From a point Z, 60 m north of X, a man walks 60√3m eastwards to another point Y. Find the bearing of Y from X.

A.

0.30^o

B.

045^o

C.

060^o

D.

090^o

View answer & explanation

Correct answer is C

tan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\)

\(\theta\) = tan \(\frac{1}{3}\) = 60^o

Bearing of y from x = 060^o

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