07 NCM 11 NSW SB TXT - mathsbooks.net Century Year 11/07_Trigonometry (Measurment... · TRIGONOMETRY 247 Example 2 Find the length of the hypotenuse h correct to 2 significant figures

TRIGONOMETRY 245

77TrigonometryMEASUREMENT

Read any book on the history of astronomy and you will find that the earliest discoveries and theories were made by the ancient Greeks between 600 BC and AD 200, with no major breakthroughs occurring again until Copernicus’ time in the sixteenth century. The Greek astronomer Hipparchus could calculate the distances to the Moon and the Sun as well as predict the exact lengths of the month and year. He already knew about six of the nine planets, and in 129 BC became the first astronomer to chart a star catalogue, a map showing the positions of over 850 stars. How was Hipparchus able to achieve all this more than 2000 years ago?

Hipparchus laid the foundations of a new branch of mathematics, one that uses angles, triangles and circles to calculate lengths and distances that cannot be measured physically. This new mathematics is called trigonometry, from the Greek words trigon and metron, meaning ‘triangle’ and ‘measure’ respectively. Trigonometry is used widely today to calculate immeasurable lengths and distances—in engineering, surveying, navigation, astronomy, electronics and construction.

In this chapter you will learn how to:� use Pythagoras’ theorem to calculate an unknown side in any right-angled triangle� use Pythagoras’ theorem to prove that a triangle is right-angled� use Pythagoras’ theorem to a solve a variety of real-life problems� use the sine, cosine and tangent ratios to calculate unknown sides and angles in right-

angled triangles� understand and use degrees, minutes and seconds as measures of angle size� understand and use direction bearings and angles of elevation and depression� use trigonometry to solve a variety of real-life problems.

246 NEW CENTURY MATHS GENERAL: PRELIMINARY

PYTHAGORAS’ THEOREMThe Greek mathematician Pythagoras (580–496 BC) is credited with discovering the following rule about the sides of a right-angled triangle:

c2 = a2 + b2 (hypotenuse)2 = (side)2 + (other side)2

The square of the hypotenuse is equal to the sum of the squares of the other two sides.

The hypotenuse is the longest side of a right-angledtriangle and is always opposite the right angle.

The triangle-labelling conventionNote that in the diagram above, the angles of the triangle are labelled by capital letters A, B, C, while the sides are labelled by lower case letters a, b, c. Also, the side with the lower case letter (e.g. a) is always opposite the angle with the corresponding capital letter (e.g. A). This is a convention (accepted rule or agreement) of triangle geometry.

Example 1Determine whether these triangles are right-angled.

Solution(a) The longest side is 8 cm.

c2 = 82

= 64a2 + b2 = 42 + 62

= 16 + 36= 52

c2 ≠ a2 + b2

Pythagoras’ theorem does not work.∴ This triangle is not right-angled.

The right angle is opposite the longest side.

A

C B

b

a

c

Hypotenuse

(a) (b)

8 cm

A

B

C

4 cm

6 cm

2.8 cm

10 cm9.6 cm

QP

R

(b) The longest side is 10 cm.p2 = 102

= 100q2 + r2 = 9.62 + 2.82

= 92.16 + 7.84= 100

p2 = q2 + r2

Pythagoras’ theorem works.∴ This triangle is right-angled.

10 cm

TRIGONOMETRY 247

Example 2Find the length of the hypotenuse h correct to 2 significant figures.

Solutionh2 = 3.12 + 7.82

= 9.61 + 60.84= 70.45

h = = 8.393 …≈ 8.4 m

We can tell h ≈ 8.4 m is a reasonable answer by observing the diagram, which is drawn to scale. Also, 8.4 m is the longest side but not longer than the sum of 3.1 m and 7.8 m.

Example 3Find the length of x.

Solutionx is not the hypotenuse but one of the shorter sides.

252 = x2 + 242

x2 = 252 − 242

= 625 − 576= 49

x = = 7 m

From the diagram, the answer x = 7 m looks reasonable.

1. Write Pythagoras’ theorem for these right-angled triangles.

3.1 m

7.8 m

h

70.45

25 m 24 m

x

It may be easier to remember that we subtract when finding a shorter side:

a2 = c2 − b2

49

When using Pythagoras’ theorem:1. Draw a rough diagram if one is not provided.2. Decide whether the hypotenuse or a shorter side needs to be found.3. Check whether the answer looks reasonable: compare it to the diagram.4. Make sure that the hypotenuse is the longest side, or a shorter side is not longer

than the hypotenuse.

Exercise 7-01: Pythagoras’ theorem

(a) (b) (c)

n

p

m

i

k j

cd

e


2. Determine whether each triangle is right-angled.

3. Find the value of the pronumeral in each of these triangles.

4. Find the value of the pronumeral in each of these triangles, correct to 2 decimal places.

(d) (e) (f)

x

y

z t

r

s

u v

w

(a) (b) (c)

(d) (e) (f)

20 cm

15 cm

25 cm

4 cm

5 cm 6 cm5 cm

8 cm

11 cm

2.4 m

4 m3.2 m 2.4 cm

7 cm

7.5 cm

2 m

2.9 m

2.1 m

(a) (b) (c)

(d) (e) (f)

3.2 cm

6 cm d

2.4 m

1.8 m r

x

4 cm5.8 cm35 m

37 m

w

41 cm

9 cm

a

7.2 m

3 mk

(a) (b) (c)

(d) (e) (f)

8.1 m

3.5 m

v

q10 cm

7 cm

15 cm

24 cm

p4 cm

4 cm

h

5.5 m

9.4 mk

5 m

2 mb

TRIGONOMETRY 249

5. Find the value of the pronumeral in each of these figures, correct to 3 significant figures.

APPLICATIONS OF PYTHAGORAS’ THEOREMExample 4A boat sailed 30 nautical miles due south, then 25 nautical miles due east. How far is it from its starting point, correct to 2 decimal places?

SolutionLet d be the boat’s distance from its starting point.

d2 = 302 + 252

= 900 + 625= 1525

d == 39.0512 …≈ 39.05 nautical miles

Example 5A surveyor measured the following offsets on this block of land. Calculate the perimeter of the block to the nearest 0.1 m. (All dimensions are in metres.)

SolutionBy Pythagoras’ theorem:

AB2 = 72 + 332

= 1138 AB = 33.7342 … m

(a) (b) (c)

(d) (e) (f)

11 cm7 cm

10 cm

z

140 cm61 cm

80 cm c

46 m

85 m

96 m

w

10 m15 m 24 m

a

16 cm

87 cm

63 cm

y

18 cm

48 cm

t

N

W E

S

30

25

d

1525

A

D

E

B

C

18

7

25

20

13

38

33

25 + 20 = 45

38 − 33 = 5

B

CDo not round off in the middle of a calculation. Round off at the end.

BC2 = 452 + 52

= 2050BC = 45.2769 … m


CD2 = 382 + 132 = 1613

CD = 40.1621 … m

EA2 = 322 + 182

= 1348EA = 36.7151 … m

Perimeter = AB + BC + CD + DE + EA= 193.4781 … ≈ 193.5 m

1. A gate has dimensions 3.8 m by 1.8 m. What is the length of its diagonal brace (to the nearest centimetre)?

2. Vanessa calculated the distance across the lake by taking the measurements shown. What was the distance (to the nearest metre)?

3. TV screen sizes are described by the lengths of their diagonals. Calculate the diagonals of the following screens to the nearest centimetre.(a) small: 28 cm by 20 cm(b) medium: 41 cm by 30 cm(c) large: 53 cm by 40 cm

4. A 6 m ladder leans against a house so that its base is 2 m out from the bottom of the house. How far up the house does the ladder reach (to the nearest centimetre)?

5. A yacht leaves Newcastle and sails 160 nautical miles due north. It turns and sails due east until it is directly 200 nautical miles from Newcastle. How far east did it sail?

6. A chairlift takes skiers up to the top of a hill 270 m high and 1050 m away. How long is its cable (correct to 2 significant figures)?

DE2 = 332 + 182

= 1413DE = 37.5898 … m

Exercise 7-02: Applications of Pythagoras’ theorem

3.8 m

1.8 m

482 m140 m

d

d

h

2 m

6 m

1050 m

270 m

TRIGONOMETRY 251

7. A playground slide is made up of two right triangles. Find, correct to the nearest centimetre:(a) h, the height of the slide(b) l, the length of the slide

8. A passenger ramp of length 3.6 m connects a pier to a boat. When the horizontal distance between the pier and the boat is 2.9 m, how much higher is the boat than the pier (correct to 1 decimal place)?

