 A large tent is constructed in the shape of a rectangular-based pyramid ABCDE with the vertex E vertically above G, the midpoint of BC, as shown in the diagram on the right. Calculate:

a the angle that the triangular face AED makes with the base of the pyramid

b the angle between the lines DE and AD.

Additional Exercise 1 .6

16 cm 10 cm

12 cm Y

C

Z W

X

B

O A

45°

9 cm

15 cm

Q S

R

P x y

10 m 40 m

A

G C

D E

F B

a b

7 A straight stick 2 m long is stuck in the ground and is inclined at 60° to the horizontal, pointing in an easterly direction. Calculate the length and direction of the shadow of the stick when the sun is due north at an elevation of 30°. (Hint: Imagine a piece of string hanging vertically from the top of the stick to the ground. Consider the shadow of the string and its relation to the shadow of the stick.)

8 The diagram on the right shows the frame for a double-hipped roof. Calculate:

a the angle that the triangular face

RSU makes with the horizontal

b the length of the rafter VW

c the angle the sloping roof PTUS makes with the horizontal

d the length of the rafter UY

e the length of the rafter UR.

9 A hillside is considered to have a uniform slope of 22° with the horizontal as shown in the diagram on the right. A hiker, starting from point E, walks due north up the hillside for 120 m to point C, then walks due west to point B. The hiker then walks 170 m back to the starting point. Calculate the angle of inclination of BE to the horizontal and the total distance covered by the hiker.

10 Two groups of people set off to spotlight possums at night. One group spots a possum. The tree that has the possum is at a bearing of N28°W. The angle of elevation of their spotlight beam is 38°. At this instant the other group, which is 100 m west of the first group, sights the same tree at a bearing of N29°E. How high up the tree is the possum?

11 Elena sees that the angle of elevation to the top of a tower, 80 m high, which is due south of her position is 3l7°. The top of another tower, 60 m high, which is due west of her is at an angle of elevation of 28°. Calculate the horizontal distance between the two towers.

12 Miguel is 90 m due west of Consuella on a level road. He sees that the angle of elevation of a kite flown by their son is 30°, while Consuella sees that it is due north at an angle of elevation of 38°. What is the height of the kite?

6 m

12 m

8.4 m

R

S

T U

V

W

X _Y

Z

P

9 m Q

N

22° 120 m

170 m

22° A

B C

D

■ The side of a triangle is usually named by a lower-case letter, and the opposite vertex is named by the same capital letter.

■ Angles are measured in degrees, minutes and seconds. There are 360° in a full turn, 60′ in 1° and 60″ in 1′.

■ We use sin, cos and tan to solve right-angled triangles. The side opposite the right angle is called the hypotenuse. The side between the angle we want and the right angle is the

adjacent side, and the side opposite the angle we want is the opposite side.

sin A = cos A = tan A =

■ The bearing of the object is given as the 3-digit clockwise angle the observer turns from due north in order to face the object. A bearing can also be given as the angle from north or south towards either east or west.

■ The sine rule may be used to solve non-right-angled triangles where a side and an opposite angle are known.

In any ∆ABC = =

■ The cosine rule may be used to solve triangles if all three sides are known, or in cases where two sides and the enclosed angle are known.

In any ∆ABC a2= _b2+ _c2− _{2bc cos A or cos A} =

■ The angle of elevation of an object is the angle an observer must look upwards from the horizontal to see the object.

■ The angle of depression of an object is measured downwards from the horizontal.

■ When solving triangles:

– Look for solutions using right-angled triangles first.

– Use the sine rule where you know one angle and the opposite side. Be careful of

ambiguous cases.

– Use the cosine rule where the known angle is enclosed by two known sides or the three sides are known. After using the cosine rule once, the sine rule may be useful for other angles and sides.

– If the known side is in a different triangle from an unknown side, the side common to the triangles should usually be found as an intermediate step.

■ To solve three-dimensional problems:

1 A line box drawing may help to visualise the problem.

2 Make a 3D drawing including triangles and all relevant information. 3 Label points on the drawing.

4 ‘Unfold’ the diagram and draw the triangles flat. 5 Label the unknowns and intermediate sides and angles.

