1
Label the sides relative to the angle you know.
have adjacent, want opposite
tangent
the commonest case
have hypotenuse, want opposite
sine
ladders, ramps, wires
have hypotenuse, want adjacent
cosine
shadows along the ground
Opposite and adjacent are relative to the angle, not fixed to the picture. Mark the
angle first, then label the three sides from where it sits.
2
Elevation looks up; depression looks down; both measure from the horizontal.
string 60 m is the hypotenuse
the slanted side
height is opposite the 40°
so use sine
h = 60 sin 40° = 60(0.6428)
substitute
≈ 38.6 m
less than the string, as it must be
The height can never exceed the string length. A sanity check of that kind catches a
misapplied tangent immediately, since tangent would have given 50.3 m.
Tangent is the right ratio when the distance along the GROUND is known instead: standing
30 m from a pole, with an angle of elevation of 50° to its top, the height is opposite
and the 30 m adjacent, so h = 30 tan 50° = 30(1.1918) ≈ 35.8 m.
3
Bearings are measured clockwise from north.
N 30° E
bearing 030°
30° east of north
due east
bearing 090°
a quarter turn
S 45° W
bearing 225°
180 + 45
Bearings start at north and turn clockwise
That is the opposite of standard position, which starts at the positive x-axis and
turns anticlockwise. Sketch the compass before converting, or every angle will be
reflected.
4
Non-right triangles need the sine or cosine law.
a side with its opposite angle
sine law
AAS, ASA, SSA
SAS or SSS
cosine law
no matched pair to start from
right angle present
plain trigonometry
simplest of all
Look for a side and the angle opposite it. If you have such a pair, the sine law
starts; if not, only the cosine law can open the problem.
5
Two sides and the angle between them: the cosine law, then the area.
With no side–angle pair, the cosine law finds the third side. The same two sides and angle give the area directly.
In triangle ABC, a = 6, b = 10 and C = 60°. Find c and the area.
c² = a² + b² − 2ab cos C
the cosine law
= 36 + 100 − 2(6)(10)(0.5) = 76
cos 60° = 0.5
c = √76 ≈ 8.7
take the root last
area = ½ab sin C = ½(6)(10)(0.866) ≈ 26.0
the same two sides and included angle
Check: c lies between 10 − 6 and 10 + 6, as any third side must.
Forgetting the −2ab cos C term
36 + 100 = 136 gives c ≈ 11.7, which treats the triangle as right-angled at C. Only a 90° angle makes the last term vanish.
6
A side with its opposite angle: the sine law, and the ambiguous case.
The sine law starts from a matched pair: a side and the angle facing it.
A = 40°, B = 65°, a = 12. Find b.
bsin B = asin A
the sine law
b = 12 sin 65°sin 40°
multiply by sin B
= 12 × 0.90630.6428 ≈ 16.9
b faces the larger angle, so it is longer ✓
Two sides and an angle not between them: A = 30°, a = 6, b = 10
sin B = 10 sin 30°6 = 56 ≈ 0.833
the sine law, solved for sin B
B ≈ 56.4° or B ≈ 180° − 56.4° = 123.6°
sine is positive in quadrants I and II
30° + 123.6° < 180°, so both fit
each leaves room for C
two different triangles
the ambiguous case
Whenever SSA is given, test the obtuse partner of B. If it fits with A under 180°, there are two answers.
7
Variations: depression, an unknown angle, and a bearing.
The right-triangle ratios again, set up from three different descriptions.
Angle of depression: a 40 m cliff, boat at 25° below the horizontal
the angle at the boat is also 25°
alternate angles with the horizontal
tan 25° = 40d, so d = 40 ÷ tan 25°
the cliff is opposite, d adjacent
d = 40 ÷ 0.4663 ≈ 85.8 m
divide, do not multiply
The depression angle at the top equals the elevation angle at the bottom; put it at the boat to label the sides.
The angle is the unknown: an 8 m ladder reaching 7 m up a wall
sin θ = 78 = 0.875
opposite over hypotenuse
θ = sin⁻¹(0.875) ≈ 61.0°
the inverse sine
Finding an angle uses the inverse function on the calculator, not the ratio itself.
A bearing: 30 km on a bearing of 060°
the angle from north is 60°
bearings start at north
east = 30 sin 60° ≈ 26.0 km
opposite the 60° angle
north = 30 cos 60° = 15 km
adjacent, along north
Measured from north, sine gives the east part and cosine the north part — the reverse of standard position.
8
Anything that repeats can be modelled by a sine wave.
total rise and fall 3 m
amplitude is 1.5 m
period 12 s, so 2πb = 12
b = π6
h(t) = 1.5 sin(πt6)
measured from the mid level
h(3) = 1.5 sin(π2) = 1.5 m
highest after 3 s ✓
The tower is about 35 m tall
The amplitude is half the total movement, because the wave swings equally
above and below the midline. A buoy moving through 3 m in total has an amplitude of
1.5 m, and reading that as 3 doubles every later answer.
Remember the flow
Draw and label from the angle · Pick the ratio by what you have and want · Bearings clockwise from north · Sine law needs a matched pair · Amplitude is half the total swing
Check every answer for plausibility: a height cannot exceed the hypotenuse.
On the exam
Tests give a short word problem — a tower, ladder, ship or triangle — and ask for a length, an angle or an area, often to the nearest whole number. The fastest route: sketch it, mark the angle, choose sin, cos or tan by what you have and want, and switch to the sine or cosine law when there is no right angle. A 10-second check: a leg must be shorter than the hypotenuse, and the longest side must face the largest angle.