9. Nicole is moving house and wants to pack a metre-long umbrella into a cube-shaped cardboard box of length 60 cm. Can she pack the umbrella along the length:(a) EH? (b) EG?(c) AH? (d) AG?

10. This diagram shows a boy flying a kite. How high is the kite above the ground (correct to 1 decimal place)?

11. ABCD is a rhombus with diagonals of length 12 cm and 18 cm. Calculate the perimeter of ABCD correct to 2 decimal places.

12. Jackie wants to use an old tennis-ball can as a pencil case. If this can has a diameter of 7.5 cm and a height of 20 cm, what is the length of the longest pencil that will fit inside the can (to the nearest millimetre)?

5 m1 m

3.5 mh

l

Pier

2.9 m

h

3.6 m

E H

A

B C

GF

D

1.1 m

17 m

25 m

h

D B

A

C

20 cm

7.5 cm


13. Find the pitch line p of the roof of this house (to the nearest centimetre).

14. Calculate the perimeter of this field diagram correct to 2 decimal places. All measurements are in metres.

15. Mount Everest, the highest mountain in the world, is 8.86 km above sea level.(a) If the Earth has a radius of 6400 km, what is the

distance d (to the nearest kilometre) to the visible horizon from the top of the mountain?

(b) The formula d = 8

used in Chapter 1 also gives the distance to the horizon in kilometres, where h is the height in metres. Use this formula to calculate the distance d that can be seen from the top of Mount Everest (to the nearest kilometre) and compare it with your answer from part (a).

16. Calculate the perimeters of these figures to the nearest 0.1 m.

17. Use a ruler to construct a right-angled triangle with:(a) two shorter sides 2.5 cm and 6 cm, and measure the length of the hypotenuse

(b) a hypotenuse of length cm

18. If a tentpole is 2 m high and the rope is 2.4 m long, how far from the base of the pole should the rope be pegged (correct to 2 significant figures)?

p

7.8 m

2.7 m 2.7 m3.4 m

7

3

4

10

19

29

22

6400 km

6400

km

8.86 k

m

d

h5---

(a) (b)20 cm

15 cm12 cm 70 m

50 m

20

TRIGONOMETRY 253

19. A wire supporting a post is 17 m in length. It is attached to the ground 8 m from the base of the post. Find how far up the post the wire reaches.

20. When opened out, the top of a stepladder is 2.5 m above the ground and its feet are 0.9 m apart. How long is the stepladder (correct to 2 decimal places)?

21. The length of a TV screen is 1.3 times its height. If a TV screen has a diagonal of length 51 cm, what are its dimensions (to the nearest centimetre)?

22. A small fruit-juice Tetrapak has dimensions 64 mm × 39 mm × 105 mm. Show that the length of the longest straw that can fit into the pack is approximately 129 mm.

INVESTIGATING THE TANGENT RATIOExample 6Grant found the height of a flagpole by making some measurements on the ground. He walked out 25 m from the base of the pole and used a device called a clinometer to measure the angle (of elevation) to the top of the pole. He made a scale drawing of his measurements.

(a) What scale did Grant use for his diagram?(b) Can you calculate or estimate the height of the flagpole from the diagram?

Solution(a) 1 cm : 5 m or 1 : 500 (b) 21 m

8 m

17 m

2.5 m

0.9 m

64 mm

105 mm

39 mm

25 m

40°


Trigonometry is the mathematics of using angles and triangles to calculate lengths and distances that are either difficult or impossible to measure. It uses the three sides of a right-angled triangle, each of which has a special name related to a particular angle in the triangle:� opposite side: the side directly facing the angle, not joined to the angle;� adjacent side: the side leading to the right angle (‘adjacent’ means ‘next to’);� hypotenuse: the longest side (this has already been introduced with Pythagoras’

theorem).

The diagram shows the three sides related to angle θ in the triangle.

In trigonometry, three ratios are used: sine, cosine and tangent. We will first investigate the tangent ratio, abbreviated tan.

The tangent of an angle is the ratio of the length of the opposite side to the length of the adjacent side.

Example 7For ΔXYZ, write:(a) tan ∠Y(b) tan ∠Z

Solution(a) tan ∠Y = = (b) tan ∠Z = =

Note that the opposite and adjacent sides depend on the angle being referred to. For example, YX = 9 is the opposite side for ∠Z but the adjacent side for ∠Y. However, the hypotenuse(YZ = 41) is always the same (longest) side and does not depend on the angle.

Example 8Calculate tan θ as a decimal correct to 4 decimal places.

Solutiontan θ =

=

= 6.461 53 … ≈ 6.4615

θ

Hypotenuse

Opposite side

Adjacent side

tan θ = oppositeadjacent--------------------

θ

Opposite

Adjacent

X Z

41

40

9

Y

oppositeadjacent-------------------- 40

9------ opposite

adjacent-------------------- 9

40------

P

R

13

85

84θ

Q

oppositeadjacent--------------------

8413------

TRIGONOMETRY 255

For similar triangles, the ratio is the same for each matching angle. That is, for any

angle size, the tangent ratio is a constant, regardless of the size of the triangle.

Tangent values can also be found using a scientific calculator with a key.

Example 9Use a calculator to find tan 35° correct to 4 decimal places.

SolutionDisplay

35

So tan 35° ≈ 0.7002

Tangent values can be expressed as fractions or decimals.

1. Write tan θ as a fraction for each of these triangles.

2. Calculate tan θ correct to 4 decimal places for each of these triangles.


tan θ = = ab--- c

d---θ

θ

c

d

ab

tan

Make sure your calculator is in degrees mode (DEG) or your answer will be incorrect.tan 0.700207…

Exercise 7-03: Investigating the tangent ratio

(a) (b) (c)

(d) (e) (f)

θp

mn

a

b

c

θ

4.8

25.2

θ

θ

t

r

s

2.8

4.5

5.3

θ

θ3

45

(a) (b) (c)

(d) (e) (f)

θ

15 cm

7 cm

θ

3.1 cm

6.9 cm

4 m

9.6 m

10.4 m

θ

42 cm56 cm

70 cm

θ

5 m

2 m

θ 6 m

2.5 m

θ

6.5 m


3. These right-angled triangles each have an angle of 30°, so they are similar triangles. For each triangle, find the value of tan 30° correct to 4 decimal places by measuring the length of the side opposite 30° and dividing it by the length of the adjacent side.

4. These right-angled triangles each have an angle of 55°. Calculate tan 55° to 4 decimal places for each triangle by dividing the length of the opposite side by the adjacent side.

(a) (b)

(c) (d)

30°

4 cm

opp

30°

6 cm

opp

30°

7 cm

opp

30°

9 cm

opp

(a) (b) (c)

55°

3 cm

opp

55°

5 cm

opp

55°

6 cm

opp

TRIGONOMETRY 257

5. Draw three right-angled triangles with an angle of 42° and use them to approximate tan 42° correct to 4 decimal places.

6. Use your calculator to check these values:(a) tan 30° ≈ 0.5774 (b) tan 55° ≈ 1.4281 (c) tan 42° ≈ 0.9004

7. Use your calculator to find (correct to 4 decimal places):(a) tan 80° (b) tan 22° (c) tan 7°(d) tan 76.5° (e) tan 12° (f) tan 46°(g) tan 33° (h) tan 8.9° (i) tan 63.02°

1. What should tan 45° be? Don’t use the calculator.

2. Draw a right-angled triangle with an angle of 45°. What’s so special about it?

USING TAN TO FIND A MISSING SIDEDegrees, minutes and secondsAngle size is measured in degrees (°). However, 1 degree is divided into 60 minutes (60′) and 1 minute into 60 seconds (60″).

For example, ‘22 degrees and 45 minutes’ is written 22°45′ and ‘50 degrees, 8 minutes and 49 seconds’ is written 50°8′49″. To enter angle sizes involving degrees and minutes into the

calculator, we use the or degrees-minutes-seconds key.

