Cambridge 12 Standard 1 - CH 4 - Right-Angled Triangles
Cambridge 12 Standard 1 - CH 4 - Right-Angled Triangles
Cambridge 12 Standard 1 - CH 4 - Right-Angled Triangles
triangles
Syllabus topic — M3 Right-angled triangles
This topic is focused on solving problems involving right-angled triangles in a variety
of contexts.
Outcomes
• Find unknown sides using Pythagoras’ theorem.
• Solve problems using Pythagoras’ theorem.
• Define the trigonometric ratios – sine, cosine and tangent.
• Use a calculator in trigonometry with angles to the nearest minute.
• Find unknown sides using trigonometry.
• Find unknown angles using trigonometry.
• Using trigonometry to solve practical problems.
• Solve problems involving compass and true bearings.
• Solve problems involving angles of elevation and depression.
Knowledge check
The Interactive Textbook provides a test of prior knowledge for this chapter, and
may direct you to revision from the previous years’ work.
4A Pythagoras’ theorem
Pythagoras’ theorem links the sides of a right-angled triangle. In a right-angled Hypotenuse
triangle the side opposite the right angle is called the hypotenuse. The hypotenuse (longest side)
is always the longest side.
PYTHAGORAS’ THEOREM
Pythagoras’ theorem is used to find a missing side of a right-angled triangle if two of the sides are
given. It can also be used to prove that a triangle is right angled.
12 mm 5 mm
SOLUTI ON :
1 Write Pythagoras’ theorem. h 2 = a2 + b2
2 Substitute the length of the sides. 122 = x2 + 52
3 Make x2 the subject. x2 = 122 − 52
4 Take the square root to find x. x = √122 − 52
5 Express the answer to correct one decimal place. ≈ 10.9 mm
Exercise 4A
Example 1 1 Find the length of the hypotenuse, correct to one decimal place.
a b 5 cm c
h cm
6 cm
12 cm 24 mm h mm
8 cm h cm
10 mm
d e f 54 cm
h cm
2.5 cm
h mm 20 mm
63.2 cm h cm
4.2 cm
10 mm
a 15 cm b 15 cm c
x cm
12 cm
21 cm x mm
x cm 13 mm
6 mm
d e 2.3 cm f
x cm 14.1 cm
x cm
24 mm 32 mm 9.5 cm
4.8 cm
x mm
3 Calculate the length of the side marked with the pronumeral. (Answer to the nearest
millimetre.)
a y mm b c 16 mm
30 mm
63 mm
42 mm x mm 28 mm
35 mm
a mm
d 10 mm e 33 mm f
m mm
8 mm
d mm
52 mm 12 mm
27 mm b mm
g h 3.3 mm i 15.7 mm
22.3 mm k mm
y mm 2.3 mm
26 mm x mm
10 mm
4 Find, correct to one decimal place, the length of the diagonal of a rectangle with dimensions
7.5 metres by 5.0 metres.
5.0 m
7.5 m
7 cm y cm x cm
4 cm 6 cm
SOLUTI ON :
1 Draw a diagram and label the information from the
question.
200 m xm
2 Label the hypotenuse with x. This represents the
distance of the helicopter from the landing pad.
320 m
3 Write Pythagoras’ theorem. x2 = a2 + b2
4 Substitute the length of the sides. = 3202 + 2002
5 Take the square root to find x. x = √3202 + 2002
6 Express the answer correct to two decimal places. ≈ 377.36 m
7 Write your answer in words. ∴ The helicopter is 377.36 metres
from the landing pad.
Exercise 4B
Example 3 1 A farm gate that is 1.6 m high is supported by a diagonal bar of
length 3.0 m. Find the width of the gate, correct to one decimal 3.0 m
1.6 m
place.
2 A ladder rests against a brick wall as shown in the diagram on the right.
The base of the ladder is 2.0 m from the wall, and reaches 3.4 m up the
3.4 m
wall. Find the length of the ladder, correct to one decimal place.
