3 Complete the table below, giving answers correct to 1 decimal place.
D
E C ∠ BAC = 30°
∠ BAD = 45°
∠ BAE = 60°
30° h B
A
θ opposite side adjacent side hypotenuse h-o- -ha- -ao-
(o) (a) (h)
∆BAC 30°
∆BAD 45°
∆BAE 60°
4 a In question 1, are the respective ratios -ho- , h-a- , -ao- the same for each triangle?
b In question 2, are the respective ratios -ho- , h-a- , -ao- the same for each triangle?
5 a Using a protractor and ruler, construct a right-angled triangle with an angle of 35° and a
base of 5 cm. By measurement, calculate the value of each of the ratios given, correct to
1 decimal place.
i -ho- ii -ha- iii a-o-
b Construct another right-angled triangle with an 35°
angle of 35° and a base of 10 cm and calculate base
again the value of the ratios:
i -ho- ii h-a- iii a-o-
c What conclusion can you draw from the results in a and b?
13:03 | The Trigonometric Ratios
In questions 1 and 2 of Exercise 13:02, each of the three triangles had angles of the same size.
When this happens the triangles are said to be similar.
Similar right-angled triangles can always be superimposed to produce a diagram like the one below.
381CHAPTER 13 TRIGONOMETRY
Check the results in the table below.
Oθ A C These similar
B D E triangles overlap!
1 dec. pl.
F
θ h-o- h-a- a-o- I Answers are
∆AOB 26·5° 0·4 0·9 0·5 given correct to
∆COD 26·5° 0·4 0·9 0·5 one decimal place.
∆EOF 26·5° 0·4 0·9 0·5
• Each of the right-angled triangles in this diagram (ie AOB, COD, EOF) is similar to the others,
since the corresponding angles in each are the same. The angle at O is obviously the same for
each triangle.
• The ratios h-o- , -ha- , a-o- are equal in all of the triangles.
• These ratios are called the trigonometric ratios (abbreviated to trig. ratios) and are given special
titles.
The ratio h-o- -s--i--d---e----oh---p-y--p-p---oo---st--ei--t-n-e--u--a-s-n--e--g---l-e-----θ- is called the sine ratio. It is abbreviated to sin θ.
The ratio -ha- s---i--d---e----a---dh---j-y-a--pc---eo---nt--e-t--n-t--uo---s--a-e--n---g---l-e-----θ-- is called the cosine ratio. It is abbreviated to cos θ.
The ratio -ao- s---i-s-d-i--de---e-a---do---jp--a-p-c---oe---ns--i-t-t--et--o--a---na---ng---lg-e--l-e--θ---θ-- is called the tangent ratio. It is abbreviated to tan θ.
• The three ratios have constant values for any particular angle irrespective of how big the
right-angled triangle may be.
• For any angle, the values of the ratios can be obtained from a calculator.
• As the lengths of right-angled triangles can often be surds, many trig. ratios are irrational
numbers.
382 INTERNATIONAL MATHEMATICS 4
hyp. sin θ = h----yo--p-p---op---to--e-s-n-i--tu--e-s---e- cosθ = -h---ya--p-d---oj--a-t-c-e--en---nu---ts---e- tan θ = -oa--d-p--j-p-a--o-c--s-e-i-n-t--e-t
opp.
adj.
• Because -ho- ÷ h-a- = -ho- , we also have that:
sin θ ÷ cos θ = tan θ or tan θ = c-s--io--n-s----θθ--
worked examples
1 Find sin θ, cos θ and tan θ for each triangle, a 13 bθ 7
and express each as a decimal correct to θ
5 5
three decimal places. 12
2 Find sin α, cos β and tan α. A8 B
Solutions 5 C
β
1 a sin θ = o-h---py---pp---.. cos θ = -ha---yd---pj--.-.
= -1-5--3- = 11----23- Dα
Ӏ 0·385 Ӏ 0·923
tan θ = o--a--pd---pj--.-.
= 1--5--2-
Ӏ 0·417
b First the hypotenuse must be calculated using Pythagoras’ theorem. So, then:
sin θ = ----7----- tan θ = -75- I h2 = 52 + 72 h7
74 Ӏ 1·400 = 25 + 49 5
= 74
Ӏ 0·814
ie h = 74
cos θ = ----5-----
74
Ӏ 0·581
2 ABCD is a rectangle.
Hence DC = 8, BC = 5
Also BD = 89 (Pythagoras’ theorem)
∴ sin α = -BB---DC--- cos β = -AB---D-D-- tan α = -DB----CC--
= ----5----- = ----5----- = 8-5-
89 89
383CHAPTER 13 TRIGONOMETRY
This should help you remember!
sin = opp./hyp. cos = adj./hyp. tan = opp./adj.
↓ ↓↓ ↓↓↓ ↓ ↓↓
S OH C AH T OA
↓ ↓↓ ↓↓↓ ↓ ↓↓
Some Old Hobos Can’t Always Hide Their Old Age
Exercise 13:03
1 Find sin θ, cos θ and tan θ in these triangles (as a simple fraction).
a b 12 5 c 25
24
5 θ θ
3 13 7
θ
4
2 Evaluate sin A, cos A and tan A for each triangle. Give your answers in decimal form correct to
3 decimal places.
aB bC 15 A c A
17 6·5 2·5
86 8 C
A 10 C B 6·0 B
3 Find the unknown side using Pythagoras’ theorem and then find sin θ and cos θ in decimal
form.
a 12 b 3 c
θ 6 θ2
9θ
2
4 Use Pythagoras’ theorem to find side YZ, then state the value of tan X.
a Y bX cY Z
2
10 5 61 13 X
X 8Z Y Z
384 INTERNATIONAL MATHEMATICS 4
5 Complete the statements below.
a (90°–θ ) b c
5
3 3 5 4 60°
2
θ (90°– θ )
34 30°
12
θ
4
sin θ = . . . cos θ = . . . sin 60° = . . .
cos 30° = . . .
cos (90° − θ) = . . . sin (90° − θ) = . . .
6 For the triangle on the right, complete the following: B
a sin θ = . . . b cos θ = . . . c a
C
cos (90° − θ) = . . . sin (90° − θ) = . . .
A (90°– )
b
7 Find the value of x, given that:
a cos 25° = sin x° b sin 60° = cos x° c cos 10° = sin x°
8 a For the triangle shown, write down the value of: C
53
i sin A ii cos A A 4B
iii tan A iv sin A ÷ cos A
b Does tan A = -cs--oi--n-s----A-A-- ?
9 a Use Pythagoras’ theorem to find the value of the missing 2 cm 1 cm
side as a surd. Hence find the value of sin 30°, cos 30° 30°
and tan 30°. (Leave your answer as a surd.)
b By rationalising the denominator, arrange the values for
sin 30°, cos 30° and tan 30° in ascending order.
10 a If sin A = 41-- , find the values of cos A and tan A.
b It is known that cos θ = -43- . What is the value of sin θ ?
11 a Find i sin A ii sin C b Find i sin θ ii cos α
D6C 20
4 30
A 8B
385CHAPTER 13 TRIGONOMETRY
12 a By finding tan θ in two different m 5
triangles, find the value of m. 3 4
b Find x and y. AE y B
(Note: DE = x and 5x C
CE = y.)
D 10
c Find:
i sin θ ii cos θ
iii m iv n 4m
2
v sin 2θ
5n
vi Show that sin 2θ = 2 × sin θ × cos θ
(90°– ) a
c
b
sin θ = cos (90° − θ)
cos θ = sin (90° − θ)
• Trigonometry is used in
surveying to calculate
lengths and areas.
Challenge worksheet 13:03 The range of values of the trig. ratios
386 INTERNATIONAL MATHEMATICS 4
13:04 | Trig. Ratios and the Calculator
As we have already found, the values of the trig. ratios are constant for any particular angle and
these values can be found from a calculator. You can also use the calculator to find an angle when
you are given the ratio.
Finding a ratio given the angle
To find tan 31°, ensure your calculator is operating in ‘degrees’ and then press:
tan 31 =
The calculator should give tan 31° = 0·600 860 6, correct to 7 decimal places.
Degrees and minutes
So far the angles have all been in whole degrees. 1 degree = 60 minutes
One degree, however, can be divided into 1º = 60′, [ 1′ = 610º]
60 minutes.
F30ormeixnaumtepsl.e,T3h1is21--w° owuoludldbeewquraitlte3n1 degrees and
as:
31°30′.
We can now find the trigonometric ratios of angles
given to the nearest minute by using the calculator
as shown in the examples below.
worked examples
Find:
1 sin 25°41′ 2 tan 79°05′
Give your answers correct to 4 decimal places.