6 Use the methods of right-angled triangles, sine and cosine rules to find the unknowns. 7 Write the solution to the problem in an appropriate form.

opposite hypotenuse

--- adjacent hypotenuse

--- opposite adjacent

---a A

sin

--- b

B

sin

--- c

C

sin

---b2_+c2_–a2

---Knowledge and procedures

1 Write down the trigonometric ratios in terms of the sides of a right-angled triangle.

2 Find the missing sides of the following triangles.

3 Find the length of the unknown side or angle marked in each of the following triangles.

4 A lighthouse, L, is 40 km away from a harbour located at

K in the diagram on the right. The lighthouse is on a

bearing of 225°T from the harbour.

A fishing boat travels from the harbour on a bearing of 205°T until it is due south of the lighthouse.

Calculate:

a how far the fishing boat travels

b how far the fishing boat is from the lighthouse.

5 The area of the triangle shown is equal to:

A bc sin θ B ab sin θ C bc cos θ

D ab cos θ E abc tan θ

6 Solve each of the following triangles.

a ABC, ∠A = 42°, b = 36 m, a = 28 m

b FEG, ∠F = 58.2°, f = 42.5 cm, ∠E = 28.4°

c KMN, ∠M = 35°, m = 29 m, n = 19 m

7 Find the angles in a triangle with sides 8.2 m, 9.5 m and 10.6 m.

Ex 1.1

a b

15.0 m

21.3 m a

16.5 m

22.5 m

b

Ex 1.1

a b c

d e f

27.9 42°24′

29

x _α

72

m 67

θ

41

f

52°42′

64°15′ 53 47

55.2

98.1 u

Ex 1.1

N

L

M

K 205°

225°

40 km

Ex 1.2

Ex 1.3

θ

a b

c 1

2

--- 1

2

--- 1

2

---1 2

--- 1

2

---Ex 1.3

8 Find the marked angles and sides in the following triangles.

9 Find the marked side in each of the following.

Modelling and problem solving

10 A roof truss is being made to the design shown on the right. Calculate the lengths of timber required, and the lengths required to make the cuts at the correct angles as shown in the detail drawings.

11 A yacht sails on a triangular course. The first leg is 18 km NE and the second leg is 15 km at 168°T. What is the third leg?

12 A person standing at point Z on the bank of a river observes a tree at point X on the opposite bank, at a bearing of 292.17°T. The person then walks 885 m down the bank to point Y and notes the tree is now at a bearing of 285.33°T. Find the distance between the tree and the initial position, Z, of the observer.

13 A coastal surveillance vessel Interceptor detects two suspect boats on the radar screen. The first is 60 km away from the Interceptor on a bearing of 062.2°T. The second is 42 km away on a bearing of 296.58°T. How far apart are the two suspect boats?

14 The top of a hill is at an angle of elevation of 7°30′. A hundred metres closer to the hill, the angle of elevation is 8°. What is the height of the hill?

15 A surveyor notes that the top of a water tower is at a bearing of 062°T. After walking 200 m directly east she finds that the tower is at a bearing of 059°10′T. The angle of elevation of the top of the tower is then 4°31′. What is the height of the tower?

16 A table is 1 m wide, 2.5 m long and 900 mm high. The 5 cm thick top is supported on legs splayed outwards along the diagonals by 8° from the vertical. The legs meet the floor directly under the corners of the table. Find how far from the ends and sides the legs are attached.

Ex 1.4

a b c

22.3 m

37.4 m 48.5°

a

θ

38 mm

44 mm

34 mm α

β φ

103° _{54 m}

48 m

b

Ex 1.5

a b

164 cm 27° a

61° 28.1° 19.5 m

b

54.2°

57.3°

17.6 m

6500 mm

18° 90

90

y

x

90 90

w z

Ex 1.1

Ex 1.2

Ex 1.3

292.17°T 285.33°T

885 m X

Y Z

Ex 1.4

Ex 1.5

Ex 1.6