This note covers right-angled trigonometry, angles of elevation and depression, three-figure bearings, the sine rule (including the ambiguous case), the cosine rule and the area formula for a triangle. These tools let you find distances and angles that cannot be measured directly, which is why they are used in navigation, surveying and engineering.
Key ideas
Right-angled triangles. Label the sides relative to the angle \(\theta\) you are using: the hypotenuse is opposite the right angle, the opposite side faces \(\theta\), and the adjacent side touches \(\theta\) (but is not the hypotenuse). Then
\[ \sin\theta = \dfrac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \dfrac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \dfrac{\text{opp}}{\text{adj}}. \]
Use the inverse functions \(\sin^{-1}\), \(\cos^{-1}\), \(\tan^{-1}\) on the GDC to find an angle from two sides. Check that the GDC is in degree mode. Pythagoras' theorem, \(a^2 + b^2 = c^2\), links the three sides.
Angles of elevation and depression. Both are measured from the horizontal. The angle of elevation is the angle up from the horizontal to an object. The angle of depression is the angle down from the horizontal to an object. The angle of depression from A to B equals the angle of elevation from B to A.
Bearings. A bearing is an angle measured clockwise from north, written with three figures, such as 045° or 270°. To find a bearing, draw a north line at the starting point. The back bearing (the bearing of the first point from the second) differs by 180°: add 180° if the bearing is less than 180°, and subtract 180° if it is 180° or more. Parallel north lines give co-interior angles that add to 180°, which helps to find angles in a bearing diagram.
Non-right-angled triangles. Label the triangle with side \(a\) opposite angle \(A\), and so on.
- Sine rule: \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}\). Use it when you know a side and its opposite angle plus one more side or angle.
- Cosine rule: \(a^2 = b^2 + c^2 - 2bc\cos A\), or \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\). Use it when you know two sides and the included angle, or all three sides.
- Area: \(\text{Area} = \dfrac{1}{2}ab\sin C\), where \(C\) is the angle between sides \(a\) and \(b\).
The ambiguous case. If you know two sides and a non-included angle (for example \(a\), \(b\) and \(A\)), the sine rule gives \(\sin B\). Since \(\sin B = \sin(180^\circ - B)\), there may be two possible values for \(B\): an acute one and an obtuse one. Accept both only if each gives a valid triangle, which means the three angles must total 180° with all angles positive. This happens when the given angle is acute and the side opposite it is shorter than the other given side. If the side opposite the given angle is the longer one, only the acute angle works.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\) | Right-angled triangles only. |
| \(\dfrac{a}{\sin A} = \dfrac{b}{\sin B}\) | Sine rule. Needs a matching side and angle pair. |
| \(a^2 = b^2 + c^2 - 2bc\cos A\) | Cosine rule for a side. Two sides and the included angle. |
| \(\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}\) | Cosine rule for an angle. Three sides known. |
| \(\text{Area} = \dfrac{1}{2}ab\sin C\) | Area of any triangle from two sides and the included angle. |
| Bearing | Clockwise angle from north, written with three figures. |
| Angle of elevation / depression | Angle above / below the horizontal, measured from the observer. |
Worked examples
Example 1 An observer stands on a cliff 85 m above the sea. A boat is seen at an angle of depression of 12°. (a) How far is the boat from the foot of the cliff? (b) From a point on level ground 40 m from the base of a tower, the angle of elevation of the top of the tower is 32°. How tall is the tower?
(a) The angle of depression is 12°, so the angle at the boat, between the sea and the line of sight, is also 12° (alternate angles). The cliff height is opposite this angle and the distance \(d\) is adjacent.
\(\tan 12^\circ = \dfrac{85}{d}\), so \(d = \dfrac{85}{\tan 12^\circ} = \dfrac{85}{0.21256} = 399.9\ldots\) m.
(b) The height \(h\) is opposite the 32° angle and 40 m is adjacent. \(\tan 32^\circ = \dfrac{h}{40}\), so \(h = 40\tan 32^\circ = 24.99\ldots\) m.
Answer: (a) 400 m; (b) 25.0 m.
