WAEC / WASSCE
WAEC Bearings and Trigonometry Practice Questions, Solved
WAEC bearings questions use three-figure bearings measured clockwise from north, and usually combine a journey diagram with Pythagoras, the sine rule or the cosine rule. Angles of elevation and depression appear almost every year too. Below are five original questions in WASSCE style, each solved with the diagram described in words and the marking explained.
Bearings and trigonometry make up one of the most reliable Section B questions on the WASSCE core maths paper, and also one of the most failed, usually because candidates skip the diagram. Every solution below starts by describing the diagram you should draw, because in WAEC marking the sketch itself often carries a mark and a wrong sketch poisons everything after it. These five questions are original, written in the WASSCE style at exam difficulty, not reproduced from past papers. Work each one with a pencil, ruler and calculator, drawing north lines at every point before touching any formula.
What WAEC asks on bearings and trigonometry
Bearings questions follow a pattern: a journey with two or three legs, each on a stated bearing, then find the distance and bearing of the finish from the start. If the legs meet at a right angle you use Pythagoras and basic tan; if not, the cosine rule finds the distance and the sine rule finds the angle for the bearing. Separately, the paper loves angles of elevation and depression, a mast, a cliff, a boat, which are single right triangles once drawn. Pure triangle questions also appear: given two sides and the included angle, find the third side and the area using half ab sin C. The trigonometry itself is never deep; the marks are for setting up the right triangle.
Diagrams first, three figures always
Write every bearing with three figures: 065 degrees, not 65 degrees. Draw a north arrow at each point of the journey, because the back bearing rule, add 180 degrees if the bearing is under 180, subtract 180 if it is over, only makes sense on a diagram with parallel north lines. When a question mixes bearings with the cosine rule, the hardest step is finding the angle inside the triangle from the two bearings; do it on the diagram with alternate angles rather than in your head. For elevation and depression, remember the angle of depression from the top equals the angle of elevation from the bottom. Check answers for sense: a bearing must be between 000 and 360 and a distance can never be negative.
Worked questions, step by step
The bearing of Q from P is 065 degrees and the bearing of R from P is 158 degrees. Find (a) the bearing of P from Q, (b) the bearing of P from R, (c) the angle QPR.
- Back bearing rule: if a bearing is less than 180 degrees, add 180; if more, subtract 180.
- Bearing of P from Q: 065 plus 180 equals 245 degrees.
- Bearing of P from R: 158 plus 180 equals 338 degrees.
- Angle QPR is the angle between the two directions at P: 158 minus 65 equals 93 degrees.
Answer: (a) 245 degrees (b) 338 degrees (c) 93 degrees
Where marks slip: A1 for each back bearing, M1 A1 for the angle between the bearings. Writing 65 instead of 065 is untidy rather than fatal, but a bearing over 360 shows the rule was applied twice and scores zero.
Try one yourself: The bearing of B from A is 132 degrees. Find the bearing of A from B. [Answer: 312 degrees]
A man walks 5 km due east from A to B, then 12 km due north from B to C. Calculate (a) the distance AC, (b) the bearing of C from A, correct to the nearest degree.
- The path forms a right angle at B, with AC as the hypotenuse.
- AC equals sqrt(5^2 plus 12^2) equals sqrt(25 plus 144) equals sqrt(169) equals 13 km.
- From A, C lies 5 km east and 12 km north. The angle from north satisfies tan of the angle equals 5 divided by 12 equals 0.4167.
- The angle is 22.6 degrees, so the bearing of C from A is 023 degrees to the nearest degree.
Answer: (a) 13 km (b) 023 degrees
Where marks slip: B1 for the sketch, M1 A1 for Pythagoras, M1 A1 for the bearing. The trap is using tan as 12 over 5, which gives the angle from east, not from north; the bearing must be measured from north, clockwise.
