O lvl E Maths: Trigonometry Bearings

In the diagram, the bearing of A from B is 320° and the bearing of B from C is 240°. Given that AB = BC = 10 km, calculate
(i) ∡BCA,
(ii) the bearing of C from A.

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Answer:

(i) ∡QCB = 240° - 180° = 60°
∡NBC = 60° (alternate angles)
∡NBA = 360° - 320° = 40° (angles at a point)
∡ABC = 40° + 160° = 100°
∡BCA = (180° - 100°) / 2 = 40° (angles in an isosceles triangle)


(ii) ∡ACQ = 40° + 60° = 100°
Bearing of C from A = 100° (alternate angles)

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