Example 2 A ship leaves port P and sails 14 km on a bearing of 060° to a buoy A. It then sails 9 km on a bearing of 130° to a lighthouse Q. Find (a) the distance PQ, (b) the bearing of Q from P.
(a) At A, the back bearing to P is \(060^\circ + 180^\circ = 240^\circ\). The bearing to Q is 130°. The angle PAQ is \(240^\circ - 130^\circ = 110^\circ\).
Use the cosine rule with the included angle: \(PQ^2 = 14^2 + 9^2 - 2(14)(9)\cos 110^\circ = 196 + 81 + 86.19 = 363.19\).
So \(PQ = 19.06\ldots\) km.
(b) Use the sine rule to find angle \(APQ\): \(\dfrac{\sin(APQ)}{9} = \dfrac{\sin 110^\circ}{19.057}\), so \(\sin(APQ) = 0.44377\) and \(APQ = 26.3^\circ\). It is acute, since the side opposite it (9 km) is shorter than the side opposite the 110° angle.
The ship turned clockwise at A, so Q is clockwise from line PA when seen from P. The bearing is \(60^\circ + 26.3^\circ = 86.3^\circ\).
Answer: (a) 19.1 km; (b) 086° (to the nearest degree).
Example 3 In triangle ABC, angle \(A = 38^\circ\), \(a = 7\) cm and \(b = 10\) cm. Find the possible values of angle \(B\), and the area of each possible triangle.
Sine rule: \(\sin B = \dfrac{b\sin A}{a} = \dfrac{10\sin 38^\circ}{7} = 0.87953\).
So \(B = 61.6^\circ\) or \(B = 180^\circ - 61.6^\circ = 118.4^\circ\). Both are valid since \(a \lt b\) and \(A\) is acute: \(38^\circ + 118.4^\circ = 156.4^\circ \lt 180^\circ\).
Case 1, \(B = 61.58^\circ\): \(C = 180^\circ - 38^\circ - 61.58^\circ = 80.42^\circ\). Area \(= \dfrac{1}{2}ab\sin C = \dfrac{1}{2}(7)(10)\sin 80.42^\circ = 34.5\) cm².
Case 2, \(B = 118.42^\circ\): \(C = 180^\circ - 38^\circ - 118.42^\circ = 23.58^\circ\). Area \(= \dfrac{1}{2}(7)(10)\sin 23.58^\circ = 14.0\) cm².
Answer: \(B = 61.6^\circ\) (area 34.5 cm²) or \(B = 118.4^\circ\) (area 14.0 cm²).
Common errors
- Measuring a bearing anticlockwise or from east, or writing it with fewer than three figures (such as 60° instead of 060°).
- Using the sine rule when you have two sides and the included angle. The cosine rule is needed there.
- Forgetting the second, obtuse solution in the ambiguous case, or including it when it does not make a valid triangle.
- Using \(\dfrac{1}{2}ab\sin C\) with an angle that is not between the two sides used.
- Working in radian mode when angles are in degrees, or rounding intermediate values and losing accuracy.
Exam tip: Draw a clear diagram with a north line at each point and the angles you know written on it. Most mistakes in bearings problems come from working with the wrong angle.
Check yourself
- A ladder of length 12 m makes an angle of 50° with level ground. How high up the wall does it reach?
Answer
The height is opposite the 50° angle and 12 m is the hypotenuse. \(h = 12\sin 50^\circ = 9.19\) m. - A triangle has sides 7 cm, 9 cm and 12 cm. Find its largest angle.
Answer
The largest angle is opposite 12 cm. \(\cos\theta = \dfrac{7^2 + 9^2 - 12^2}{2(7)(9)} = \dfrac{-14}{126} = -0.1111\), so \(\theta = 96.4^\circ\). - Find the area of a triangle with two sides 8 cm and 11 cm and an included angle of 64°.
Answer
\(\text{Area} = \dfrac{1}{2}(8)(11)\sin 64^\circ = 44 \times 0.89879 = 39.5\) cm². - The bearing of B from A is 125°. What is the bearing of A from B?
Answer
The back bearing is \(125^\circ + 180^\circ = 305^\circ\).