Example 10Calculate correct to 4 decimal places:(a) tan 37°30′ (b) tan 64°22′

SolutionDisplay

(a) 37 30 or

For calculators with direct algebraic logic (DAL), enter first:

37 30

So tan 37°30′ ≈ 0.7673.

(b) 64 22 or

So tan 64°22′ ≈ 2.0840.

Think: What should tan 45° be?

1° = 60′ (1 degree = 60 minutes)1′ = 60″ (1 minute = 60 seconds)

° ′ ″ DMS

° ′ ″ ° ′ ″ 37°30′00″ 37.5

tan 0.767326…

tan

tan ° ′ ″ ° ′ ″ =

° ′ ″ ° ′ ″ 64°22′00″ 64.366666 …

tan 2.084048…


Finding a missing sideIn the previous section, we discovered that for a given angle (say 42°), the tangent ratio

is a constant value (tan 42° ≈ 0.9004), regardless of the size of the triangle.

We can use this fact to calculate unknown lengths and distances involving right-angled triangles. We return to Grant’s flagpole problem in Example 6 (page 253).

Example 11Find h, the height of the flagpole, correct to 1 decimal place.

Solutiontan 40° =

tan 40° × 25 = × 25 Multiplying both sides by 25

25 tan 40° = h

h = 25 tan 40° Enter 25 40 .

= 20.9774 …≈ 21.0 m

From the diagram, h ≈ 21.0 m seems a reasonable answer.

Example 12Find x correct to 4 significant figures.

Solutiontan 59°27′ =

x = 7 tan 59°27′ Enter 7 59 27 .= 11.8599 …≈ 11.86 m

Again from the diagram, x ≈ 11.86 m seems a reasonable answer.

Do you notice that when the angle is less than 45° the opposite side is shorter than the adjacent side, but when the angle is greater than 45° the opposite side is taller?

Why do you think?


25 m

40°

h

h25------

h25------

× tan =

x

7 m

59°27′

x7---

× ° ′ ″ ° ′ ″ tan =

Think: The over-45s are taller

TRIGONOMETRY 259

Example 13Find k correct to 2 decimal places.

Solutiontan 36° =

Note that the missing side k appears in the denominator (bottom) of the equation.

tan 36° × k = × k Multiplying both sides by k

k tan 36° = 15

k =

= 20.6457…≈ 20.65 m

36°k

15 m15k

------

If the missing side appears in the denominator, we divide by tan to find the answer:

tan 36° = → k = 15k

------ 15tan 36°-----------------

15k

------

15tan 36°-----------------

Just for the record

WHY 360 DEGREES? WHY 60 MINUTES?In modern times, we are accustomed to working with a base 10 system of numerals, measurement and currency. There are 100 centimetres in a metre, 1000 grams in a kilogram and 100 cents in a dollar. So why are there 90° in a right angle, 360° in a revolution, 60 minutes in an hour and 60 seconds in a minute?

In 2000 BC, the Babylonians used a base 60 or sexagesimal system of numeration. Why did the the Babylonians choose 60?� It is a rounder, more convenient number as it has more factors than 10. For example,

you cannot divide 10 evenly by 3 or 4, but 60 is divisible by 2, 3, 4, 5, 6, 10, 15, 20and 30.

� 6 × 60 = 360, which was the Babylonian approximation of the number of days in a year. They defined a revolution as being 360°, so that each day the Earth would travel 1° around the Sun. A right angle, being a quarter-revolution, thus became 360 ÷ 4 = 90°.

It is amazing that the Babylonian influence on our measurement system has lasted over 4000 years. Our units for measuring time still follow a base 60 system:

1 minute = 60 seconds1 hour = 60 minutes1 day = 24 hours

1 month ≈ 30 days1 year = 12 months

Many imperial units of measurement and currency also used multiples of 6.

1. Find out how many:(a) inches in a foot (b) feet in a yard(a) (c) feet in a fathom(d) in a dozen (e) in a gross (f) pence in a shilling

2. Why did students at primary school once learn their ‘times tables’ up to 12 × 12?


1. Evaluate these expressions correct to 2 decimal places.(a) tan 25°7′ (b) tan 71°36′ (c) tan 48°49′(d) 7 tan 28° (e) 6 tan 4°20′ (f) 18 tan 67°30′

(g) (h) (i)

(j) 10 tan 45° (k) 3.7 tan 18°42′ (l)

2. Find the values of the pronumerals in these triangles correct to 2 decimal places.

3. A triangle PQR has ∠P = 90°, ∠R = 65° and q = 14 cm. Find the length of side r correct to 2 decimal places.

4. The angle of elevation of a chimney from a distance of 64 m from its base is 56°. Calculate the height of the chimney to the nearest metre.

Exercise 7-04: Using tan to find a missing side

7tan 63°----------------- 4.6

tan 32°----------------- 13

tan 28°17′-------------------------

11.2tan 77°6′----------------------

(a) (b) (c)

(d) (e) (f)

31º4.5 m

f

49°18 cm

d

x

9 m12.5°

13 m

k

24º54°12′

8 m

p63.4°

r

17 cm

28°31′y

22 m

66.7°45 mm

b

8.8 m

h58°

70°59′3 m

t

q

7.2 m

15°46′ 55.3°

38 mm

c

(g) (h) (i)

(j) (k) (l)

14 cm

65°P

Q

R

64 m

56°

h

TRIGONOMETRY 261

5. Find the width w of the river correct to 1 decimal place.

6. From a point 45 m from the base of a tree, Julie observed an angle of elevation of 32°. Calculate the height of the tree correct to 1 decimal place.

7. A bungy jumper leaps off at an angle of 7° to the vertical. If he is swinging 16.3 m off-centre, what vertical distance has he dropped (correct to 1 decimal place)?

8. The angle of elevation to the top of an office block from a point 100 m from its base is 73°. Find its height to the nearest metre.

9. From the top of a cliff the angle of depression of a yacht 790 m out to sea is 13°. Find the height of the cliff to the nearest metre.

20 m

65°

w

45 m

32°

h

7°

d

16.3 m

Horizontal means flat, across. Vertical means up, down.

Horizon Vertical blinds

100 m

73°

h

790 m

13°

h


10. What is the distance between the netballer and the post if her shooting angle is 10°?

11. A stairwell is inclined at 34° to the ground floor and has a horizontal length of 4.5 m. Ronnie wants to place a bookcase of length 1.45 m underneath the stairwell. What is the height h of the tallest bookcase that can fit under the stairwell? Write your answer correct to 2 decimal places.

Equipment: clinometer, trundle wheel, metre ruler or height chart, calculator.

Choose a tall object outside such as a tree, flagpole, football goalpost, tower or top of a building. We will use a clinometer and trundle wheel to measure its height.

1. Copy the following table.

Distance, d Angle of elevation, θ Height, h = d tan θ Revised height, h + x

10 m

15 m

20 m

30 m

50 m

10°2.25 m

1.85 m

d

1.45 m

34°

h

4.5 m

Modelling activity: Using a clinometer to measure heights

TRIGONOMETRY 263

2. Use the trundle wheel to measure 10 m from the base of the object and stand at that point.

3. Use the clinometer to view and measure the angle of elevation θ to the top of the object. Write this value under Angle of elevation in the table.

4. tan θ = , so h = d tan θ. Find h correct to

2 decimal places and write it under Height.

5. Note from the diagram that h is not exactly the height of the object. The height of the observer must be taken into account, as the angle of elevation was measured at eye level. Measure x, the height of the observer’s eye (above the ground) in metres correct to 2 decimal places.

6. Calculate x + h and write the answer under Revised height in the table. This is the height of the object.

7. Repeat this experiment for other values of d, at different distances from the base of your object, using the values given in the table or some of your own.

8. How do your calculated results compare with each other? They are meant to be the same, but why might the answers be different?

9. How could we come up with a single accurate value for the height of the object, using these results?

Use the clinometer to measure and calculate the heights of other objects around your school.

USING TAN TO FIND A MISSING ANGLEThis table shows the values of tangent ratio for angles from 0° to 90° in intervals of 5°.

Before calculators were invented, the values for tan θ were printed for all angles from 0°0′ to 89°59′ in tables like the one above. Now, suppose we are given the value for tan θ. Is it possible to work backwards to find the matching angle size θ?

Fortunately, the calculator has a ‘reverse’ feature called the inverse tan or tan−1 function for finding an angle after its tangent ratio is entered. This function is activated by keying

or . For example, if tan 74° ≈ 3.4874, then entering 3.4874

and pressing will give the original angle 74°.