Solutions
Two methods are shown, one for each solution. Choose the one that best suits your calculator.
Method 1: I Warning:
1 For calculators with a Degrees/Minutes/Seconds button. Your calculator may
work differently to
This is usually marked in either of two ways. the one used here.
DMS or °′″
Press: sin 25 DMS 41 =
The calculator gives 0·433 396 953.
Method 2:
2 We convert 79°05′ into decimal degrees by realising that 05′ is -6-5--0- of one degree.
Press tan ( 79 + 5 ÷ 60 ) =
The calculator gives 5·184 803 521.
387CHAPTER 13 TRIGONOMETRY
Finding an angle, given the ratio
If the value of the trigonometric ratio is known and you want to find the size of the angle to the
nearest minute, follow the steps in the examples below.
worked examples
1 If sin θ = 0·632, find θ to the nearest minute. What if I
2 If cos θ = 0·2954, find θ to the nearest minute. want to
find the
Solutions angle?
Note: One minute may be divided further, into
60 seconds, and this fact will be used to
round off answers to the nearest minute.
Again two methods are shown that correspond to the
two methods on the previous page.
1 If sin θ = 0·632 press: 2nd F sin 0·632 =
The calculator now displays 39·197 833 53°. To
convert this to degrees/minutes/seconds
mode, press DMS . The calculator gives 39°11′52·2″.
∴ θ = 39°12′ (to the nearest minute)
2 If cos θ = 0·2954, press 2nd F cos 0·2954 =
The answer on the screen is 72·818 475 degrees. The alternative method of converting
this to degrees and minutes is to find what 0·818 475 of one degree is, in minutes;
ie 0·818 475 × 60′, which gives an answer of 49·1085 minutes, ie 49′ (to the nearest minute).
∴ θ = 72°49′.
Exercise 13:04
1 Using the degrees/minutes/seconds button on your calculator, write each of the following in
degrees and minutes, giving answers correct to the nearest minute.
a 16·5° b 38·25° c 73·9° d 305·75°
e 40·23° f 100·66° g 12·016° h 238·845°
2 Write in degrees, correct to 3 decimal places where necessary.
a 17°45′ b 48°16′ c 125°43′ d 88°37′
h 36°53′
e 320°15′ f 70°54′ g 241°29′
3 Use your calculator to find the value of the following, correct to 4 decimal places.
a sin 30° b cos 30° c tan 30° d sin 71°
e cos 58° f tan 63° g sin 7° h cos 85°
4 Find the size of θ (to the nearest degree) where θ is acute.
a sin θ = 0·259 b sin θ = 0·934 c sin θ = 0·619
f cos θ = 0·9
d cos θ = 0·222 e cos θ = 0·317 i tan θ = 3
g tan θ = 1·2 h tan θ = 0·816
388 INTERNATIONAL MATHEMATICS 4
5 Find, correct to 3 decimal places, the following ratios.
a sin 30°10′ b sin 62°45′ c cos 52°30′ d cos 83°03′
h tan 72·57°
e tan 61·25° f tan 79·36° g sin 17·8°
6 Find θ, to the nearest minute, given that θ is acute.
a sin θ = 0·6 b sin θ = 0·43 c sin θ = 0·645
f cos θ = 0·5216
d cos θ = 0·2 e cos θ = 0·031 i tan θ = 2·67
g tan θ = 1·3 h tan θ = 0·625
7 Redo question 6, this time giving answers in degrees correct to 2 decimal places.
8 What is the value of a-o-, h-o-, -ha- for each of the following triangles, correct to 3 dec. pl.?
abc
58°
28°15'
30°
9 Find the value of -1-x--0- for each of the following, correct to 3 decimal places. 47°10'
abc 10
10 x x
60°
10
28° x
10 a If -1-x--0- = cos 60° , find the value of x.
b If a = 3 sin 40° + 4 cos 30°, find the value of a correct to 3 decimal places.
c By substituting values for A and B, find if sin A + sin B = sin (A + B).
d If sin A = 21-- and sin B = 1-3- find A + B.
e Jim thinks that if you double the size of an angle you double its sine, that is
sin 2A = 2 × sin A. Is Jim correct?
• Trigonometry is used
in many branches of
science.
389CHAPTER 13 TRIGONOMETRY
Practical Activity 13:04 | The exact values for the trig. ratios 30°, 60° and 45°
∆ABC is an equilateral triangle of side 2 units. A
AD is perpendicular to BC.
1 Copy the diagram and write in the size of BD and ∠BAD.
2 Using Pythagoras’ theorem, calculate the length of AD 22
as a surd.
3 Now, from ∆ABD, write down the values of sin, cos and 60° D C
tan for 30° and 60°. B
F ∆DEF is a right-angled isosceles triangle.
The two equal sides are 1 unit in length.
4 Why is ∠EDF equal to 45°?
1 5 What is the length of DF as a surd? I Leave surds in
your answers.
6 Write down the values of sin 45°,
cos 45° and tan 45°.
D 1E
Leave your answers in
surd form. Do not approximate.
2 30° 21
3
60° 45°
1 1
sin 60° = ---2--3- , sin 30° = -12- sin 45° = ---1---
2
cos 60° = 2-1- , cos 30° = ---2--3- cos 45° = ---1---
2
tan 60° = 3, tan 30° = ---1--- tan 45° = 1
3
390 INTERNATIONAL MATHEMATICS 4
13:05 | Finding an Unknown Side
For the triangle given, state: C prep qu iz
1 the hypotenuse
2 the side opposite the marked angle 13:05
3 the side adjacent to the marked angle
f A B g
Write true or false for these triangles: e
45 6
c h
a
i
bd
cos θ = -gi-
sin θ = -ac- cos θ = -ef
Find correct to 3 decimal places: 9 If tan 25° = 4-x- 10 If tan 25° = 4-x-
then x = . . . then x = . . .
7 sin 75° 8 tan 25°30′
Pythagoras’ theorem is used to find an unknown side in a right-angled triangle when the other two
sides are known. Trigonometry is used when only one side and one of the acute angles are known.
worked examples
1 Find a in these triangles, correct to 1 decimal place.
a b c
am 15 m am xm 9·2 m
29° 38° 28°
9·6 m
2 A ladder that is 8 metres long leans against a wall, and makes an angle of 21° with the wall.
How far does the ladder reach up the wall, to the nearest centimetre?
3 Find the length of a guy rope that must be used to secure a pole 12·5 m high, if the angle
the guy rope makes with the ground is 56°.
Solutions Make sure your calculator
is operating in ‘degrees’ mode.
Use the trig. button on your calculator.
1 a -1-a--5- = sin 29°
∴ a = (sin 29°) × 15 ↔ sin 29 × 15 =
= 7·272 144 3
So a = 7·3 (to 1 decimal place)
continued §§§
391CHAPTER 13 TRIGONOMETRY
b 9---a-·--6- = cos 38° If x is the hypotenuse
∴ a = (cos 38°) × 9·6 ↔ cos 38 × 9·6 = you’ll need to invert
= 7·564 903 2 each side of the equation.
= 7·6 (to 1 decimal place)
x 9·2
c 9---x-·--2- = sin 28° (Note that x is the 28°
9---x-·--2- = s---i--n---1-2----8---° denominator of the
fraction, not the
numerator.)
∴ x = s---i--n9---·-2-2---8---° ↔ 9·2 ÷ 38 sin 28 =
= 19·6 (to 1 decimal place)
2 From the information in the question, a diagram like the one 8 m 21°
to the right can be drawn. Let the height up the wall be h m.
hm
So: 8-h- = cos 21°
h = 8 × cos 21° ↔ 8 × cos 21 =
= 7·468 643 4
= 7·47 (to the nearest centimetre)
∴ The ladder reaches 7·47 m up the wall.
3 Let the length of the rope be x metres. 12·5 m xm
56°
Then: 1----2-x--·--5-- = sin 56°
so: -1---2-x--·--5-- = -s--i--n---1-5----6---°
x = s---i1--n-2---5-·--5-6---°
Ӏ 15·08
∴ The rope is 15·08 metres long (to the nearest centimetre).
392 INTERNATIONAL MATHEMATICS 4
Exercise 13:05 Foundation Worksheet 13:05
1 Find the value of the pronumeral in each triangle, Using trigonometry to find side lengths
correct to 1 decimal place. 1 Find x correct to 1 decimal place:
a 8-x- = sin 15° b 6---y-·--8-- = tan 38° . . .
a b y 2 Find x correct to 1 decimal place:
31° a 1--x--5- = cos 40° . . .
10 6
x 3 In each of the following state which trig.