2.0 m
3 The base of a ladder leaning against a wall is 1.5 m from the base of the wall.
If the ladder is 4.5 m long, find how high the top of the ladder is from the
4.5 m
ground, correct to one decimal place.
1.5 m
15 m
B C
25 m
7 Use the measurements on the diagram to determine the distance the boat is
out to sea. (Answer correct to the nearest metre.)
50 m 70 m
8 The hypotenuse of a right-angled triangle is 40 cm long and one of the shorter sides measures
20 cm. What is the length of the remaining side in the triangle? (Answer correct to two decimal
places.)
4C Trigonometric ratios
Trigonometric ratios are defined using the sides of a right-angled triangle. The hypotenuse is
opposite the right angle, the opposite side is opposite the angle θ and the adjacent side is the
remaining side.
Hypotenuse
Opposite
θ
Adjacent
The opposite and adjacent sides are located in relation to the position of angle θ . If θ was in the
other angle, the sides would swap their labels. The letter θ is the Greek letter theta. It is commonly
used to label an angle.
What are the values of the hypotenuse, the opposite side and the adjacent
θ
side of the triangle shown? 3 5
4
SOLUTI ON :
1 Hypotenuse is opposite the right angle. Hypotenuse is 5 (h = 5)
2 Opposite side is opposite the angle θ . Opposite side is 4 (o = 5)
3 Adjacent side is beside the angle θ , but not the Adjacent side is 3 (a = 5)
hypotenuse.
TRIGONOMETRIC RATIOS
6 3
4 2
2 1
30° 30° 30°
The three triangles drawn above show the ratio of the opposite side to the hypotenuse as 0.5.
(2 4 6)
1 2 3
, or . This is called the sine ratio. All right-angled triangles with an angle of 30° have a sine
ratio of 0.5. If the angle is not 30° the ratio will be different, but any two right-angled triangles with
the same angle will have the same value for their sine ratio.
Similarly, the three triangles drawn below show the ratio of the opposite side to the adjacent side as 1
(1 2 3)
1 2 3
, or . This is called the tangent ratio. All right-angled triangles with an angle of 45° have a
tangent ratio of 1.
3
2
1
45° 45° 45°
1 2 3
The ratio of the opposite side to the hypotenuse (sine ratio), the ratio of the adjacent side to the
hypotenuse (cosine ratio) and the ratio of the opposite side to the adjacent side (tangent ratio) will
always be constant irrespective of the size of the right-angled triangle.
Find the sine, cosine and tangent ratios for angle θ in the triangle shown. 15
θ
8
17
SOLUTI ON :
1 Name the sides of the right-angled triangle. a (15), o (8), h (17)
o
2 Write the sine ratio (SOH). sin θ =
h
3 Substitute the values for the opposite side and the 8
hypotenuse. =
17
a
4 Write the cosine ratio (CAH). cos θ =
h
5 Substitute the values for the adjacent side and the
15
hypotenuse. =
17
o
6 Write the tangent ratio (TOA). tan θ =
a
7 Substitute the values for the adjacent side and the
8
opposite side. =
15
6
Find sin θ in simplest form given tan θ = .
8
SOLUTI ON :
1 Draw a triangle and label the opposite and adjacent
sides.
6
2 Find the hypotenuse using Pythagoras’ theorem.
θ
3 Substitute the length of the sides into Pythagoras’ 8
theorem. h 2 = 62 + 82
4 Take the square root to find the hypotenuse (h). h = √62 + 82
= 10
5 Evaluate.
o
6 Write the sine ratio (SOH). sin θ =
h
7 Substitute the values for the opposite side and the 6
=
hypotenuse. 10
8 Simplify the ratio. 3
=
5
Exercise 4C
Example 4 1 State the values of the hypotenuse, opposite side and adjacent side in each triangle.
a 8 b 13 c 4
θ 3
6 θ
10 12
θ 5 5
d 39 e 9 f θ
θ 30
18
36
15 12
15 24
2 State the values of the hypotenuse, opposite side and adjacent side in each triangle.
a x b c
c f
e
θ a
y θ
z
b d
θ
Example 5 3 Write the ratios for sin θ , cosθ and tan θ for each triangle in question 1.