θ tan θ θ tan θ θ tan θ

0° 0 35° 0.7002 70° 2.7475

5° 0.0875 40° 0.8391 75° 3.7321

10° 0.1763 45° 1 80° 5.6713

15° 0.2679 50° 1.1918 85° 11.4301

20° 0.3640 55° 1.4281 90° undefined

25° 0.4663 60° 1.7321

30° 0.5774 65° 2.1445

d = 10 m

x x

h

θhd---

SHIFT tan INV tan

SHIFT tan


Example 14Find θ to the nearest degree if:(a) tan θ = 2.6051 (b) tan θ =

Solution Display

(a) 2.6051

(For calculators with DAL: 2.6051 .)So θ ≈ 69°.Check: tan 69° ≈ 2.6051

(b) 4 7

(For calculators with DAL: 4 7 .)So θ ≈ 30°.Check: tan 30° ≈ 0.5774, ≈ 0.5714

Rounding an angle to the nearest degree or minuteAs there are 60 minutes in a degree and 60 seconds in a minute, when rounding off angles to the nearest degree or minute, we use 30 as the halfway mark.

Example 15Write 36°22′49″ correct to:(a) the nearest degree (b) the nearest minute

Solution(a) 36° 49″ ≈ 36° to the nearest degree because 22 minutes is less than 30 minutes.

(b) 36°22′ ≈ 36°23′ to the nearest minute because 49 seconds is more than 30 seconds.

Example 16Find B to the nearest minute if tan B = 1.6732.

Solution Display

(a) 1.6732

To convert 59.135 047 …° to degrees, minutes, seconds, use the degrees–minutes–seconds

key: or

This means θ ≈ 59°8′6.17″, which rounded to the nearest minute is θ ≈ 59°8′.Check: tan 59°8′ ≈ 1.6731 ≈ 1.6732

Example 17Find the size of angle α to the nearest minute.

Solution tan α =

α = 16.6992 …°≈ 16°41′57.28″ ≈ 16°42′

From the diagram, α = 16°42′ seems a reasonable answer.

47---

SHIFT tan 69.000084…

SHIFT tan =

ab/c SHIFT tan 29.74488131

SHIFT tan ab/c =

47---

22′49″

SHIFT tan 59.135047…

SHIFT ° ′ ″ SHIFT DMS 59°8′6.17″

6.17 seconds is less than 30 seconds, so we round down to 59°8′.

3 m

10 m

α3

10------

TRIGONOMETRY 265

1. Round these angles to the nearest minute.(a) 37°45′21″ (b) 18°29′7″ (c) 50°41′52″(d) 63°8′45″ (e) 24°39′33″ (f) 28°5′48″

2. Find θ to the nearest degree.(a) tan θ = 0.7508 (b) tan θ = (c) tan θ = 8.1962

(d) tan θ = (e) tan θ = 0.4129 (f) tan θ =

3. Find A to the nearest minute.(a) tan A = 3.2692 (b) tan A = (c) tan A = 0.8

(d) tan A = (e) tan A = 2.3718 (f) tan A = 5

4. For each of these triangles, find θ to the nearest degree.

5. For each of these triangles, find φ to the nearest minute.

6. ΔABC has ∠B = 90°, a = 6 cm and c = 4.5 cm. Find ∠A to the nearest degree.

7. A playground slide has a rise (vertical height) of 2.2 m and a run (horizontal length) of 4.8 m. What angle (to the nearest degree) does it make with the ground?

Exercise 7-05: Using tan to find a missing angle

513------

14--- 11

3------

1.73.5-------

92---

(a) (b) (c)

(d) (e) (f)

68 cm

40 cm

3 m

8 m

θ 21 cm

27 cmθ

15 cm

15 cm

θ 3.8 m

11.1 mθ

2.6 m

7.2 m

θ

θ

(a) (b) (c)

(d) (e) (f)

12 m

13 mφ

3.1 cm

15.7 cm

φ

φ14 mm

63 mm

9 m

10 m

φ

φ45 mm

120 mm φ

16.2 m

27.6 m

4.8 m

2.2 m


8. What angle do the Sun’s rays make to the horizontal when a tower 105 m tall has a shadow of 4.2 m? Answer to the nearest minute.

9. Find the angle of inclination θ of this ramp in a multistorey car park, to the nearest degree.

10. A hill has a gradient of , which means that it rises 2 m

for every 15 m horizontally. Find the angle the hill makes to the horizontal, to the nearest minute.

11. Lisa stands near a flagpole of height 4.5 m. The flagpole has a shadow of length 8 m while Lisa’s shadow is 3 m. Find:(a) θ, the angle of elevation of the Sun,

to the nearest minute(b) Lisa’s height to the nearest

centimetre using:(i) the value of θ found in part (a)

(ii) similar triangles

12. When opened out, the legs of a stepladder are 0.9 m apart and its top is 2.5 m above the ground. What is the angle between the legs (to the nearest degree)?

THE SINE RATIOThe sine ratio, abbreviated sin, involves the opposite side and hypotenuse of a right-angled triangle.

The sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse.

105 m

4.2 m

θ

15 m

3.2 m

θ

α2 m

15 m

215------

3 m8 m

θ

4.5 m

h

sin θ = oppositehypotenuse---------------------------

θ

OppositeHypotenuse

TRIGONOMETRY 267

Example 18Find x correct to 2 decimal places.

Solutionsin 41° =

=

x = 11 sin 41°= 7.2166 …≈ 7.22 m

Example 19 (a) Write sin φ as a fraction.(b) Hence find φ to the nearest minute.

Solution

(a) sin φ = =

(b) φ = 67.3801 …° Enter 12 13 .

≈ 67°23′ Enter or

1. Evaluate these expressions correct to 2 decimal places.

(a) 16 sin 48.8° (b) (c) 7 sin 25°44′

2. Find β to the nearest degree if:

(a) sin β = (b) sin β = 0.9815 (c) sin β =

3. Write an expression for sin θ for each of these triangles.

41°

11 mx

oppositehypotenuse---------------------------

x11------

5 m

12 m 13 m

φ

Use to get inverse sine (or sin−1) on your calculator.

SHIFT sin opposite

hypotenuse--------------------------- 12

13------

ab/c SHIFT sin

SHIFT ° ′ ˝ SHIFT DMS

Exercise 7-06: The sine ratio

25sin 10°----------------

819------ 5

10------

(a) (b) (c)

(d) (e) (f)

k

pm

θ

6

3.26.8

θ

t

wy

θ

24

25

7θ

d

e

f

θ 10

24

26

θ


4. Find the value of the pronumeral in each of these triangles correct to 2 decimal places.

5. Find θ in these triangles, to the nearest degree.

6. Find α in these triangles, to the nearest minute.

7. A 5 m ladder leaning against a wall makes an angle of 74° with the ground. How far does the ladder reach up the wall? Answer to the nearest centimetre.

8. In ΔHIJ, ∠J = 90°, ∠I = 26° and HJ = 5.1 m. Find the length of HI correct to 2 significant figures.

9. The vertical distance between the top and bottom of a ramp is 1.7 m. If the ramp is inclined at 12° with the ground, find the length of the ramp to the nearest centimetre.

10. A plane reached a height of 171 m after flying 500 m in a straight line. At what angle to the horizontal was the plane flying (to the nearest degree)?

11. Caitlin is on a swing of length 1.6 m. When she is 0.7 m away from the swing’s frame, what angle does the swing make to the vertical? Answer to the nearest degree.

(a) (b) (c)

(d) (e) (f)

h

14 m

63°11 m

50.6°

u 34 mmx

27°

p 25 cm

62°50′

24°3′

a3 m 45°

d

6.8 m

(a) (b) (c)8 cm

12 cm

θ

10.9 m

6.1 m

θ

58 mm

92 mm

θ

(a) (b) (c)

16 cm

39 cm

α 7 m15 m

α

α

29 cm42 cm

0.7 m

1.6 m

θ

TRIGONOMETRY 269

12. The pitch of a roof is 28°. If the peak of the roof is 5.4 m from the gutter, how much higher is the peak than the gutter?