36°
ratio needs to be used to find x and then
find it correct to 1 decimal place.
c 20 d 45° a b
29° x 10
p 3·6 62°
33°
10 x
4 In each of the following state which trig.
a ratio needs to be used to find the length
of the hypotenuse and then find it
correct to 1 decimal place.
a b
x8 10 60° x
40°
e 60° d f
9·2 15·6
g 25° 30′ 39° i y cm
8m x
xm 39° 52′
h
6 cm
am
31° 15′
12 m
j k l 4·6 m
62° 10′
nm 21° 49′ y cm
9·4 cm xm
42° 45′
21·2 m
2 Determine the value of each pronumeral, correct to 1 decimal place.
a a bc
31°
10 7 25
41° 59°
x p
d 9·2 e f 65° 11·3
k 5·3 q
52°
43°
d
393CHAPTER 13 TRIGONOMETRY
g h i
xm 7m 12 m 8m
43° 20′ 27° 50′ 53° 09′
xm xm
j xm k l 37° 12′ 9·7 m
35° 42′
5·1 m 5·9 m xm
63° 25′
xm
For questions 3 to 11 the diagrams relate to the questions below them.
3 C4 5
7m
27° A 5·3 m
B
52° 73°
4m
3 Find out everything you can about the triangle.
4 A ladder leans against a wall so that the angle it makes with the ground is 52° and its base is
4 m from the wall. How far does the ladder reach up the wall (to the nearest centimetre)?
5 A ladder leaning against a wall reaches 5·3 m up the wall when the angle between the ground
and the ladder is 73°. How long, to the nearest centimetre, is the ladder?
67 8
16·3 cm 70 m 52°
3°
37°
6 The diagonal of a rectangle is 16·3 cm long and makes an angle with the longest side of 37°.
Find the length of the rectangle, to the nearest centimetre.
7 A ship out at sea observes a lighthouse on the top of a 70 m cliff at an angle of 3°. How far out
to sea is the ship (to the nearest metre)?
8 A boat is anchored in a river that is 3·2 m deep. If the anchor rope makes an angle of 52° with
the surface of the water, how long is the rope from the surface of the water? (Answer to the
nearest centimetre.)
394 INTERNATIONAL MATHEMATICS 4
9 10 11
16 m 80°
38° 42′
9·6 cm
9 The equal sides of an isosceles triangle are 16 metres long and the apex angle is 80°.
Find, to the nearest centimetre, the length of the base.
10 The base of an isosceles triangle is 9·6 cm long and each of the base angles is 38°42′.
Find the length of each of the equal sides. (Answer correct to 3 significant figures.)
11 If the length of a child’s slippery-dip is 3·4 m and one end makes an angle of 38°42′ with the
ground, how high above the ground is the other end? (Answer to the nearest centimetre.)
For questions 12 to 20, draw a diagram first!
12 a In ∆ABC, ∠A = 90°, ∠B = 63°25′ and BC = 6 m. Find AC, correct to the nearest centimetre.
b In ∆XYZ, ∠Z = 90°, ∠X = 42°34′ and XZ = 9·2 m. Find YZ, correct to the nearest centimetre.
c In ∆ABC, ∠B = 90°, ∠A = 52° and AB = 2·7 cm. Find AC, to 1 decimal place.
d In ∆XYZ, ∠X = 90°, ∠Y = 31°20′ and XZ = 10·3 cm. Find XY, to 1 decimal place.
13 The diagonal of a square is 21·2 cm. Find the length of each side (to the nearest millimetre).
14 Find the length of the diagonal of a rectangle if the length of the rectangle is 7·5 cm and the
diagonal makes an angle of 25° with each of the longer sides. (Answer correct to the nearest
millimetre.)
15 Find the length of a rectangle if its diagonal is 34 cm long and the angle the diagonal makes
with the length is 27°50′. (Answer correct to the nearest centimetre.)
16 Find the base of an isosceles triangle if the height is 8·2 cm and the base angles are each 39°.
(Answer correct to the nearest millimetre.)
17 When the altitude of the sun is 51°47′, a vertical stick casts a shadow 45 cm long. How high,
to the nearest millimetre, is the stick?
18 A painting is hung symmetrically by means of a string passing over a nail with its ends attached
to the upper corners of the painting. If the distance between the corners is 55 cm and the angle
between the two halves of the string is 105°, find the length of the string, correct to the nearest
millimetre.
19 The vertical rise from the bottom to the top of a track that slopes uniformly at 6°54′ with the
horizontal is 36 m. Find, to 1 decimal place, the length of the track.
395CHAPTER 13 TRIGONOMETRY
20 A road rises steadily at an angle of 6°45′. What will be the vertical rise of the road for
a horizontal distance of 300 m? (Answer correct to the nearest metre.)
21 At noon a factory chimney casts a shadow when the sun’s altitude
is 85°24′. If the chimney is 65 m high, what is the length of the
shadow, to the nearest centimetre?
85° 24′
22 9·2 m Calculate the sloping area of this roof that needs to be tiled,
23° 23° given that the width of the roof is 5·4 m and its length is 9·2 m.
5·4 m Each roof section is pitched at an angle of 23°. (Answer correct
to the nearest square metre.)
23 A plane is flying at an altitude (height) of 750 metres.
A boy on the ground first observes the plane when it is
directly overhead. Thirty seconds later, the angle of 750 m
elevation of the plane from the boy is 24°14′.
a Through what distance did the plane fly in 30 seconds, 24° 14′
to the nearest metre?
b Calculate the speed of the plane in km/h, correct to 3 significant figures.
24 Calculate the area of a right-angled triangle that has a hypotenuse 8 cm long and an
angle of 50°.
25 A regular hexagon of side a units is made by joining six
equilateral triangles together, as shown in the diagram.
We want to find a formula for the area of the hexagon in C
terms of its side length, a.
Consider the area of one of the equilateral triangles. Aa B
B
a Using the exact trig. ratios on page 469, find the C
exact length of DC.
a
b What is the area of ∆ABC? 60°
AD
c What is the area of a hexagon of side a units?
d Find the area of a hexagon with a side length of:
i 2 cm ii 5 cm iii 10 cm
396 INTERNATIONAL MATHEMATICS 4
13:06 | Finding an Unknown Angle
Complete the ratios below for each triangle. prep qu iz
13 1 sin θ = 8 15 3 tan θ = 13:06
5 2 cos θ = 17 4 sin θ =
12
Given that θ is acute, find θ to the nearest degree, if:
5 tan θ = 0·635 6 sin θ = 0·2135 7 cos θ = 0·0926
If 0° р A р 90°, find A to the nearest minute if:
8 sin A = 0·52 9 tan A = 2·673 10 cos A = 0·7231
We have already seen in 13:04 that a calculator can be used to find the size of an angle if the value
of the trigonometric ratio is known.
1 Find the size of angle θ. worked examples
Answer to the nearest degree.
2 What angle, to the nearest minute, does
2 the diagonal of a rectangle make with its
length, if the dimensions of the rectangle
are 12·6 cm by 8·9 cm?
5
Solutions Remember ‘2nd F’ may
be called ‘SHIFT’ on
1 In the triangle,
some calculators.
tan θ = 52--
= 0·4 ↔ 2nd F tan 0·4 =
∴ θ = 21·801 409°
so θ = 22° (to the nearest degree).
2 Let the required angle be θ. Then: = 2nd F DMS
tan θ = 1--8--2--·-·9--6-- 2nd F tan ( 8·9 ÷ 12·6 )
8·9 cm
∴ θ = 35°14′7·59″
12·6 cm ∴ θ = 35°14′ (to the nearest minute).
397CHAPTER 13 TRIGONOMETRY
Exercise 13:06
1 Find the size of the angle marked θ in each triangle. Give your answers correct to the nearest
degree.
a b 3 c
52 2 10 7
d e6 f
10 5 20
14
6
i 1·7
g 4·7 h
2·6
8·9 6·3
8·9
2 For each, find the size of θ correct to the nearest minute. c 12 m
a b 5m
7m
9m
12 m 7m
d e 11·5 m f
6·9 m
6·2 m 4·6 m 8·2 m
10·1 m
3 Use trigonometry to find x in three different ways. 5 x°
3 4
4 a In ∆LMN, ∠M = 90°, LN = 9·2 m and LM = 8·2 m.
Find ∠L, to the nearest degree.
b In ∆PQR, ∠R = 90°, PR = 6·9 m and QR = 5·1 m.
Find ∠P, to the nearest minute.
5 a A ladder reaches 9 m up a wall and the foot of the ladder is 2 m 9m
from the base of the wall. What angle does the ladder make with 2m
the ground? (Answer correct to the nearest degree.)
b What angle will a 5 m ladder make with the ground if it is to
reach 4·4 m up a wall? (Answer correct to the nearest degree.)
