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
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
Cos θ = \(\frac{adj}{hyp}\)
= \(\frac{300}{600}\)
= 0.5
θ = Cos - 10.5
= 60
∠RPQ = ∠PQs
So the bearing of P from Q is 180 + 60 = 240\(^o\)
Answer is D
Cos θ = \(\frac{adj}{hyp}\)
= \(\frac{300}{600}\)
= 0.5
θ = Cos - 10.5
= 60
∠RPQ = ∠PQs
So the bearing of P from Q is 180 + 60 = 240\(^o\)
Answer is D