4 Write the ratios for sin θ, cosθ and tan θ for each triangle in question 2.
a 26 b
θ
12
10 θ 20
24
16
10 12
i i
26 20
24 16
ii ii
26 12
10 16
iii iii
24 20
6 Find the sine, cosine and tangent ratios in simplest form for each angle.
a i angle θ 11 b i angle θ 32
θ
ii angle ϕ ϕ ii angle ϕ
60 40
24
61 ϕ
θ
c i angle θ 45 d i angle θ
θ
ii angle ϕ θ ii angle ϕ r p
24 ϕ
51 ϕ
q
7 Find the sine and cosine ratios in simplest form for angle A and B for each triangle.
a A C 20
b A
34
16 15
25
C 30 B B
Example 6 8 Draw a right-angled triangle for each of the following trigonometric ratios and
i find the length of the third side
ii find the other two trigonometric ratios in simplest form.
3
a tan θ =
4
8
b sin θ =
10
7
c cos θ =
25
3
2
a Measure the length of the hypotenuse, adjacent and opposite sides in each triangle.
b What is the value of the cosine ratio for 60° in both triangles?
c What is the value of the sine ratio for 60° in both triangles?
DEGREES MINUTES
SOLUTI ON :
4
a Given sin θ = , find the value of θ to the nearest degree.
5
1
b Given cos θ = , find the value of θ to the nearest degree.
√2
SOLUTI ON :
4 4
1 Press SHIFT │ sin−1 │ │ = │ or │ exe │. a sin θ =
5 5
θ = 53.130102 … ≈ 53°
1 b cos θ = 1
2 Press SHIFT │ cos−1 │ │ = │ or │ exe │.
√2 √2
θ = 45°
Exercise 4D
1 What is the value of the following angles in minutes?
a 1° b 3° c 5° d 7°
e 10° f 15° g 20° h 60°
j °
1
i 0.5° k 0.2° l 0.25°
3
Example 7a 3 Find the value of the following trigonometric ratios, correct to two decimal places.
a sin 20° b cos 43°
c tan 65° d cos 72°
e tan 13° f sin 82°
g cos 15° h tan 48°
Example 7b 4 Find the value of the following trigonometric ratios, correct to two decimal places.
a cos 63°30′ b sin 40°10′ c cos 52°45′ d tan 35°23′
e sin 22°56′ f tan 53°42′ g tan 68°2′ h cos 65°57′
Example 7c 5 Find the value of the following trigonometric ratios, correct to one decimal place.
a 4 cos 30° b 3 tan 53° c 5 sin 74° d 6 sin 82°
e 6 tan 77° f 2 cos 43° g 8 sin 12° h 9 tan 54°
6 Find the value of the following trigonometric ratios, correct to one decimal place.
a 4 sin 65°20′ b 5 tan 23°55′ c 12 cos10°41′ d 8 sin 21°9′
e 11 sin 21°30′ f 7 cos 32°40′ g 4 sin 25°12′ h 8 tan 39°24′
Example 7d 7 Find the value of the following trigonometric ratios, correct to two decimal places.
5 1 12 3
a b c d
tan 40° sin 63° cos 25° sin 42°
4 5 6 7
e f g h
cos 38°9′ tan 72°36′ sin 55°48′ cos 71°16′
Example 8a 8 Given the following trigonometric ratios, find the value of θ to the nearest degree.
a sin θ = 0.5673 b cos θ = 0.1623 c tan θ = 0.2782
d cos θ = 0.7843 e tan θ = 0.5047 f sin θ = 0.1298
Example 8b 9 Given the following trigonometric ratios, find the value of θ to the nearest minute.
a tan θ = 0.3891 b sin θ = 0.6456
c cos θ = 0.1432 d sin θ = 0.8651
e cos θ = 0.3810 f tan θ = 0.8922
Example 9a 10 Given the following trigonometric ratios, find the value of θ to the nearest degree.