398 INTERNATIONAL MATHEMATICS 4
6 The beam of a see-saw is 4·2 m long. If one end is 1·2 m above 4·2 m
the ground when the other end is resting on the ground, find
the angle the beam makes with the ground, correct to the nearest
degree.
7 1m A road is inclined so that it rises 1 m for each horizontal
distance of 8 m. What angle does the road make with the
8 m horizontal? (Answer correct to the nearest minute.)
8 At a certain time of the day, a tree 25 m high 25 m
casts a shadow 32 m long. At this time of day,
what angle do the rays of the sun make with the
ground? (Answer correct to the nearest minute.)
32 m
9 What angle does a diagonal of a rectangle make with each of the sides if the dimensions of the
rectangle are 4·7 m by 3·2 m? (Answer correct to the nearest minute.)
10 Find the angle θ in each of the following. (Answer correct to the nearest minute.)
ay b y cy (5, 3)
2
(3, 2)
3 x −1 2x
x
11 The cross-section of a roof is an isosceles triangle. 5·1 m
Find the pitch of the roof (the angle it makes with 9·6 m
the horizontal) if the width of the roof is 9·6 m and
the length of one of the pitched sections is 5·1 m.
Give your answer correct to the nearest minute.
12 Find the size of the base angles of an isosceles triangle if the length of the base is 10 cm and the
height is 8·4 cm. (Answer to the nearest minute.)
13 Find the apex angle of an isosceles triangle, if the apex angle
length of each of the equal sides is 14·3 cm and the 14·3 cm
length of the base is 20·8 cm. Give your answer to
the nearest minute.
20·8 cm
D 6C
50°
14 The diagram shows a trapezium.
EB
a If BC = 8, find θ. 15
b If CE = 8, find θ. A
Challenge worksheet 13:06 Trigonometry and the limit of an area
13:06 Shooting for a goal
399CHAPTER 13 TRIGONOMETRY
13:07 | Miscellaneous Exercises
Before continuing with further trigonometric examples there is some general information that
should be mentioned.
Angles of elevation and depression line of sight
angle of elevation
When looking upwards towards an object, the angle of horizontal
elevation is defined as the angle between the line of sight
and the horizontal. horizontal
angle of depression
When looking downwards towards an object, the angle of line of sight
depression is defined as the angle between the line of sight
and the horizontal.
worked examples
1 The angle of elevation of the top of a vertical cliff is observed to be 23° from a boat
180 m from the base to the cliff. What is the height of the cliff? (Answer correct to
1 decimal place.)
2 An observer stands on the top of a 40-metre cliff to observe a boat that is 650 metres out
from the base of the cliff. What is the angle of depression from the observer to the boat?
(Answer to the nearest minute.)
Solutions h metres
1 For this example, the diagram would look like 23°
the one on the right. 180 metres
Let the height of the cliff be h metres.
Then: -1---8h---0- = tan 23°
ie h = (tan 23°) × 180
= 76·405 467 (from calculator)
∴ Height of cliff = 76·4 m (to 1 decimal place).
2A D Note: The angle of depression ∠DAB = ∠ABC
angle of (alternate angles and parallel lines).
40 m depression =
C 650 m B
ie tan θ = -6-4--5-0--0- 2nd F tan ( 40 ÷ 650 ) = DMS
θ = 3°31′17·23″
= 3°31′ (to the nearest minute).
400 INTERNATIONAL MATHEMATICS 4
Compass bearings
The direction of a point Y from an original point X This has great
is known as the bearing of Y from X. This is mainly bearing on
expressed in one of two ways. Examine the diagram
below. trigonometry!
The bearing of Y from X
N can be given as:
1 150° (the angle between
the interval XY and
the north line measured
W X 150° E in a clockwise direction),
or,
30° 2 south 30° east (S30°E).
Y
Sometimes, only letters are used. So SE (or south-east)
S is halfway between south (180°) and east (90°); that is,
135° or S45°F.
Other examples would look like these. N N
Y
N
40° 15′
Y
60°
WX EW X 245° 09′ E WX E
Y 65° 09′ 319° 45′
SS S
060° or N60°E 245°09′ or S65°09′W 319°45′ or N40°15′W
worked examples
1 If the town of Bartley is 5 km north and 3 km
west of Kelly Valley, find the bearing of Bartley
from Kelly Valley.
2 Two people start walking from the same point.
The first walks due east for 3·5 km and the
second walks in the direction 123° until the
second person is due south of the first person.
How far did the second person walk (to the
nearest metre)?
continued §§§
401CHAPTER 13 TRIGONOMETRY
Solutions N Bartley 3 km
1 The diagram for this question
would look like the one on the
right.
Let the angle indicated in the W E 5 km
diagram be θ.
329°
Thus: tan θ = 35--
= 0·6
So: θ = 31° (to the nearest S Kelly Valley
degree)
So the bearing of Bartley from Kelly Valley would be N31°W or simply 329°.
2 This diagram shows the information in the North
question above.
Since ∠SAB = ∠CBA West 123° C East
(alternate angles, AS // CB) A 3·5 km
then ∠CBA = 57° Check out this step! 57°
So: 3---x-·--5- = sin 57° x
ie 3---x-·--5- = s---i--n---1-5----7---°
S B Finish
x = -s--i--n3---·-5-5---7---° South
= 4·173 km
Press: 3·5 ÷ sin 57 =
Exercise 13:07 63° Foundation Worksheet 13:07
35 m
1 The angle of elevation of the top of a Angles of elevation and
tower from a point 35 m from the base depression, and bearings
of the tower was measured with a 1 Find the bearing of B from A
clinometer and found to be 63°. Find
the height of the tower, correct to if B is 6 km north and 3 km
1 decimal place. east of A.
2 9° North 3 B
6
A East
2 From a lighthouse 105 m
above the sea the angle of
depression of a boat is 2°.
How far is the boat from
the shore?
800 m LH 2°
105 m
The angle of depression of a boat 800 m out to sea from the top of
x boat
a vertical cliff is 9°. Find the height of the cliff, to the nearest metre.
402 INTERNATIONAL MATHEMATICS 4
3 From the top of a cliff 72 m high, the angle of depression of a boat is 12°47′. How far is the
boat from the base of the cliff? (Answer to the nearest metre.)
4 A vertical shadow stick has a height of 1·8 m. If the angle of elevation of the sun is 42°, what is
the length of the shadow at that time, correct to 1 decimal place?
5 Find the angle of elevation of the top of a vertical tower from a point 25 m from its base, if the
height of the tower is 40 m. (Answer to the nearest degree.)
6 From a lighthouse 70 m above sea level a ship, 1·2 km out to sea, is observed. What is the
angle of depression from the lighthouse to the ship? (Answer to the nearest minute.)
7 A kite is on the end of a string 80 metres long. If the vertical height of the kite, above the
ground, is 69 metres, find the angle of elevation of the kite from the person holding the string.
(Assume the string is a straight line, and answer to the nearest minute.)
8 A cyclist travels 15 km in the direction N15°27′E. How far has
he travelled in a northerly direction (to the nearest metre)?
9 A ship sails from P to Q a distance of 150 km on a course of
120°30′. How far is P north of Q? Also, how far is Q east of P?
(Answer to the nearest kilometre.)
10 Two towns, A and B, are 9 km apart and the bearing of B from
A is 320°. Find how far B is west of A (to the nearest kilometre).
11 Two cars leave from the same starting point, one in a direction due west, the second in a
direction with a bearing of 195°. After travelling 15 km, the first car is due north of the second.
How far has the second car travelled (to the nearest kilometre)?
12 An aircraft flew 10 km south and then 6 km west. What is its bearing from its starting point?
(Answer to the nearest degree.)
N
13 A, B and C are three towns. A lies 7 km north-east of B, and B lies 12·5 km A
north-west of C. Find the bearing of A from C. Also, how far is A from C? 45°
(Answer to the nearest metre.) B
14 A ship is 5 nautical miles from a wharf on a bearing of 321°, and a C
lighthouse is 11·5 nautical miles from the wharf on a bearing of 231°.
Find the bearing of the ship from the lighthouse. (Answer correct to
the nearest minute.)
15 The bearings from a point P of two landmarks X and Y are 35° and 125° and their distances
from P are 420 m and 950 m respectively. Find the bearing of Y from X (to the nearest minute).
16 X is due north of Y and 2 km distant. Z is due east of Y and has a bearing of S35°12′E from X.
How far, to the nearest metre, is Z from X?
403CHAPTER 13 TRIGONOMETRY
17 A wire is stretched from point A on the top of 21·3 m A
a building 21·3 m high, to point B on the top of 20° 15′
a shorter building, 15·6 m high. The angle of B
depression from A to B is 20°15′. 15·6 m
a What is the horizontal distance between the
buildings (to the nearest centimetre)?
b How long is the wire (to the nearest centimetre)?
