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°
  • B 045°
  • C 060°
  • D 090°
jamb 1991mathematics

The correct answer is C. 060°

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

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

Bearing of y from x = 060

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