13. The distance between Mars and Earth is 78 341 200 km. If the angle between Mars’ centre and surface, measured from Earth, is 0.0025°, calculate r, Mars’ radius, to the nearest kilometre.

Equipment: metre ruler, tape measure, blackboard protractor, calculator.

Safety experts claim that a ladder leaning against a wall should make an angle with the ground no smaller than 60° and no greater than 80°.

1. What do you think would happen if this angle dropped below 60°?

2. What do you think would happen if this angle increased above 80°?

3. Suppose you had a ladder of length 4 m. Calculate (correct to 1 decimal place):(a) the highest distance the ladder could reach up the wall safely(b) the lowest distance the ladder could reach up the wall safely

4. Use a metre ruler to act as a model for a ladder. Find a wall against which to lean theruler at different angles to the ground. Copy the following table:

5. Using a blackboard protractor placed at ground level, lean the metre ruler against the wall at an angle of inclination of 10° and measure the height h it reaches up the wall, to the nearest 0.01 m. Write your answer in the table.

θ Sin θ Height, h Safe? (Yes/No)

10°

20°

30°

40°

50°

60°

70°

80°

28°

5.4 mh

78 341 200 km0.0025°

Earth

Mars

r

Modelling activity: The safe angle of a ladder

h

θ

h80

20

40

60100

120

140

160

1 m


6. sin θ = , so h = sin θ. Calculate the value of sin θ correct to 4 decimal places and write

it in the table. How does it compare with your measured value for h? Are the values thesame?

7. Judge whether a ladder leaning at this angle would be safe and write Yes or No in theSafe? column. How could you judge whether a ladder would be safe at this angle?

8. Repeat steps 5–7 for the other angles of inclination and complete the table.

9. Do your results agree with the claimed safety angle range of 60° to 80°?

hl---

Just for the record

DEGREES, RADIANS OR GRADS?When measuring the size of angles, we use degrees (°), so it is important to set your calculator in degrees mode (DEG) when performing trigonometric calculations; otherwise your answers will be incorrect. However, calculators also recognise two other modes for measuring angle size: radians (RAD) and grads (GRAD).

Radian measureRadian measure uses the circumference of a circle (of radius 1 unit) to measure angle size. Because there are 360° in a circle and the circumference of a circle is C = 2πr,

360° = 2 × π × 1= 2π radians (≈ 6.283 radians)

For smaller angles, radian measure is the length of the arc opposite the angle, a fraction of the circumference.

180° = π radians (≈ 3.141)

90° = radians (≈ 1.571)

60° = radians (≈ 1.047)

Grad measureGrads are easier to explain. We are used to working with 90° in a right angle, but some European cartographers (mapmakers) prefer to use a base 10 system and divide a right angle into 100 equal parts. So the grad was invented.

1 right angle = 100 grads1 revolution = 400 grads

1 degree = 1 grads

1. Convert 45° to: (a) radians (in terms of π)(b) radians (correct to 2 decimal places)(c) grads

2. Investigate the radians and grads modes on your calculator.

360°

1 unit

2π units

1 unit

π3 units

60°π2---

π3---

19---

TRIGONOMETRY 271

THE COSINE RATIOThe cosine ratio, abbreviated cos, involves the adjacent side and hypotenuse of a right-angled triangle.

The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.

Example 20(a) Write cos φ as a fraction.(b) Hence find φ to the nearest minute.

Solution(a) cos φ = =

(b) φ = 67.3801 …° Enter 5 13 .

≈ 67°23′ Enter or

Note that these are the same triangle and angle as in Example 19 on page 267.

Study tips

WHEN TO STUDY

Are you a morning or a night person? Different people concentrate better at different times of day. When do you do your best work? In the morning, in the afternoon, after dinner, or late at night? Identify your peak performance period.

You should aim to work around the same time each day so that studying becomes a habit. There will be times when you can’t study due to commitments such as sport, work, chores or going out, but it’s important to decide when to study, so that you can develop a regular routine and your family can know when you should not be interrupted.

Considerations are:� How much study should you do each week?� Will you work more heavily during the week or on weekends?� When do you feel like studying? When don’t you?� Do you want to set one night or day to be study-free?

Students often ask ‘How many hours should I study each night?’, but the quality of study time is more important than the quantity. It is a question of the effort and commitment required, not the number of hours. The amount of time needed for study is the amount of time necessary for you to fulfil all of your study tasks and demands. It’s up to you!

cos θ = adjacenthypotenuse---------------------------

θAdjacent

Hypotenuse

5 m

12 m 13 m

φ

adjacenthypotenuse--------------------------- 5

13------

Use to get inverse cos

(or cos−1) on your calculator.SHIFT cos

ab/c SHIFT cos

SHIFT ° ′ ˝ SHIFT DMS


Example 21Find w correct to 2 significant figures.

Solutioncos 52°17′ = =

w =

= 10.6251 …≈ 11 m

1. Evaluate these expressions correct to 2 decimal places.

(a) (b) 12 cos 72°6′ (c)

2. Find φ to the nearest degree if:

(a) cos φ = (b) cos φ = 0.3368 (c) cos φ =

3. Write an expression for cos θ for each of these triangles.

4. Calculate the values of the pronumerals in these triangles correct to 2 decimal places.

52°17′

w

6.5 m

Remember: When the pronumeral is in the denominator, we divide to find the answer.

adjacenthypotenuse--------------------------- 6.5

w-------

6.5cos 52° 17′---------------------------

Exercise 7-07: The cosine ratio

8cos 31.5°---------------------- 11

cos 12° 49′---------------------------

6.520------- 3

6---

(a) (b) (c)

(d) (e) (f)

k

pm

θ

6

3.26.8

θ

t

wy

θ

24

25

7θ

d

e

f

θ 10

24

26

θ

(a) (b) (c)

(d) (e) (f)

11 m

50°

u

16.5°

48 mm

t

29°16′

17 m

k

4 m

58.4°

d

5 cm

a

63°

21 cmy

41°20′

TRIGONOMETRY 273

5. Find θ in these triangles, to the nearest degree.

6. Find α in these triangles, to the nearest minute.

7. The Leaning Tower of Pisa is 55 m tall but leans at 6° from the vertical. How high is its top above the ground, to the nearest centimetre?

8. An escalator ramp has a length of 32 m and covers a horizontal distance of 29 m. What is its angle of inclination θ (to the nearest minute)?

9. A hot-air balloon is anchored by a 50 m towline but the wind pushes it 29 m off-centre.(a) How high is the balloon above the ground (to

the nearest metre)?(b) What angle θ does the towline make with the

ground (to the nearest degree)?

10. In ΔMNP, ∠N = 90°, m = 75 mm and n = 83 mm. Find ∠P to the nearest minute.

11. A wheelchair ramp is inclined at 18° to the horizontal. How long is it if it links two levels 8.7 m apart?

12. A rectangular field of width 38 m has a diagonal path that makes an angle of 40° with that side. Find the length of the path to the nearest 0.1 m.

13. A pedestrian skyway linking two buildings is 35 m long and inclined at 14° to the horizontal. How far apart are the two buildings? Answer correct to 1 decimal place.

(a) (b) (c)

8 m18 m

θ35 cm

40 cm

θ

θ19 m

2 m

(a) (b) (c)

10.7 m

14.2 m

α

13 cm

33 cm

α

8 mm

40 mm

α

55 mh

6°

29 m

32 mθ

29 mθ

50 mh

8.7 m

18°

14°d

35 m


MIXED PROBLEMSWhen solving trigonometric problems involving right-angled triangles, it is important to identify which trigonometric ratio is appropriate: sine, cosine or tangent.

Some students memorise the initials of each ratio—SOH, CAH, TOA—and pronounce it as ‘so car toe-ah’. Others remember a phrase like ‘Some One Has Created A Helpful Trig Organisation Aid’ or ‘Some Old Hens Can’t Always Hide Their Own Age’, where the first letter of each word corresponds with the initials. Find the mnemonic (memory aid) that best suits you.

Example 22Find ∠N to the nearest degree.

Solution

sin N =

N = 24.9947 …≈ 25°

Example 23In ΔPQR, ∠Q = 47.8°, ∠R = 90° and p = 3.5 m. Find the length of q correct to 3 significant figures.