18 PQ is a diameter of the circle, centre O, as shown with R Q
∠PRQ = 90°. If the radius of the circle is 6 cm, find, PO
to the nearest millimetre, the length of the chord PR,
given that ∠PQR = 40°.
19 A tangent of length 16 cm is drawn to a circle O T
of radius 7·5 cm from an external point T.
What is the angle, marked θ in the diagram,
that this tangent subtends at the centre of
the circle?
20 The diagonals of a rhombus are 11 cm and 7·6 cm. Find the angles, to the nearest degree,
of the rhombus.
21 Find the acute angle, to the nearest minute, between the diagonals of a rectangle that has sides
of 8 cm and 14 cm.
22 The eaves of a roof sloping at 23° overhang the walls, the edge of the
roof being 75 cm from the top of the wall. The top of the wall is
5·4 metres above the ground. What is the height above the ground
of the edge of the roof, to the nearest centimetre?
23 The arms of a pair of compasses are each 12 cm long. To what angle (to the nearest minute)
must they be opened to draw a circle of 4 cm radius? How far from the paper will the joint be,
if the compasses are held upright? (Answer to the nearest millimetre.)
24 Find the exact value of x in each of the following. c
ab
x
10 x 2
30° 45° 30°
12
x
404 INTERNATIONAL MATHEMATICS 4
25 a A rectangle is 10 cm long. The angle between the 10 m
diagonal and the length is 30°. What is the exact 60°
area of the rectangle?
5 cm
b A pole is to be supported by three guy wires.
The wires are to be fixed 10 m from the base of
the pole and must form an angle of 60° with the
ground (which is horizontal). What will be the
exact length of each guy wire?
c Find the exact value of x in the diagram.
4 3 cm 45°
60°
x cm
13:08 | Problems Involving Two
Right Triangles
Some problems can be solved by the consideration No, you’re wrong! I’m more
of two right-angled triangles within the problem. I’m right! right than you!
Examine the following two problems carefully and
then attempt Exercise 13:08.
worked examples
Example 1 P
1 A pole PT stands on the top of a building BT. From a
T point A, located 80 m from B, the angles of elevation of
the top of the building and the top of the pole are 43°
and 52° respectively. Find the height of the pole, PT,
correct to the nearest metre.
52° B
43°
A 80 m
Solution 1
1 To find the length of the pole PT, the lengths PB and TB are calculated and of course
PT = PB − TB.
In ∆PBA, P-8---0B-- = tan 52° In ∆TBA, -T8---0P-- = tan 43°
PB = 80 tan 52° TB = 80 tan 43°
Now PT = PB − TB
= 80 tan 52° − 80 tan 43°
= 80 (tan 52° − tan 43°) continued §§§
= 28 m (to the nearest metre)
405CHAPTER 13 TRIGONOMETRY
Example 2 N Q R
25° 6·7 km
P, Q and R are three villages. Q is 5 km and N25°E from P. 5 km
R is east of Q and is 6·7 km from P. What is the bearing of P
R from P, to the nearest degree?
Solution 2
To find the bearing of R from P, we need to find the size of
angle NPR. In ∆NPR we know the length of PR, but we need
to know one of the other sides, either NR or NP. Side NP
can be calculated using ∆NPQ.
In ∆NPQ: -N--5--P-- = cos 25°
ie NP = 5 cos 25°
In ∆NPR: cos ∠NPR = -6N---·-P-7-
= 5-----c---6o---s·--7--2---5---°-
= 0·676 349
∴ ∠NPR = 47° (to the nearest degree)
∴ The bearing of R from P is N47°E.
Exercise 13:08 Foundation Worksheet 13:08
1 The top of a 20-metre tower is observed from two positions, Problems with more than
A and B, each in line with, but on opposite sides of, the one triangle
tower. If the angle of elevation from A is 27° and from 1A
B is 35°, how far is point A from point B (to the nearest metre)?
B 30° x 2y0° C
D
a Use ∆ABD to find x.
b Use ∆ADC to find y.
20 m 2 ay
x
25°
35° 27° 30°
B A 20
B Use the fact that a = y − x to
find the value of a.
2 In triangle ABC, BD is perpendicular to
AC. Given that AB = 13 m, BD = 11 m
and DC = 10 m, find, to the nearest A D C
degree, the size of angle ABC.
3 Two points, P and Q, are in line with the foot of a tower 25 m high. T
The angle of depression from the top of the tower to P is 43° and to 43°
57°
Q is 57°. How far apart are the points? (Answer to the nearest metre.)
25 m
B QP
406 INTERNATIONAL MATHEMATICS 4
4 A plane is flying at an altitude of 900 m. From a
point P on the ground, the angle of elevation to the
900 m 68° 30′ plane was 68°30′ and 20 seconds later the angle of
25° 12′ elevation from P had changed to 25°12′. How far had
the plane flown in that time, and what was its speed,
to the nearest kilometre per hour? (Find the distance
to the nearest metre.)
5 Find x in each diagram. Give answers correct to 2 decimal places. Find a different
ab side first.
50 62° x
3·6 3·7
57° 40°
x d
c x
25° 40′ 40° 20′
x 2·5
6 aA b 3·5
2·5
5
50° 12′
B 3C 4 D
Find ∠CAD to the nearest minute. Find θ to the nearest minute.
7 a From ∆XWY, show that Y
XW = z cos X.
b From ∆ZWY, show that z x
ZW = x cos Z.
c Hence show that Z
y = z cos X + x cos Z. X W y
8 a Show that AM = c sin B. A
b Hence show that the area cb
of ∆ABC = 21-- ac sin B.
B MC 15
a 12·6
9 Two ladders are the same distance from the base of a wall. 407CHAPTER 13 TRIGONOMETRY
The longer ladder is 15 m long and makes an angle of 58°
with the ground. If the shorter ladder is 12·6 m long, what
angle does it make with the ground? (Answer to the nearest
degree.)
10 N A, B and C are three towns where A and B are due north of C.
N From a position X on a map, A has a bearing of N27°E and B
A has a bearing of N67°E. Town C is due east of X and 7·5 km
from it. Find the distance, correct to 1 decimal place, between
B A and B.
XC
11 Find the exact value of CE given that AE = 16. 16 C
30° E
A
45°
DB
12 In ∆ABC, AB = 12, C
∠CAB = 60° and
∠CBA = 75°. Find A 60° 12 75°
as exact values: B
a AC
b BC
c area of ∆ABC
un spot Fun Spot 13:08 | What small rivers flow into the Nile?
f
313:08
1
----2-3-
-43-
-13-
-12-
-14-
32
----1--
2
Work out the answer to each question and put the letter for that part in the box that is above
the correct answer.
What are the exact values of:
I cos 60° E (cos 30°)2
L (sin 30°)2 J tan 60°
V sin 60° N (tan 30°)2
S sin 45° E c---o---s--3--4---5---°-
U (sin 30°)2 + (cos 30°)2
Challenge worksheet 13:08 Three-dimensional problems
408 INTERNATIONAL MATHEMATICS 4
Mathematical Terms 13 mathematical ter
adjacent side (to a given angle) hypotenuse ms
• The side of a triangle which together with • The longest side in a right-angled triangle. 13
• The side which is not one of the arms of
the hypotenuse forms the arms of a given
angle. the right-angle in a right-angled triangle.
hypotenuse opposite side (to a given angle)
• The side of a triangle which is not one of
adjacent side
the arms of the given angle.
angle of depression
• When looking down, the angle between hypotenuse opposite
side
the line of sight and the horizontal.
similar triangles
horizontal • Two triangles that have the same shape but
angle of depression a different size.
line of sight • Triangles that can be changed into each
angle of elevation other by either an enlargement or
• When looking up, the angle between the reduction.
• Triangles that have matching angles equal.
line of sight and the horizontal. • Triangles where the ratio of matching sides
is constant.
line of sight sine ratio (of an angle θ)
angle of elevation
• The ratio -s--i--d---e----oh---p-y--p-p---oo---st--ei--t-n-e--u--a-s-n--e--g---l-e-----θ-
horizontal
• Abbreviated to sin θ.
bearing North
• An angle used to measure 120° tangent ratio (of an angle θ)
the direction of a line from A • The ratio s---i-s-d-i--de---e-a---do---jp--a-p-c---oe---ns--i-t-t--et--o--a---na---ng---lg-e--l-e--θ---θ--
north. • Abbreviated to tan θ.
• Bearings can be recorded in
two ways. trigonometric (trig.) ratios
eg 120° or S60°E • A collective name for different ratios of the
side lengths of right-angled triangles.
