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JAMB Mathematics 2018 Paper

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Question : 59

Total: 62

From a point R, 300m north of P, a man walks eastwards to a place; Q which is 600m from P. Find the bearing of P from Q correct to the nearest degree

026 $^{\circ}$
045 $^{\circ}$
210 $^{\circ}$
240 $^{\circ}$

Solution:

Cos θ = $\frac{\text{adj}}{\text{hyp}}$

= $\frac{300}{600}$

= 0.5

θ = Cos - 10.5

= 60

∠ RPQ = ∠ PQs

So the bearing of P from Q is 180 + 60 = 240 $^{\circ}$

Answer is D

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