3 1
a tan θ = b sin θ =
4 2
5 1
c cos θ = d cos θ =
8 4
3 1
e sin θ = f tan θ = 1
5 3
Example 9b 11 Given the following trigonometric ratios, find the value of θ to the nearest degree.
√3 1
a sin θ = b tan θ =
2 √5
√5 4
c cos θ = d tan θ =
6 √6
√3 1
e cos θ = f sin θ =
2 √2
12 Given the following trigonometric ratios, find the value of θ to the nearest minute.
2 √3
a cos θ = b sin θ =
√7 4
√5 √7
c tan θ = d sin θ =
12 7
√2 3
e tan θ = f cos θ =
7 √11
13 Given that sin θ = 0.4 and angle θ is less than 90°, find the value of:
a θ to the nearest degree
b cos θ correct to one decimal place
c tan θ correct to two decimal places.
14 Given that cos θ = 0.8 and angle θ is less than 90°, find the value of:
a θ to the nearest degree
b sin θ correct to one decimal place
c tan θ correct to two decimal places.
15 Given that tan θ = 2.1 and angle θ is less than 90°, find the value of:
a θ to the nearest minute
b sin θ correct to three decimal places
c cos θ correct to four decimal places.
1 Name the sides of the triangle – h for hypotenuse, o for opposite and a for adjacent.
2 Use the given side and unknown side x to determine the trigonometric ratio. The mnemonic
SOH CAH TOA helps with this step.
3 Rearrange the equation to make the unknown side x the subject.
4 Use the calculator to find x. Remember to check the calculator is set up for degrees.
5 Write the answer to the specified level of accuracy.
Find the length of the unknown side x in the triangle shown. Answer correct to
three decimal places. x
SOLUTI ON : 35°
20
1 Name the sides of the right-angled triangle. a (20), o (x), h
o
2 Determine the ratio (TOA). tan θ =
a
x
3 Substitute the known values. tan 35° =
20
4 Multiply both sides of the equation by 20. x = 20 × tan 35°
5 Press 20 │ tan │ 35 │ = │ = 14.004150...
6 Write the answer correct to three decimal places. ≈ 14.004
Find the length of the unknown side x in the triangle shown. Answer correct
to two decimal places. 25 x
SOLUTI ON : 34°
1 Name the sides of the right-angled triangle. a, o (x), h (25)
o
2 Determine the ratio (SOH). sin θ =
h
x
3 Substitute the known values. sin 34° =
25
4 Multiply both sides of the equation by 25. x = 25 × sin 34°
5 Press 25 │ sin │ 34 │ = │ = 13.979822...
6 Write the answer correct to two decimal places. ≈ 13.98
To solve these types of equations, multiply both sides by x. Then divide both sides by the
trigonometric expression (sin 60°) to make x the subject.
40°
12
SOLUTI ON :
1 Name the sides of the right-angled triangle. a (12), o, h (x)
a
2 Determine the ratio (CAH). cos θ =
h
3 Substitute the known values.
12
cos 40° =
x
4 Multiply both sides of the equation by x. x × cos 40° = 12
12
5 Divide both sides by cos 40°. x=
cos 40°
6 Press 12 ÷ │ cos │ 40│ = │or │ exe │ x = 15.664 887 47
7 Write the answer correct to two decimal ≈ 15.66
places.