• The ratios have constant values for any
particular angle.
cosine ratio (of an angle θ) trigonometry
• The ratio s---i--d---e----a--h-d---yj--a-p-c--o-e---nt--e-t--n-t--o-u---s-a--e-n---g---l-e-----θ-- • A branch of mathematics, part of which
• Abbreviated to cos θ.
deals with the calculation of the sides and
angles of triangles.
1 The trigonometric ratios 4 Bearings 1 Mathematical terms 13
2 Finding sides 5 Bearings 2
3 Finding angles
409CHAPTER 13 TRIGONOMETRY
diagnostic test Diagnostic Test 13: | Trigonometry
13 • Each section of the diagnostic test has similar items that test a certain question type.
• Errors made will indicate areas of weakness.
• Each weakness should be treated by going back to the section listed.
1 Evaluate, correct to 4 decimal places: Section
13:04
a tan 75° b sin 23° c cos 68·3° d tan 48·25° 13:04
13:04
2 Evaluate, correct to 3 decimal places:
a sin 25°30′ b tan 59°09′ c cos 173°21′ d sin 342°12′ 13:01
3 If 0° р θ р 90°, find θ, to the nearest minute, given that:
a cos θ = 0·639 b sin θ = 0·741 c tan θ = 0·071
d tan θ = 3·46
4 Name the side asked for in each triangle with respect to θ.
aA Bb Qc Y
CP RX Z
adjacent side hypotenuse opposite side
5 State, as a fraction, the value of the ratio asked for. 13:03
13:05
ab c5 13:05
13 5 8 17 34
12 15
sin θ = . . . tan θ = . . . cos θ = . . .
6 Find x, correct to 1 decimal place. x 31°
ab 15
9
x
29°
cx d x
21° 30′ 8
16 43° 15′
7 Find a, correct to 1 decimal place.
ab a
43°
a 62°
15
6
410 INTERNATIONAL MATHEMATICS 4
c a d 50° 51′ Section
39° 07′ 13:06
9 7a
8 Evaluate θ, to the nearest minute.
ab
8m 5m 9m 17 m
c 6m d
5m 10 m
12 m
Chapter 13 | Revision Assignment assignm ent
1 Find the value of the pronumeral correct to b 13A
2 sig. figs.
ax 9·4 16·8
23°57′ c
14·7 5·1
b 5·6 4·8
17°37′ 3 a A ship’s captain measures the angle of
elevation of a lighthouse as 4°. If he
x knows that the lighthouse is 105 m above
the sea, how far is he from the coast (to
cx the nearest 100 m)?
43°05′
21·7
2 Find θ to the nearest minute. 105 m
a
4° x
7·2
3·9
411CHAPTER 13 TRIGONOMETRY
b A plane flies at a speed of 650 km/h. 5 From A, the T x
It starts from town A and flies on a bearing of a
bearing of 120° for 3 hours. At that
time, how far is it tower, T, is h
i south of A? 330°. From B,
ii east of A?
which is 10 km B
4 From the top, T, of a 135-metre cliff, the north of A, the
angles of depression of two cabins at A and
B are 23° and 42° respectively. How far bearing of the
apart are A and B, assuming that A, B and
X, the foot of the cliff, are collinear? tower is 290°. 10 km
(Answer to the nearest metre.)
a In the
diagram A
show that
i x = h tan 70°
ii x = (h + 10)tan 30°
T b Use the equations above to find x.
23° c Find the distance of the tower from B.
42°
A BX
Trigonometry ratios
assignment Chapter 13 | Working Mathematically
13B 1 Use ID Card 6 on page xviii to identify: 5 A solid is formed
a 2 b 3 c 4 d 10 e 11 from a cube by
f 12 g 14 h 15 i 16 j 17 cutting off the
corners in such
2 Use ID Card 7 on page xix to identify: a way that the
a 5 b 8 c 9 d 10 e 11 vertices of the
f 12 g 18 h 22 i 23 j 24 new solid will be
at the midpoints
3 Why is the 8 of the edges of the
diagram original cube.
shown 40° 20° If each of the new edges is a units long,
impossible? what is the surface area of the solid?
4 a If 6 men can do a piece of work in 8 6 Two shops sell the same drink for the same
days, in what time will 18 men do it, price per bottle. Shop A offers a 10%
working at the same rate? discount, while shop B offers 13 bottles for
the price of 12. Which shop offers the
b If 14 men can do a piece of work in 12 better discount if 12 bottles are bought?
days, how many men will be needed to
do the work in 21 days, working at the
same rate?
412 INTERNATIONAL MATHEMATICS 4
14
Vectors
What’s our vector Victor?
How should I know and who are
all these other birds following me?
Chapter Contents 14:04 Solving problems using vectors
Mathematical Terms, Diagnostic Test,
14:01 What is a vector? Revision Assignment
14:02 Column vectors and vector operations
Investigation: Operating on vectors
14:03 Magnitude of a vector
Practical Activity: Magnitude of a vector
Learning Outcomes
Students will be able to:
• Understand the definition of a vector.
• Identify relationships between vectors.
• Perform operations with vectors.
• Calculate the magnitude of a vector.
• Solve problems using vectors.
Areas of Interaction
Approaches to Learning (Knowledge Acquisition, Reflection), Human Ingenuity,
Environments
413
14:01 | What is a Vector?
In Chapter 8 you studied different aspects of intervals on the Cartesian plane.
D B All the intervals shown on this Cartesian plane
have one thing in common:
A
G They all have the same length:
C F AB = (5 – 1)2 + (5 – 2)2
E
=5
CD = (6 – 3)2 + (−5 – –1)2
H =5
EF = (1 – –2)2 + (−3 – 1)2
=5
GH = 5
Although the lengths of these intervals are the same, they are all distinct from one another.
I A vector is different from an interval because it not only has length, which is called its magnitude,
but also direction.
In order to represent this, an arrow is used to show the direction of a vector.
→d I D On the grid, five vectors are shown.
F All have magnitude represented by their length.
All have direction, represented by the arrow.
K →b
G Vectors can either be named by their endpoints
B →→ →→→
→a →e C →c
For example, AB , CD , EF , GH , IK
(Note how the arrow above the letters gives the
direction of the vector.)
AH Or by a single lower case letter either written
with an arrow above it, or in bold type.
E For example, →a , →b , →c , →d , →e or a, b, c, d, e
Some points to note on the vectors shown in the grid above:
• vector →a = vector →b since they both have the same magnitude and direction.
• vector →c = 2 × vector →a since →c has the same direction as vector →a but twice the magnitude.
• vector →e = −vector →a since it has the same magnitude as vector →a but goes the opposite
• vector →d = − -12- × vector →a direction.
since it has half the magnitude of vector →a and goes the opposite
direction.
So: →a = →b , →c = 2 →a , →e = − →a and →d = − -21- →a
Of course, you can use the other forms of notation instead if you wish.
414 INTERNATIONAL MATHEMATICS 4
Exercise 14:01 These vectors
have magnitude
1 Express each of the following in terms of w→ , →x and →y . and direction.
→y
x→ →n →
l
w→ m→
a → b m→ c →n
l
2 Write the relationship between the →a
→e
following pairs of vectors. →b
→d
a →a and →h b →a and →f →f
→h →k
c →b and → d →b and →k →c
→ →g
l
l
e →e and →c f →d and →g
3 On a grid, draw the following vectors from those given in the diagram in question 2.
a →2-a- b −2 →e c − 2-3- →h →
d 2--3--f
415CHAPTER 14 VECTORS
14:02 | Column Vectors and
Vector Operations
When we want to refer to a vector without drawing it, we can write it as a column vector.
To do this we must count how many units the vector goes horizontally and how many it goes
vertically.
For example, consider the vectors shown in the diagram. B m→ →n
A →k
→ →
For the vector AB to travel from the start of the vector to l
the finish it goes 1 horizontally and 3 vertically.
So → = 1 horizontal units on top
3 vertical units on the bottom
AB
The numbers in the brackets are called the components.
Vector m→ goes 4 units horizontally and −1 unit vertically
So m→ = 4
–1
Vector →n goes −2 units horizontally and −3 unit vertically
So →n = – 2 which can be written − 2
– 3 3
Vector →k goes −5 units horizontally and 0 unit vertically
So →k = –5
0
→
Vector l goes −2 units horizontally and −4 unit vertically
So → = – 2 which can be written − 2 or −2 1
– 4 4 2
l
inve stigation Investigation 14:02 | Operating on vectors
14:02
Please use the Assessment Grid on the page 418 to help you understand what is required for
this Investigation.
