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S4 Maths sine and cosine rule paper 3: bearings and elevations

S4 Maths sine and cosine rule paper 3: bearings and elevations

  • 2026-11-10

S4 Maths sine and cosine rule paper 3: bearings and elevations

Ten applied items on bearings, boat distances, lighthouses and hill heights; sketch first, then set up the equations.

Direct answers Items 01 to 03 cover bearings and perpendicular components, 04 to 06 two elevation equations and plot areas, 07 to 09 Heron, an opposite-bank distance and two solutions, and 10 a wrong bearing difference.

01 Paper details (check answers afterwards)

Item Detail
Questions 10
Suggested time 25
Difficulty mix Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter Chapter: S4 Maths Sine and Cosine Rule
Level S4

02 Questions (10 items)

  1. FoundationFrom A, walk 8 km east to B, then 6 km north to C. Find AC and the bearing of C from A.
  2. FoundationTwo boats leave the same pier, one at 12 km/h on a bearing of 040 and the other at 15 km/h on 130. How far apart are they after one hour?
  3. FoundationWhat is the angle between the directions with bearings 030 and 150?
  4. IntermediateFrom A the angle of elevation of a lighthouse top is 30 degrees. After walking 50 m towards it the elevation is 45 degrees. Find the height of the lighthouse.
  5. IntermediateFrom P the elevation of a hilltop is 25 degrees. Walking 120 m horizontally towards the hill to Q, the elevation is 40 degrees. Find the height.
  6. IntermediateA triangular plot has sides 50 m and 70 m with an included angle of 65 degrees. Find its area.
  7. ChallengeA triangular plot has sides 40 m, 50 m and 60 m. Find its area.
  8. ChallengePoints A and B are 200 m apart on a bank and C is on the opposite bank. Angle CAB = 65 degrees and angle CBA = 48 degrees. Find AC.
  9. ChallengeIn a triangle a = 9 cm, b = 12 cm and A = 35. Find all possible values of B.
  10. Error redoFor bearings 040 and 100 a student says the angle between them is 140 degrees, because 180 – 40. What is wrong, and what is the correct angle?

03 Answers and one-line notes

# Answer One-line note
1 10.0 km, bearing 053.1 degrees East and north are perpendicular, so AC = sqrt(64 + 36) = 10.0 km; tan(theta) = 6 / 8 gives 36.9 degrees, and the bearing is 90 – 36.9 = 53.1 degrees.
2 19.2 km The angle between the bearings is 130 – 40 = 90 degrees, so the distance is sqrt(144 + 225) = 19.2 km.
3 120 degrees Subtract the bearings: 150 – 30 = 120 degrees; bearings are measured clockwise from north and are not polar angles.
4 68.30 m With height h, h / tan30 – h / tan45 = 50, so h(1.732 – 1) = 50 and h = 68.30 m.
5 126 m h(cot25 – cot40) = 120 with cot25 = 2.145 and cot40 = 1.192, a difference of 0.953, so h = 126 m.
6 1 586 m² Area = 0.5 x 50 x 70 x sin65 = 1 586 m²; use the sine of the included angle, not of its supplement.
7 992 m² s = 75, so the area is sqrt(75 x 35 x 25 x 15) = 992 m²; with three sides given, Heron is the direct route.
8 161 m Angle ACB = 180 – 65 – 48 = 67 degrees, so AC = 200 x sin48 / sin67 = 161 m.
9 49.9 or 130.1 degrees sinB = 12 x sin35 / 9 = 0.765, so B is 49.9 or 130.1 degrees; since b > a and b sinA = 6.9 cm is less than a, both solutions hold.
10 The bearings are simply subtracted, so the angle is 60 degrees Both bearings are measured clockwise from north, so the angle is 100 – 40 = 60 degrees; subtracting from 180 only makes sense for a supplement with north or south.

04 What to do after marking

  • After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
  • Repeat the same item two days later; move to the next paper only when both attempts are correct.
  • Parents only need the answer and note columns; there is no need to explain every item.

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