Exercise 4E
Example 10 1 Find the length of the unknown side x in each triangle, correct to two decimal places.
a 15 b c 35
31° x
24°
x x
23
42°
d 9 e 76 f 34
43° x
20°
60°
x
x
g h 60 i 7
55
x 48.4°
26.7°
x x
21.1°
Example 11 2 Find the length of the unknown side x in each triangle, correct to two decimal places.
a 47 b c 31
x
58°45′ 67 44°35′
x
x 26°20′
d e 29 f 5
38°9′ 35°5′ 18°8′
x
27 x
x
Example 12 3 Find the length of the unknown side x in each triangle, correct to two decimal places.
a 34 b x c x
48° 63°
33°
77
6
x
d x e 90 f 11
47°
70°
42 x x
44°
4 Find the length of the unknown side x in each triangle, correct to one decimal place.
a x b c x
87
25°6′ 68°15′
29 x 47
16°50′
d x e f
32°1′
28°7′ 49°53′
12 51
x x 47
5 Find the length of the unknown side x in each triangle, correct to three decimal places.
a b 21.7° c
12.5°
80.9°
x x 64
39 x
42
1 Name the sides of the triangle – h for hypotenuse, o for opposite and a for adjacent.
2 Use the given sides and unknown angle θ to determine the trigonometric ratio. The
mnemonic SOH CAH TOA helps with this step.
3 Rearrange the equation to make the unknown angle θ the subject.
4 Use the calculator to find θ . Remember to check the calculator is set up for degrees.
5 Write the answer to the specified level of accuracy.
SOLU TI ON:
1 Name the sides of the right-angled triangle. a h (24), o (16), a
o
2 Determine the ratio (SOH). sin θ =
h
16
3 Substitute the known values. sin θ =
24
(24)
16
4 Make θ the subject of the equation. θ = sin−1
(6.8)
4.2
10 Make θ the subject of the equation. θ = cos−1
Exercise 4F
Example 13 1 Find the unknown angle θ in each triangle. Answer correct to the nearest degree.
a 55 b 12 c 57
θ 26
θ θ
40
13
d θ 38 e 26 f θ
15
19
33 16
θ
g 37 h θ i 57
9
θ 5 40
42 θ
2 Find the unknown angle θ in each triangle. Answer correct to the nearest degree.
50.6
d 9.8 e f
3.3
98.3
8.3 θ
θ
θ 4.5
67.5
3 Find the unknown angle θ in each triangle. Answer correct to the nearest minute.
a 8 b c 22
θ 19 θ
5
16 θ
25
61 e f 13
d
θ 43
7 θ
66
θ
47
4 Find the unknown angle θ in each triangle. Answer correct to the nearest degree.
a b 1 23 c
2 14
4 12 7 θ
8 θ
4 35
θ
3 34
5 Find the unknown angle θ in each triangle. Answer correct to the nearest degree.
a 3 b 17 c 29
θ
θ θ
20
4 15 21
5 8
6 Find the angle θ and ϕ in each triangle. Answer correct to the nearest minute.
a 26 b 30 c 8
θ 16 θ
ϕ θ
10 ϕ
24 34 6
10
ϕ
A vertical tent pole is supported by a rope tied to the top of the pole
and to a peg on the ground. The rope is 3 m in length and makes an
angle of 29° to the horizontal. What is the height of the tent pole?
Answer correct to two decimal places.
3m
29°
SOLUTI ON :
1 Draw a diagram and label the required height as x.
3m
x
29°
4.5 m
4m
SOLUTI ON :
1 Draw a diagram and label the required angle as θ.
4.5 m
θ
4m
2 Name the sides of the right-angled triangle. a (4 m), o, h (4.5 m)
a
3 Determine the ratio (CAH). cos θ =
h
4
4 Substitute the known values. cos θ =
4.5
(4.5)
4
5 Make θ the subject of the equation. θ = cos−1
Exercise 4G
Example 14 1 A balloon is tied to a string 25 m long. The other end of the string is secured
by a peg to the surface of a level sports field. The wind blows so that the 25 m
string forms a straight line making an angle of 37° with the ground. Find
the height of the balloon above the ground. Answer correct to one decimal 37°
place.
2 A pole is supported by a wire that runs from the top of the pole to a point on
the level ground 5 m from the base of the pole. The wire makes an angle of
42° with the ground. Find the height of the pole, correct to two decimal
places.