Each grid shows pairs of vectors, A→B and B→C joined end to end. This can represent a journey
from A to C. Complete the table showing the column vectors for both A→B and B→C and the
vector A→C (not drawn) which represents a shortcut from A to C bypassing B.
1 2
B B
A A
C
C
416 INTERNATIONAL MATHEMATICS 4
3 C 4 A
C
A B B
C B
5 6
A
A
BC
A→B B→C A→C
1 Is there a pattern forming between the
values in the column vectors for A→B and
2 B→C and the values in the column vector
3 for A→C ?
4 Write this pattern down as a general rule if
5 A→B = a and B→C = c
b d
6
Test out your general rule with three pairs of vectors of your own. →b
Now consider the two vectors →a and →b in this
grid. Suppose →a represents the direction and
distance an aeroplane travelled on one part of
a trip and →b the direction and distance it
travelled on the next part of its trip.
Unfortunately the navigator has been sloppy →a
and has not started the second part from
where the first finished.
How would you go about representing the
direction and distance they would have
travelled if it went from the start to the finish
in a straight line (ie taking a shortcut)?
Once you have a vector that represents what the journey would have been in a straight line, it
is possible to calculate how far from the starting point the plane finishes. It is also possible to
calculate in what direction it could have flown to get there in a straight line.
Assuming that 1 unit on the grid represents 100 km, calculate these two values.
In what other real-life situations could vectors be used?
417CHAPTER 14 VECTORS
Assessment Grid for Investigation 14:02 | Operating on vectors
The following is a sample assessment grid for this investigation. You should carefully read the
criteria before beginning the investigation so that you know what is required.
Assessment Criteria (B, C, D) for this investigation Achieved
0
a None of the descriptors below has been achieved. 1
2
b With some help, mathematical techniques have been applied 3
and the patterns in the table have been recognised. 4
Criterion B 5
Investigating Patterns Mathematical techniques have been selected and applied and
c the patterns in the table have been recognised and a general 6
rule suggested. 7
8
Mathematical techniques have been selected and applied and 0
1
d the patterns in the table have been recognised and described 2
as a general rule. Further examples are given in an effort to 3
provide justification. 4
Criterion C Further to (d), the rule that is identified is explained fully 5
Communication in Mathematics e and its application in the final problem is used as part of the
6
justification.
0
a None of the descriptors below has been achieved. 1
2
b There is a basic use of mathematical language and 3
representation. Lines of reasoning are hard to follow.
4
There is sufficient use of mathematical language and
5
c representation. Lines of reasoning are clear but not always
logical. Moving between diagrams and column vectors is 6
done with some success.
There is good use of mathematical language and
d representation. Lines of reasoning are clear, logical and
concise. Moving between diagrams and column vectors is
done effectively.
a None of the descriptors below has been achieved.
Criterion D Attempts have been made to explain whether the results in
Reflection in Mathematics b the table make sense and the importance of the findings in
a real-life context using the final problem.
A correct but brief explanation whether the results in the
c table make sense is given. A description of the importance
of the findings in a real-life context using the final problem
is given.
A critical explanation whether the results in the table make
sense is given. A detailed description of the importance of
d the findings in a real-life context using the final problem
is given. The significance of the findings has been
demonstrated in the solution to the final problem.
418 INTERNATIONAL MATHEMATICS 4
Vector Operations and Vector Geometry
In Investigation 14:02, you should have seen that to combine two vectors you place them end to
end, so that it is possible to travel from the beginning of the first vector to the end of the last vector.
Combining them in this way and drawing the shortcut from the start to the finish in a straight line
is the same as adding the components of the column vectors.
For example, if the original vectors →a and →b are to Resultant vector
be combined, we move →b so that it starts where →a →a + →b
ends; shown in blue.
When the start of the two vectors is joined to the
end of the two vectors, the resultant vector is the →a
sum of the two original vectors →a and →b , shown
in red. →b
When adding vectors on a diagram, they are
placed head to tail. The resultant vector, which →b
represents the sum, goes from the tail of the first
vector to the head of the last.
Alternatively the column vectors can be used:
For example, →a = 4 and →b = 4 , so →a + →b = 4 + 4 = 8
–6 2 – 6 + 2 –4
Check this result from the diagram.
worked examples
1 Vectors →a , →b and →c are shown in the diagram.
Add vectors
head to tail.
By using the diagram, draw the resultant vector for →a + →b + →c .
Solution →b
→c
Move →b and →c so that all three vectors are
arranged head to tail.
The resultant vector starts at the tail of →a and →a
goes straight to the end of →c . →a + →b + →c
So →a + →b + →c = 3
–1
To check, use the column vectors:
→a + →b + →c = 1 + 5 + – 3 = 3
4 –1 – 4 –1
continued §§§
419CHAPTER 14 VECTORS
2 The vectors →a and →b are shown in the diagram. →a →b
Use another vector diagram to show the resultant
vector of →a − →b .
Solution →b →a
→a →b
We can only add vectors so to do this problem,
instead of subtracting →b we add −→b .
Since →a − →b = →a + (−→b ).
So →a and −→b are placed head to tail to get the
resultant vector.
3 Use the diagram to express the following in
terms of →a , →b and →c .
B →b C
a D→A b A→D c A→C →a →c
d C→A e D→B f B→D
Solution A D
a D→A = −12 →b
Alternatively, to get from D to A we could
travel −→c then −→b then −→a so that
D→A = − →c − →b − →a
b A→D = − D→A
so A→D = −(−12→b ) = 12→b
or A→D = −(− →c − →b − →a ) = →a + →b + →c
c A→C = →a + →b , in other words, to get from A to C you must travel along →a then →b
d C→A = − A→C = −(→a + →b ) = − →a − →b
e D→B = − →c − →b
f B→D = − D→B = −(− →c − →b ) = →b + →c
420 INTERNATIONAL MATHEMATICS 4
Exercise 14:02
1 Represent the following vectors on a grid.
a →a = 5 b A→B = – 2 c d = – 3
3 1 – 4
d m→ = 6 e Y→X = 0 f f = 3
–4 5 0
2 From the vectors in question 1, calculate the resultant column vector for the following:
a →a + m→ b →a + d c A→B + Y→X d A→B − Y→X
e A→B + X→Y f Y→X − f g A→B − m→ h B→A − m→
i →a + m→ + f j 3(A→B + m→ )
3 By using the vectors shown in the grid and writing the vectors in column form, find the
resultant vector of the following.
a →a + m b P→Q − →c Q
c W→X + d d Q→P + →c →a →c
e m−d f →d − →a P
m
X
d
W
4 Express the vector →v in terms of other vectors →b
shown on the grid.
x→
v→ →a →t
5 Given that X and Y are the midpoints of AC and AB respectively, X A
express the following in terms of the vectors C→X and A→B . C Y
B
a C→A b X→A c A→Y d X→Y e B→C
6 Looking at the results for (d) and (e) from question 1, what conclusions can you draw about
the lines XY and CB?
421CHAPTER 14 VECTORS
7 Using the diagram, express the following in terms of A→B and C→A . A B
D C
a C→B b A→D c D→B D
C
8 Using the diagram, express the following in terms of →a and →b . →b
a A→B b C→D c B→D d →BE AB
Same direction →a E
and same size
gives equal
vectors.
9 The quadrilateral ABCD is made up of three equilateral triangles. By C
x
Express the following in terms of the vectors x and y. AED
a A→E b E→B c C→D d B→D
10 Show, with the aid of a diagram, how the vector →r = –3 can be formed by combining
the following vectors: 5
→p = 3 →s = –1 →q = 0
1 2 –2
→u = 6 →t = 4
–1 2
422 INTERNATIONAL MATHEMATICS 4
14:03 | Magnitude of a Vector
Practical Activity 14:03 | Magnitude of a vector
1 For each of the vectors shown on the grid, complete a right triangle as in the example
vector →a .
2 Complete the table below as for the example vector →a .
Vector Column Magnitude →a D →b c
vector 22 + 32 = 13 units
→a
→b 2
3
c →f
D→E g
→f
E
g
You should have discovered that for any vector →v = a the magnitude of the vector is given
b
by a2 + b2 .
When writing the magnitude of the vector →v we write →v sometimes called the modulus of
the vector.
I So the magnitude or modulus →v of the Pythagoras
a strikes again.
vector →v = b is given by →v = a2 + b2
423CHAPTER 14 VECTORS
Exercise 14:03
1 Find the magnitude of the vectors shown in the grid.
→c
→a
→b
→f
→d
→e
2 Find the magnitude of the following vectors.
a →a = 15 b →b = –6 c →c = – 20 d →d = 12 e →e = – 2
8 8 – 21 – 5 10
3 Evaluate the following if A→B = 5 , →t = –1 , →u = 1 .
–3 6 11
a |A→B + →t | b |→u + A→B | c |→t + →u | d |2→u − →t | e |→t − A→B |
424 INTERNATIONAL MATHEMATICS 4
14:04 | Solving Problems Using Vectors
Vectors are often used to represent objects that are in motion. The length of the vector represents
the magnitude of the motion (distance or speed), and the direction of the vector, the direction of
the object.