SolutionFirst draw a diagram.

tan 47.8° =

q = 3.5 tan 47.8°= 3.8599 …≈ 3.86 m

sin θ = (SOH)

cos θ = (CAH)

tan θ = (TOA)

oppositehypotenuse---------------------------

adjacenthypotenuse---------------------------


θAdjacent

Hypotenuse

Opposite

N

6 m14.2 m

M

L

614.2----------

The sides 6 m and 14.2 m are the opposite side and hypotenuse respectively, so use sin.

P

QRp = 3.5 m

q

47.8° The sides q and p are the opposite and adjacent sides respectively, so use tan.

q3.5-------

TRIGONOMETRY 275

Example 24Find d correct to 1 decimal place.

Solutioncos 33°40′ =

d =

= 7.2091 …≈ 7.2 km

1. For ∠I in the triangle, which trigonometric ratio(sin, cos, tan) is:

(a) ? (b) ? (c) ?

2. Draw a right-angled triangle ΔJKL where ∠L = 90° and label the sides j, k, l. Write the expression for:(a) sin J (b) cos J (c) tan J(d) sin K (e) cos K (c) tan K

3. Find the value of the pronumeral in each of these triangles correct to 2 decimal places.

33°40′

6 km

d

The sides 6 km and d are the adjacent side and hypotenuse respectively, so use cos.

6d---

6cos 33° 40′---------------------------

Exercise 7-08: Mixed problems

G H

I

hg

i

ig--- i

h--- g

h---

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

37°

16 cm e

46°p

14 m6.5 mb

18°

40 mm

x

27°34′

17.2 mh

53°7′

40.5°

20 m

k

66.8°

q

19.6 cm

12 mm

r 9°

14.7 m

t

78°49′


4. Find φ to the nearest degree.

5. Find β to the nearest minute.

6. Find θ to the nearest degree in two different ways.

7. (a) Find θ to the nearest minute.(b) Find h correct to 2 decimal places using a trigonometric

ratio and the value of θ found in part (a).(c) Find h using Pythagoras’ theorem.

8. Find the sizes of all unknown sides and angles of this triangle (to the nearest metre or degree).

9. In ΔWXY, ∠W = 90°, WX = 18.6 m and YX = 25.4 m. Find ∠X to the nearest degree.

10. Find the pitch θ of the roof of this house (to the nearest degree).

(a) (b) (c)

(d) (e) (f)

11 m

13 m

φ

6 cm

11 cm

φφ

75 mm62 mm

3.6 m9 m

φ

4.1 cm

13.7 cmφ

φ18 mm

30 mm

(a) (b) (c)

25 m

7 m

β

8.6 m

19.3 mβ

β

55 mm

45 mm

θ

24 mm 51 mm

45 mm

1.4 cm

4.8 cmh

θ

3 m 7 m

C E

D

7.8 m

2.7 m 2.7 m3.4 m

θ

TRIGONOMETRY 277

11. If tan θ = , draw a right-angled triangle and use Pythagoras’ theorem to find sin θ and

cos θ as fractions.

12. If cos θ = , find sin θ and tan θ as fractions.

13. A 9 m ladder reaches a window 7 m above the ground exactly. What angle (to the nearest minute) does the ladder make with the ground?

14. An Olympic-sized pool inclines at 2°17′ to the horizontal. How far does a person need to walk from the shallow end to stand where the water has a depth of 1.65 m? Answer correct to 3 significant figures.

15. The drawbridge in front of this castle is 5 m long and is pulled by chains from 6 m above ground level. What angle do the chains make with the ground (to the nearest degree)?

16. In ΔRST, ∠T = 90°, ∠R = 38°49′ and s = 55 cm. Find r to the nearest 0.1 cm.

17. A hockey player is standing directly in front of the centre of the goal. He is told that he can score as long as he keeps within a 48° angle. What is his distance from the goal line (correct to 2 decimal places)?

18. A flying fox is constructed between the top of a tree and a large rock. It has a length of 34 m and angle of inclination of 38°. What is the (horizontal) distance between the tree and the rock (correct to 1 decimal place)?

19. Find the perpendicular height of an equilateral triangle (correct to 2 decimal places) if its sides have length 5 cm. Then find the area of the triangle.

1235------

1525------

2°17′

d1.65 m

Cθ

A

B

6 m

5 m

d

3.6 m

48°

38°

d

34 m


BEARINGSBearings use angles to show the direction of one location from a given point. For example, the bearing of Sydney from Narrabri is 166°. This means that, from Narrabri, the direction of Sydney is 166° measured clockwise from north.

True bearings, also called three-figure bearings, are written as 3-digit angles ranging from 000° to 360°. This numbering system is illustrated on the two diagrams below.

Example 25Write the bearings of locations P, Q, R and T from point O.

SolutionP is 050° from O.Q is 90° + 30° = 120° from O.R is 270° − 55° = 215° from O.T is 270° + 18° = 288° from O.

Example 26Sketch points X and Y on a compass rose if X has a bearing of 345° from O and Y has a bearing of 200° from O.

SolutionX is between 270° and 360° so it is in the northwest (NW) quadrant.

Y is between 180° and 270° so it is in the southwest (SW) quadrant.

Sydney

Narrabri

PacificOcean

N

166°

0°

180°

90°270°

80°70°

60°

50°

40°

30°

20°10°

280°

290°

300°

310°

320° 330° 340° 350°

100°

110°

120°

130°

140°

150°

160°170°

260°250°

240°230°

220°210° 200° 190°

(000°)

S

W E(090°)(270°)

N

(180°)

NE(045°)

NW(315°)

SE(135°)

SW(225°)

P

Q

R

T

O 30°

50°

55°

18°

S

W E

N

Y

X

O

20°

75°

S (180°)

(270°) W E (090°)

N (360°)

TRIGONOMETRY 279

1. In which quadrant (NE, SE, SW, NW) will these bearings lie?(a) 260° (b) 145° (c) 073°(d) 102° (e) 340° (f) 237°

2. Write the bearing of each point from O.

3. What is another name for the bearing:(a) 315°? (b) 225°? (c) 135°?

4. Sketch the following bearings on a compass rose.(a) 352° (b) 250° (c) 166°(d) 078° (e) 196° (f) 303°(g) 147° (h) 270° (i) 025°

5. Petersville (P) is 80 km due east of Queensfield (Q). Rossmore (R) is on a bearing of 055° from Queensfield and 300° from Petersville. Which diagram represents this?

Exercise 7-09: Bearings

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(j) (k) (l)

38°

82°B

N

O

N

O

N

O

N

O

N

O

N

O

N

O

N

O

N

O

N

O

N

O

A

47°

C

45°

67°

15°

7°

L

F

64°

I

E

G

D

36°20′

J K

52°45′

22°35′

26°

N

O

H

A B C

Q35° 30°

R

P80 km

Q55° 60°

R

P80 km

Q35° 60°

R

P80 km


6. A surveyor travelling due east needs to make a detour on a bearing of 128° to avoid a flooded area. After detouring 8 km, he turns and travels NE until he rejoins the original line of travel.

(a) Copy the above diagram and mark on it all the information given.(b) Find the size of each angle in ΔABC.

7. What is the bearing of Sydney (say, Central Station) from your town or suburb?

PROBLEMS INVOLVING BEARINGSExample 27Max walked for 8.5 km on a bearing of 145° from base camp.(a) How far south is he from camp (correct to 1 decimal place)?(b) What is the bearing of the camp from his current position?

Solution(a) Let B be base camp and M be the position of Max.

Draw the bearing of 145° in the SE quadrant. Let d be Max’s distance south of camp.M is 180° − 145° = 35° from due south of camp.

cos 35° =

d = 8.5 cos 35°= 6.9627 …≈ 7.0 km

(b) Draw another compass rose at M.Mark the angle that is alternate to 35°.So base camp is 360° − 35° = 325° from Max (M).

Example 28Kerr’s Creek is 8 km west and 3 km north of Bobtown. Find its bearing from Bobtown to the nearest minute.

SolutionLet θ be the angle shown in the diagram.

tan θ =

θ = 20.5560 …°≈ 20°33′

Bearing of KC from B = 270° + 20°33′= 290°33′

East

B

CA

M

35°

145°

N

d

B

8.5 kmd8.5-------

M

35°

NB

N

35°

When drawing diagrams for bearing problems, take note where the bearing is from: this is where the compass rose is centred.

