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Question
A bird is flying from A towards B at an angle of 35°, a point 30 km away from A. At B it changes its course of flight and heads towards C on a bearing of 48° and distance 32 km away. How far is C to the North of B?
(sin 55° = 0.8192, cos 55° = 0.5736, sin 42° = 0.6691, cos 42° = 0.7431)
Solution
Distance from C to the North of B is CD
In the right ∆BCD,
sin 42° = `"CD"/"BC"`
0.6691 = `"BD"/32`
∴ CD = 0.6691 × 32
= 21.41 km
Distance of C to the North B is 21.41 km
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