42°
5m
3 Ann noticed a tree was directly opposite her on the far bank of the Tree
river. After she walked 50 m along the side of the river, she found
her line of sight to the tree made an angle of 39° with the river bank.
Find the width of the river, to the nearest metre. 39°
50 m Ann
6 A 3.5 m ladder has its foot 2.5 m out from the base of a wall. What angle does
the ladder make with the ground? Answer correct to the nearest degree.
3.5 m
2.5 m
7 A plane maintains a flight path of 19° with the horizontal after it takes off. It travels for 4 km
along the flight path. Find, correct to one decimal place:
a the horizontal distance of the plane from its take-off point
b the height of the plane above ground level.
9 A shooter 80 m from a target and level with it, aims 2 m above the bullseye and hits it. What is
the angle, to the nearest minute, that his rifle is inclined to the line of sight from his eye to the
target?
2m
shooter
target
80 m
10 A rope needs to be fixed with one end attached to the top of a 6 m vertical pole and the other
end pegged at an angle of 65° with the level ground. Find the required length of rope. Answer
correct to one decimal place.
6m
65°
11 Two ladders are the same distance up the wall. The shorter ladder
is 5 m long and makes an angle of 50° with the ground. The longer
ladder is 7 m long. Find: 7m 5m
a the distance the ladders are up the wall, correct to two decimal
50°
places
b the angle the longer ladder makes with the ground, correct to the nearest degree.
12 A pole is supported by a wire that runs from the top of the pole to a point on the level ground
7.2 m from the base of the pole. The height of the pole is 5.6 m. Find the angle, to the nearest
degree, that the wire makes with the ground.
Horizontal
Angle of θ Angle of
θ elevation depression
Horizontal
41°
7 km
2 Determine the ratio (TOA). o
tan θ =
a
x
3 Substitute the known values. tan 41° =
7
4 Multiply both sides of the equation by 7. 7 × tan 41° = x
x = 7 × tan 41°
5 Write the answer correct to two decimal places. x = 6.085 007 165
≈ 6.09
6 Write the answer in words. The height of the volcanic plume
was 6.09 km.
The top of a cliff is 85 m above sea level. Minh saw a tall ship.
He estimated the angle of depression to be 17°.
a How far was the ship from the base of the cliff? Answer to
the nearest metre.
b How far is the ship in a straight line from the top of the cliff?
Answer to the nearest metre.
SOLUTI ON :
1 Draw a diagram and label the distance to the base of the a 17°
cliff as x and the distance to the top of the cliff as y. 85 m y
17°
x
o
2 Determine the ratio (TOA). tan θ =
a
85
3 Substitute the known values. tan 17° =
x
4 Multiply both sides of the equation by x. x × tan 17° = 85
85
5 Divide both sides by tan 17°. x=
tan 17°
= 278.022...
6 Write the answer correct to nearest metre. ≈ 278 m
7 Write the answer in words. ∴ The ship is 278 metres
from the base of the cliff.
8 Determine the ratio (SOH). o
b sin θ =
h
85
9 Substitute the known values. sin 17° =
y
10 Multiply both sides of the equation by y. y × sin 17° = 85
11 Divide both sides by sin 17°. 85
y=
sin 17°
12 Write the answer correct to nearest metre. = 290.7258...
≈ 291 m
13 Write the answer in words. ∴ The ship is 291 metres
from the top of the cliff.
Exercise 4H
Example 16 1 Luke walked 400 m away from the base of a tall building, on level ground.
He measured the angle of elevation to the top of the building to be 62°. Find the
height of the building. Answer correct to the nearest metre.
62°
400 m
2 The angle of depression from the top of a TV tower to a satellite dish near 59°
its base is 59°. The dish is 70 m from the centre of the tower’s base on flat x
land. Find the height of the tower. Answer correct to one decimal place.
