Question on: JAMB Mathematics - 1991
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
0.30o
B
045o
C
060o
D
090o
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Correct Option: C
tan \(\theta\) = \(\frac{60\sqrt{3}}{60}\) = \(\sqrt{3}\)
\(\theta\) = tan \(\frac{1}{3}\) = 60o
Bearing of y from x = 060o
\(\theta\) = tan \(\frac{1}{3}\) = 60o
Bearing of y from x = 060o
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