B

N

3 kmθ

8 km

KC

(270°)

38---

TRIGONOMETRY 281

1. Amy drives at a bearing of 210° for 110 km.(a) How far west is she from her starting point?(b) How far south is she?Express answers to the nearest kilometre.

2. Selena walks 2670 m north, then turns and walks 1200 m west.(a) Find the value of θ to the nearest minute.(b) Hence calculate Selena’s bearing from her starting point.(c) How far is she from her starting point (to the nearest 10 m)?

3. A ship sails approximately SE for 20 km and finds itself 15 km east of its starting point. What is its bearing from its starting point (to the nearest degree)?

4. Springfield and Summer Bay are 12 km apart and the bearing of Summer Bay from Springfield is 285°. How far is Summer Bay west of Springfield, correct to 2 decimal places?

5. Two cars leave from the same place: one in a direction due east, the other on a bearing of 165°. After travelling 8 km, the first car is due north of the second car. How far has the second car travelled, to the nearest kilometre?

6. A plane takes off flying NW for 6 km, then turns and flies due east until it is due north of the airport. How far is it from the airport now, correct to 2 decimal places?

7. A hot air balloon travelled 15 km due east but drifted 1.8 km south. What was its bearing from its starting point (to the nearest degree)?

8. A bushwalker leaves the camp, hiking 25 km at a bearing of 148°21′. How far south of the camp is he, to the nearest metre?

9. The town of Woop Woop is 22 km from Porpoise Spit, on a bearing of 340°. The town of Baccabirk is 48 km from Porpoise Spit, on a bearing of 070°.(a) Calculate the size of θ to the

nearest degree.(b) Hence find the bearing of Woop

Woop from Baccabirk. (Hint: alternate angles.)

Exercise 7-10: Problems involving bearings

110 km

N

2670 m

N1200 m

θ

20 km

N

15 km

N

12 km

Springfield

SummerBay

WoopWoop

N N

Baccabirk

Porpoise Spit

θ


10. Leo, standing at a point P, measures the bearing of a TV tower H to be 332°. A big tree T, 4.9 km due west of Leo, is directly south of the tower. Calculate the distance between the tower and the tree to the nearest metre.

11. A yacht sails on a bearing of 105° from Jervis Bay and travels 8 km. How far will it now have to sail along a bearing of 060° so that it is due east of Jervis Bay? Answer to the nearest metre.

12. Kingsley is 79 km due west of Macksville and Laurieton has a bearing of 207° from Macksville. If Laurieton is due south of Kingsley, calculate to the nearest kilometre the distance between:(a) Kingsley and Laurieton(b) Laurieton and Macksville.

13. Two ships P and Q leave port at the same time. P sails on a course NE while Q sails at a bearing of 124°. After P sails 152 km, it is due north of Q.(a) What is the distance between P and Q?(b) How far has Q sailed?Express your answers correct to 2 decimal places.

14. After flying for 2 hours on a bearing of 306°17′, an aircraft is 600 km north of its starting point. Find the distance travelled and the average speed of the plane. Answer correct to 2 decimal places.

15. A triathlete cycles 20 km on a bearing of 200°, from the end of a swim leg to the finish line.(a) How far to the south (to the nearest kilometre) has the triathlete cycled?(b) What is the bearing of the end of the swim leg from the finish line?

16. Castlegrove is 20 km west of the straight road that runs north–south from Arronville to Burton’s Ridge. From Castlegrove, the bearings of Arronville and Burton’s Ridge are 050° and 120° respectively. How long is the road between Arronville and Burton’s Ridge, to the nearest 0.1 km?

17. A hiking group walks from Sandy Flats to Black Mountain, a distance of 23.4 km in the direction 102°. They then turn and hike due north to Rivers End, then due west back to Sandy Flats. What was the total length of their journey, to the nearest 0.1 km?

18. A triangular orienteering course starts at Alpha and passes through the checkpoints Bravo and Charlie before finishing back at Alpha. Bravo is 8.5 km due east of Alpha and Charlie is 10.5 km due south of Bravo.(a) Draw a diagram showing this information.(b) Calculate the length of the circuit to the nearest 0.1 km.(c) Find the bearing of Charlie from Alpha to the nearest minute.(d) Find the bearing of Alpha from Charlie to the nearest minute.

K M

L

TRIGONOMETRY 283

APPLICATIONS OF TRIGONOMETRYAngles of elevation and depressionThe angle of elevation is the angle of looking up, measured from the horizontal.

The angle of depression is the angle of looking down, measured from the horizontal.

For both angles the observer needs to tilt their head to see the object.

Example 29Joanne stands 40 m away from the base of a 25 m tower. What is her angle of elevation to the top of the tower, to the nearest degree?

Solutiontan θ =

θ = 32.0053 …≈ 32°

Example 30The angle of depression of a ship from the top of a 220 m vertical cliff is 32°. Find the distance of the ship from the base of the cliff correct to 1 decimal place.

SolutionAngle of depression = 32°, so the adjacent complementary angle is 90° − 32° = 58°.

tan 58° =

d = 220 tan 58°= 352.0735 …≈ 352.1 m

Horizontal

Horizontal

Memory aid: When you feel elevated, things are ‘looking up’. When you feel depressed, things are ‘looking down’.

θ40 m

25 m

2540------

d

220 m

32°58°

32°

d220---------


Alternative solution:

By alternate angles, the angle of elevation of the clifftop from the boat is also 32°.

tan 32° =

d =

= 352.0735 …≈ 352.1 m

1. From the ground, Carolyn sees the top of a 180 m tower at an angle of elevation of 49°. How far is she from the base of the tower, to the nearest metre?

2. From her apartment 130 m high, Justine can see the park at an angle of depression of 50°. How far is the park from the base of Justine’s apartment block, to the nearest metre?

3. A boat is anchored by a chain of length 74 m that makes an angle of 48° from the vertical. Find the depth of the water to the nearest 0.1 m.

4. A car travels 3 km along a road that rises at an angle of 12°. What is the vertical height gained by the car (to the nearest metre)?

5. A ship’s captain observes a lighthouse on top of a vertical cliff at an angle of elevation of 16.3°. If the lighthouse is 40 m above sea level, what is the distance between the ship and the bottom of the cliff, correct to 3 significant figures?

6. A ladder leaning against a wall makes an angle of 68° with the ground. If it stands 1.2 m away from the base of the wall, calculate (correct to 2 decimal places):(a) the distance the ladder reaches up the wall(b) the length of the ladder

7. When the altitude of the sun is 31°4′, a fencepost casts a shadow of length 2.7 m. How high is the fencepost, to the nearest centimetre?

8. The diagonal of a square is 12.8 cm long. Find the length of one side (correct to 2 decimal places):(a) using trigonometry(b) using Pythagoras’ theorem.

9. Liam observed a plane flying at a constant height of 800 m. At first it appeared at an angle of elevation of 33°, then it flew directly overhead. What was the distance travelled by the plane (to the nearest metre) during this period of observation?

220d

---------

220tan 32°-----------------

Angle of elevation/depression problems usually involve the tan ratio.

Exercise 7-1 1: Applications of trigonometry

2.7 m

31°4′

h

12.8

cm

d

33°

800 m

TRIGONOMETRY 285

10. In a concert hall, Bill’s seat is 20 m from the stage by line of sight and 5 m above stage level. What is the angle of depression of the stage from Bill’s seat, to the nearest degree?

11. From her office window 110 m above street level, Elaine saw Jerry walking along the street towards her. At first his angle of depression was 27°, but a minute later the angle increased to 54°. What distance did Jerry walk during the minute (to the nearest metre)?

12. Sally stands at the top of her city apartment block and observes a taller office tower 280 m away. The angle of elevation of the top of the tower is 14° and the angle of depression of the bottom of the tower is 33°. Calculate the height of the tower to the nearest metre.

13. A netball pole with a height of 3.1 m casts a shadow of length 2 m. What is the angle of elevation of the sun, to the nearest degree?

14. An aircraft pilot is flying horizontally at an altitude of 580 m. He sees the airport at an angle of depression of 15°. What is:(a) the direct distance between the plane and the airport?(b) the horizontal distance between the plane and the airport?Express your answers to the nearest metre.

15. A radar station tracking a missile measures the angle of elevation to be 28°12′ and the line-of-sight distance to be 46 km. What is the altitude (vertical height) of the missile (correct to 1 decimal place)?