70 m
Example 17 3 When Sarah looked from the top of a cliff 50 m high, she noticed a boat at 25°
an angle of depression of 25°. How far was the boat from the base of the
cliff? Answer correct to two decimal places. 50 m
5 The angle of elevation to the top of a tree is 51° at a distance of 45 m from the
point on level ground directly below the top of the tree. What is the height of
the tree? Answer correct to one decimal place.
51°
45 m
7 Jack measures the angle of elevation to the top of a tree from a point on level
ground as 35°. What is the height of the tree if Jack is 50 m from the base of
the tree? Answer to the nearest metre.
35°
50 m
800 m
θ
300 m
12 Find, to the nearest degree, the angle of elevation of a railway line that
rises 7 m for every 150 m along the track. 7m
150 m
15 The angle of elevation to the top of a tree from a point A on the ground is 25°. The point A is
22 m from the base of the tree. Find the height of the tree. Answer correct to nearest metre.
16 A plane is 460 m directly above one end of a 1200 m runway. Find the angle of depression to
the far end of the runway. Answer correct to the nearest minute.
Compass bearings
Compass bearings use the four directions of the compass: north, east, N
south and west (N, E, S and W). The NS line is vertical and the EW line NW NE
is horizontal. In-between these directions are another four directions:
north-east, south-east, south-west and north-west (NE, SE, SW and
W E
NW). Each of these directions makes an angle of 45° with the NS
and EW lines.
A direction is given using a compass bearing by stating the angle either SW SE
S
side of north or south. For example, a compass bearing of S50°W is
found by measuring an angle of 50° from the south direction towards the west side.
120°
W E
O
S OLU TI ON: S
1 Determine the quadrant of the compass bearing. a The line OA is in the north/east
quadrant.
2 Find the angle the direction makes with the vertical 30°
(north/south) line.
3 Write the compass bearing with N or S first, then the Compass bearing of A from O is
angle with the vertical line and finally either E or W. N30°E.
4 Determine the quadrant of the compass bearing. b The line OB is in the south/east
quadrant.
5 Find the angle the direction makes with the vertical 180° − 120° = 60°
(north/south) line.
6 Write the compass bearing with N or S first, then the Compass bearing of B from O is
angle with vertical line and finally either E or W. S60°E.
True bearings
A true bearing is the angle measured clockwise from north around to the required N
direction, and it is written with the letter T after the degree or minutes or seconds
symbol. True bearings are sometimes called three-figure bearings because they Clockwise
are written using three numbers or figures. For example, 120°T is the direction 120° from north
measured 120° clockwise from north. It is the same bearing as S60°E.
The smallest true bearing is 000°T and the largest true bearing is 360°T. The eight directions of the
compass have the following true bearings: north is 000°T, east is 090°T, south is 180°T, west is
270°T, north-east is 045°T, south-east is 135°T, south-west is 225°T and north-west is 315°T.
The bearings in the following diagrams are given using both methods.
N N
P
30°
170° Compass bearing Compass bearing
W E S 10°E True W E N 30°W True
330°
bearing 170°T bearing 330°T
10°
P
S S
A direction given by stating the angle A direction given by measuring the angle
either side of north or south, such as S60°E. clockwise from north to the required
direction, such as 120°T.
C S
S OL UTIO N:
1 Find the angle the bearing makes in the clockwise a 210°
direction with the north direction.
2 Write the true bearing using this angle. Add the letter ‘T’. C from O is 210°T.
3 Write the true bearing of west. b 270°T
4 Add angle between west and D to true bearing for west. 270° + 60° = 330°
5 Write the true bearing using this sum. Add the letter ‘T’. D from O is 330°T.
Exercise 4I
1 State the compass bearing and the true bearing of each of the following directions.
a NE b NW c SE d SW
Example 18, 19 2 State the compass bearing and the true bearing of each of the following directions.