You will need to use trigonometry to solve these problems.
worked examples
1 A hiker walks on a bearing of 10° for a distance of 5 km and then on a bearing of 60° for
another 5 km. How far, and in what direction is he, from his starting point?
Solution →a →b
30°
Let the vectors →a and →b represent the first and second legs of the 80°
hike respectively. r→
To add the vectors to get the resultant vector →r we need to
calculate the vertical and horizontal components of →a and →b .
If →a = AAyx , then using trigonometry
cos 80 = A--5---x and sin 80 = A--5--y- since |→a | = |→b | = 5
∴ Ax = 5 cos 80 and Ay = 5 sin 80
∴ →a = 5 cos 80
5 sin 80
Similarly, if →b = BByx then Bx = 5 cos 30 and By = 5 sin 30
∴ →b = 5 cos 30
5 sin 30
The resultant vector →r = →a + →b
= 5 cos 80 + 5 cos30
5 sin 80 + 5 sin30
= 5·20
7·42
The distance from the starting position is given by |→r | = (5·20)2 + (7·42)2
= 9·06 km
The direction from the starting point can be found by considering the components of the
resultant vector as a triangle.
tan θ = -75---··-42---02-- r→ 7.42
5.20
θ = tan−1 5-7---··-42---02--
θ = 55°
As a result, after his hike, the hiker is 9·06 km from his starting point on a bearing of 45°.
continued §§§
425CHAPTER 14 VECTORS
2 An aeroplane is flying at a ground speed of 200 km/h on a bearing of 300°. A crosswind is
blowing at 50 km/h on a bearing of 50°. Use vectors to calculate the resultant velocity and
direction of the aeroplane.
Solution w→ Wy
Wx
Here the magnitude of the vector represents the speed. 40°
→p
The vectors →p and w→ must be resolved into their vertical and horizontal parts.
30°
→p PPyx and w→ WWyx Py Px
= =
→p = – →p cos 30 ° and w→ = w→ cos 40 °
→p sin30° w→ sin 40 °
where →p = 200 km/h and w→ = 50 km/h
∴ →p = –200 cos 30° and w→ = 50 cos 40°
200 sin 30° 50 sin 40°
So the resultant vector →r which is →p + w→ = –200 cos30° + 50 cos40° 132.14
200 sin30° + 50 sin40°
θ
Note: It is important that these = –134 ·91 134.91
signs are correct so that the final 132·14
direction can be worked out.
Therefore the resulting velocity is given by →r = (–134·91)2 + (132·14)2
= 188·8 km/h
The direction of the plane is given by 270° − θ, where tan θ = 11----33---42---··--91---14-
So that θ = tan−1 1-1---33---42---··--19---41-
= 44·4°
In conclusion, the aeroplane is flying at 188·8 km/h on a bearing of 314·4°.
Exercise 14:04 N
1 A yacht sails on a bearing of 100° for a distance of 54 nautical miles 20°
and then on a bearing of 40° for a distance of 80 nautical miles. 50°
How far (to the nearest nautical mile) and in what direction
(to the nearest degree), is the yacht from its starting point?
(Remember: from the diagram, the first y component
is negative.)
426 INTERNATIONAL MATHEMATICS 4
2 Crazy Ivan has set off to cross the Gobi desert. He first walked for
35 km on a bearing of 200° and realised he was heading the wrong
way so he then walked for 20 km on a bearing of 310°. How far, 20°
and in what direction (to the nearest whole degree) is he now from
his starting point? 40°
(Remember: from the diagram, the first x and y components are
negative and the second x component is negative.)
3 A plane flew first on a bearing of 320º for 150 km and 50°
then due east for 150 km. How far, and in what
direction, is the plane from its starting point? 10°
(Remember: due east only has an x component.) 20°
4 A balloonist is at the mercy of the wind. When he first
takes off, the wind blows him on a bearing of 100º for
80 km and then on a bearing of 200º for 100 km.
What is the balloonist’s distance and direction from
the starting position?
5 A canoeist is paddling due north across a river which N
runs from east to west.
River direction Canoe
If the canoeist is paddling at 3 km/h and the river is direction
flowing at 5 km/h, find the resultant speed and direction
of the canoe.
6 A plane is flying on a bearing of 050º at a ground speed 50°
of 180 km/h. The wind is blowing on a bearing of 200º 50°
at a speed of 50 km/h. Find the resulting speed and
direction of the plane.
(Answer to the nearest whole km/h and degree.)
7 A flock of seagulls is flying on a bearing of 135° at 45°
a ground speed of 8 km/h. The wind is blowing on a 20°
bearing of 70° at a speed of 5 km/h. Find the resulting
speed and direction of the seagulls.
(Answer with speed correct to 3 significant figures and
the bearing to the nearest degree.)
427CHAPTER 14 VECTORS
8 A woman is rowing across a 600 m wide river, starting at 5 km/h B
point A, at a speed of 3 km/h. The current is moving at 5 km/h.
If the woman starts rowing straight across the river: 3 km/h
a how far downstream from point B will she finish? A
b what will be her resulting speed (3 significant figures) and
direction (nearest whole degree) if B is due north of A?
hematical terms Mathematical Terms 14
mat 14 interval magnitude of a vector
• Represented by a straight line joining two
• The length of a vector.
points.
Has only one dimension: length. Denoted by | |
For example the magnitude of vector
vector →v = x is given by |→v | = x2 + b2
• Represented by an arrow joining two y
points. modulus of a vector
Has two dimensions: length and • The same as the magnitude of a vector.
direction.
bearing
column vector • The direction in which an object is from
• Represents a vector by its horizontal and some starting position.
Always measured clockwise from north
vertical component parts. and given as a three digit number.
x
Is written in the form y
resultant vector
• The result of putting a number of vectors
head to tail in order to add them.
Goes from the tail of the first vector
to the head of the last vector.
Mathematical terms 14
428 INTERNATIONAL MATHEMATICS 4
Diagnostic Test 14: | Vectors diagnostic test
• Each section of the diagnostic test has similar items that test a certain question type. 14
• Errors made will indicate areas of weakness.
• Each weakness should be treated by going back to the section listed.
1 Express each of the following vectors in terms of →v and w→ . Section
14:01
a →a b →b c →c
d →d e →e f →f
g →g h →h i →i
v→ w→ →a
→e →h
→d →b
→f →c
→g
→i
2 Using the vectors →v and w→ shown, draw the following on grid paper. 14:01
a →a = →v + w→ b →b = →v − w→ c →c = −2 →v
d →d = −2 →v + w→ e →e = w→ − →v f →f = -21- →v − w→
v→ w→
3 Represent the vectors in the grid as column vectors. 14:02
(d)
(a) (b) (c)
(e) (f )
(g) (h) (i)
429CHAPTER 14 VECTORS
4 If →a = 3 , →b = 0 and →c = – 4 write the resultant vector for each Section
–1 5 – 4 14:02
of the following: 14:03
a →a + →b + →c b →a − →b c 2 →a − →c 14:03
d →c − 2 →b e 2(→b − →a ) f →c − →b − →a 14:04
5 If →a = 15 , →b = 6 and →c = 20 evaluate the following correct to
– 8 10 21
3 significant figures.
a |→a | b |→b | c |→c |
d |2→a | e |→a + →b | f |→c − →a |
6 Find the value of x in the following. Answer correct to 3 significant
figures where necessary.
a 3 =5 b 12 = 13 c x = 12
x x 9
d 5 =x e 17 = 28
8 x
7 Solve the following problems by first drawing 30°
vector diagrams. 20°
a A plane, flying in still air flies on a bearing of
120° for 150 km and then on a bearing of 250°
for 200 km. What distance and bearing is the
plane from its starting position? (Answer to
the nearest metre and whole degree.)
b A yacht is sailing at a still water speed of 50°
10 knots on a bearing of 330°. The ocean 30°
current is running at a speed of 3 knots on a
bearing of 040°. What is the resulting speed
and bearing of the yacht? (Answer both to the
nearest whole number.)
c A plane is flying on a bearing of 210° at 30°
a ground speed of 200 km/h. The wind is
blowing at 25 km/h on a bearing of 280°.
Find the resulting speed (correct to 1 decimal
place) and bearing (to the nearest whole
degree) of the plane.
10°
430 INTERNATIONAL MATHEMATICS 4