16. Kim is flying a glider at an altitude of 1000 m and preparing to land at Bankstown airport. She descends at an angle of depression of 15°. How far must she travel at this angle before reaching the ground? Write your answer correct to the nearest metre.

17. Find the angle of inclination of a section of a rollercoaster ride (to the nearest minute) if it is 175 m long but covers a ground distance of 75 m.

18. Standing 1.8 km horizontally from the crest of a hill, Megan sees a transmission tower on the crest. The angles of elevation to the top and bottom of the tower are 6.8°and 5° respectively. Calculate the height of the tower to the nearest metre.

19. From the top of a vertical cliff 115 m above sea level, the angle of depression of a yacht is 35°. How far is the yacht from the bottom of the cliff (correct to 3 significant figures)?

20. Maurice is rowing towards a lighthouse that is 230 m above sea level. At first the lighthouse appears at an angle of elevation of 18°, but after 3 minutes this angle increases to 22°. How far did Maurice row during this period? Answer to the nearest metre.

280 m

14°

33°

75 m

175 m

1.8 km

6.8°5°

h


Equipment: metre ruler, tape measure, watch, calculator.

The length of an object’s shadow depends on the altitude or elevation of the sun: the angle at which its rays hit the ground. When the sun is at a low angle (at the start or end of the day), the shadow is long. When the sun is at a high angle (middle of the day), the shadow is short.

1. Go outside and find a spot to measure the length of a metre ruler’s shadow at different times of day, preferably at hourly intervals. Copy the following table:

2. The altitude of the sun θ can be found using tan θ = .Use the formula to find θ to the nearest minute. Write your results in the table.

3. What patterns do you notice?

4. What happened to the altitude of the sun over the day?

5. When was the shadow:(a) the longest? (b) the shortest?

6. (a) What is the altitude of the sun when the shadow is the same length as the ruler?(b) What is the altitude of the sun when there is no shadow?Estimate the times at which these two events happened.

Time Length of shadow, L Altitude, θ (where tan θ = )

e.g. 9 am 1.12 m 41°46′

Investigation: The altitude of the sun

θθ

1L---

L m

1 m

θ

1L---

Study tips

WHERE TO STUDY?Some students study by themselves in their room or at the library. Others prefer company and study in the lounge or dining room because they need the noise and space. Some like to sit outside in the sun. You may choose different places for different types of homework.

A good place of study has:� lots of space to spread out work, such as a big desk� minimal noise, few distractions and interruptions� good lighting and ventilation, and is neither too hot nor too cold� comfortable and supportive seating.

Study is a serious business. You’ll be working for a while, so choose a place where youwon’t be easily distracted. Use a desk lamp; otherwise poor lighting will make your eyesfeel sore and tired. Open the window for some fresh air or you may feel drowsy.

TRIGONOMETRY 287

Chapter review

Trigonometry1. Pythagoras’ theorem2. Applications of Pythagoras’ theorem3. Investigating the tangent ratio4. Using tan to find a missing side5. Using tan to find a missing angle6. The sine ratio7. The cosine ratio8. Mixed problems9. Bearings

10. Problems involving bearings1 1. Applications of trigonometry

This chapter, Trigonometry, examined the measurement of right-angled triangles through Pythagoras’ theorem and the three basic trigonometric ratios—sine, cosine and tangent—and applied them to real-life situations. You also learned about seconds and minutes as fractions of a degree of angle size. In your summary, make sure you include notes and diagrams on bearings and angles of elevation/depression. Memorise the definitions of the sine, cosine and tangent ratios using the SOH, CAH, TOA mnemonic, and become skilled at calculating missing sides and angles in triangles by using the appropriate ratio.

Make a summary of this topic. Use the chapter outline above as a guide. An incomplete mind map is started below. Use your own words, symbols, diagrams, boxes and high-lighting. Make connections, look for general principles and include personal observations and reminders. Use the questions in Your say on the next page to think about your under-standing of the topic. Gain a ‘whole picture’ view of the topic and identify any weak areas.

Topic summary

Trigonometry

Degrees °Minutes ´Seconds ˝

N

S

EWO

P

Bearings

sin θ =

cos θ =

tan θ =

SOH CAH TOA

opphyp---------

adjhyp---------

oppadj---------

angle ofdepression

angle of

elevation

Pythagoras’theorem

b

ac2 = a2 + b2


� Have you satisfied the outcomes listed at the front of this chapter?� What was the most important thing that you learned?� How did you feel about the topic? Did you enjoy it?� What was new?� What are your weaknesses? What will you need to study more?� How will you revise and summarise this topic?

1. Test whether these triangles are right-angled.

2. Find the values of the pronumerals in these right-angled triangles, correct to 2 decimal places where necessary.

3. A TV screen is 34 cm long and 24 cm high. What is the length of its diagonal, to the nearest centimetre?

4. Test whether a 30 cm ruler can fit into a rectangular pencil case of dimensions 25 cm by 14 cm.

Your say: Reflecting about the topic ● ● ● ●

Chapter assignment

(a) (b) (c)72 mm

97 mm

65 mm7.5 m

4 m

8.5 m 32 cm

40 cm

28 cm

(a) (b) (c)

(d) (e) (f)

3.1 cm

5.4 cm

y

10.5 cma

23 cm

x

42 cm

8 m

15 m

k

r

6 m 11.4 m

16 m

u

10 m8.4 cm

d

34 cm

24 cm

14 cm

25 cm

TRIGONOMETRY 289

5. Find the perimeters of these figures correct to 2 decimal places.

6. After taking off, a plane flies 850 m but covers a ground distance of 840 m. How high is the plane above the ground?

7. Find the values of the pronumerals in these triangles correct to 2 decimal places.

8. Find the values of the pronumerals in these triangles to the nearest degree.

9. Find the values of the pronumerals in these triangles to the nearest minute.

10. A skateboard ramp is 40 cm high and makes an angle of 20° with the ground. How long is the ramp, to the nearest centimetre?

(a) (b)

48 mm

60 mm

38 mm

4 cm

7 cm

(a) (b) (c)

(d) (e) (f)

50° 10 cm

bp

4.5 cm

61.3°

5 m58°

x

31°25′

14 cm

r

y

8 cm

36°60°41′

x

12 cm

(a) (b) (c)11 cm

17 cmα

3 cm8 cmφ θ2 m

5 m

(a) (b) (c)

α

14 cm

10 cm

7 cm

24 cmθ

2 cm

10 cm

φ

40 cm

20°

x


11. A tree 18 m high casts a shadow of length 35 m. What angle do the rays of the sun make with the ground, to the nearest minute?

12. A 13 m ladder is placed against a building so that its foot is 3 m from the base of the building.(a) How high is the top of the ladder above the ground, to the nearest 0.1 m?(b) What angle does the ladder make with the ground, to the nearest minute?

13. ΔXYZ is a triangle with ∠Y = 90°, ∠Z = 44°39′ and XZ = 8 cm. Find YZ correct to 2 decimal places.

14. What is the bearing of Y from X?

15. Sketch on a compass rose a point that has a bearing of 157°.

16. A pigeon is perched on top of a beacon. The direct distance between the pigeon and the top of a tree is 21 m. If the tree and the tower are 8 m apart, what is the angle of depression of the top of the tree from the pigeon (to the nearest degree)?

17. A lighthouse is 500 nautical miles due north of a ship. If the ship sails on a course of 333° until it is due west of the lighthouse, how far west is it, correct to 2 decimal places?

18. Gene saw a hot air balloon flying horizontally at an angle of elevation of 31° and height 650 m. How far (to the nearest metre) must the balloon fly before it is directly above Gene?

19. A boat sails due east for 18 km, then turns and sails due south for 5 km.(a) How far is it from its starting point, correct to 2 decimal places?(b) What is its bearing from its starting point (to the nearest degree)?

20. A glider is directly above one end of a runway at a height of 430 m. The pilot sees that the angle of depression of the other end of the runway is 29°. How long is the runway, to the nearest metre?

35 m

18 m

N

15°X

Y

8 m

21 m

L

500 nauticalmiles

31°

650 m

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07 NCM 11 NSW SB TXT - mathsbooks.net Century Year 11/07_Trigonometry (Measurment... · TRIGONOMETRY 247 Example 2 Find the length of the hypotenuse h correct to 2 significant figures