N b N c N
a
A
A
52° 63°
W E W E W E
55°
A
S S S
N N f N
d e
A
A
20° 35°
W E W E W E
33°
A
S S S
g N h N i N
A
59°
A 78° 98°
A
j N N N
k l
A
240° 283° 250° A
A
3 Sketch each of these bearings on a separate diagram.
a N 10°E b S 25°W c N 60°W d S42°E
e 300° f 105° g 219° h 050°
N
Y
8 Riley travels from X to Y for 125 km on a bearing of N 32°W. 32°
a How far did Riley travel due north, to the nearest kilometre?
b How far did Riley travel due west, to the nearest kilometre? W E
X
Summary
Pythagoras’ Pythagoras’ theorem states that the square h
a
theorem of the hypotenuse is equal to the sum of the
squares of the other two sides (h2 = a2 + b2).
b
Applying 1 Read the question and underline the key terms.
Pythagoras’ 2 Draw a diagram and label the information from the question.
theorem 3 Decide whether to determine the hypotenuse or the length of a shorter side.
4 Use Pythagoras’ theorem to calculate a solution.
5 Check that the answer is reasonable and units are correct.
o
Trigonometric ratios sin θ = (SOH)
h Hypotenuse
a Opposite
cos θ = (CAH)
h
o θ
tan θ = (TOA) Adjacent
a
Using the calculator 1 degree = 60 minutes 1 minute = 60 seconds
in trigonometry 1° = 60′ 1′ = 60″
Finding an 1 Name the sides of the triangle.
unknown side 2 Use the given side and unknown side x to determine the trigonometric
ratio. The mnemonic SOH CAH TOA helps.
3 Rearrange the equation to make the unknown side x the subject then use
the calculator to find x.
Finding an 1 Name the sides of the triangle.
unknown angle 2 Use the given sides and unknown angle θ to determine the trigonometric
ratio. The mnemonic SOH CAH TOA helps.
3 Rearrange the equation to make the unknown angle θ the subject then use
the calculator to find θ .
Solving practical 1 Read the question and underline the key terms.
problems 2 Draw a diagram and label the information from the question.
3 Use trigonometry to calculate a solution.
Angles of elevation Horizontal
and depression Angle of θ Angle of
θ elevation depression
Horizontal
Bearings Compass bearing A direction given by stating the angle either side of north
or south such as S60°E.
True bearing A direction given by measuring the angle clockwise from north
to the required direction such as 120°.
Multiple-choice
Review
1 What is the length of the hypotenuse if the two other sides are 12 cm and 16 cm?
A 400 cm B 20 cm C 28 cm D 7.46 cm
(4) (5)
5 4 θ
A tan−1 B tan−1 5
(4) (5)
5 4
C tan D tan
7 What is the size of angle θ in question 6? (Answer correct to one decimal place.)
A 38.6° B 39.0° C 51.0° D 51.3°
8 What is the angle of elevation to the top of a tower 80 m tall and 100 m from the observer?
Answer in degrees correct to one decimal place.
A 51.3° B 51.4° C 38.6° D 38.7°
Short-answer
Review
1 Find the value of x, correct to two decimal places.
a xm b x mm c
21 m 27 cm 47 mm
28 m 14 mm 21 mm
x cm
x 6 mm
10 mm
24
7 Find the value of the following trigonometric ratios, correct to two decimal places.
a tan 68° b cos 13° c sin 23° d cos 82°
8 Given the following trigonometric ratios, find the value of θ to the nearest degree.
1
a cos θ = 0.4829 b sin θ = c tan θ = 0.2
3
a b c x
25 m
56° 56°
35 m 34°
x 17 mm
x
d x e f x
x
65°15´
8 63°
36
47° 24
10 Find the unknown angle θ in each triangle. Answer correct to the nearest minute.
a 17 b 34 c
θ θ 41
8
16
32 θ
11 e 10 f
d
θ
12 15
10 8
θ θ
9
11 Susan looked from the top of a cliff, 62 m high, and noticed a ship at
an angle of depression of 31°. How far was the ship from the base of
the cliff? Answer correct to one decimal place.
31°
62 m