Try one yourself: A woman walks 9 km east then 12 km north. Find how far she is from her start and the bearing of her position from it. [Answer: 15 km, approximately 037 degrees]
In triangle PQR, angle P = 52 degrees, angle Q = 68 degrees and PR = 14 cm. Calculate (a) angle R, (b) the length QR, correct to three significant figures.
- Angles in a triangle sum to 180: angle R equals 180 minus 52 minus 68 equals 60 degrees.
- Side PR is opposite angle Q, and side QR is opposite angle P.
- By the sine rule: QR divided by sin 52 equals 14 divided by sin 68.
- QR equals 14 times sin 52 divided by sin 68 equals 14 times 0.7880 divided by 0.9272 equals 11.9 cm.
Answer: (a) 60 degrees (b) 11.9 cm (3 s.f.)
Where marks slip: A1 for angle R, M1 for a correctly paired sine rule statement, A1 for 11.9. Pairing a side with the wrong opposite angle is the whole danger of the sine rule; label opposite pairs on your sketch first.
Try one yourself: In triangle PQR, angle P = 40 degrees, angle Q = 75 degrees and PR = 10 cm. Find QR. [Answer: approximately 6.7 cm]
In triangle XYZ, XY = 8 cm, XZ = 5 cm and angle X = 60 degrees. Calculate (a) the length YZ, (b) the area of the triangle, correct to three significant figures.
- YZ is opposite the known angle X, so use the cosine rule.
- YZ^2 equals 8^2 plus 5^2 minus 2 times 8 times 5 times cos 60 equals 64 plus 25 minus 80 times 0.5 equals 49.
- YZ equals 7 cm.
- Area equals half times 8 times 5 times sin 60 equals 20 times 0.8660 equals 17.3 cm^2.
Answer: (a) 7 cm (b) 17.3 cm^2 (3 s.f.)
Where marks slip: M1 for the cosine rule with correct substitution, A1 for 7, then M1 A1 for the area formula half ab sin C. The commonest slip is adding the 80 cos 60 term instead of subtracting; a distance longer than 13 cm should ring alarm bells.
Try one yourself: Two sides of a triangle are 6 cm and 10 cm with an included angle of 60 degrees. Find the third side. [Answer: approximately 8.7 cm]
From a point A on level ground, 24 m from the foot of a vertical mast, the angle of elevation of the top of the mast is 30 degrees. B is a point between A and the foot of the mast, on the same straight line, where the angle of elevation is 60 degrees. Calculate (a) the height of the mast, (b) the distance AB, each correct to three significant figures where necessary.
- Height: h equals 24 times tan 30 equals 24 times 0.5774 equals 13.9 m to three significant figures.
- From B: the horizontal distance is h divided by tan 60 equals 13.856 divided by 1.7321 equals 8.00 m.
- AB equals 24 minus 8.00 equals 16.0 m.
Answer: (a) 13.9 m (b) 16.0 m
Where marks slip: B1 for a single diagram carrying both angles, M1 A1 for the height, M1 A1 for AB. Keep the unrounded height, 13.856, in your calculator for part (b); rounding early is exactly how accuracy marks leak away.
Try one yourself: From a point 20 m from the foot of a tower, the angle of elevation of the top is 45 degrees. Find the height of the tower. [Answer: 20 m]
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Solve my question →Quiz me on this topicQuestions students ask
Must bearings always be written with three figures?
Yes, that is the convention WAEC expects: 065 degrees, not 65 degrees. Bearings are measured clockwise from north and run from 000 to 360 degrees.
When do I use the sine rule and when the cosine rule?
Cosine rule when you know two sides and the included angle, or all three sides. Sine rule when you have a matching pair, a side and its opposite angle, plus one more piece. In bearings journeys, the cosine rule usually finds the distance and the sine rule the angle for the bearing.
Are these real WAEC bearings past questions?
No, they are original questions in the WASSCE style and difficulty. The real papers are WAEC copyright, so I write fresh questions that use the same setups the exam repeats: journeys, back bearings, elevation and depression, and triangle rules.