Exams › IBPS PO › Quantitative Aptitude
From a point P on level ground, the angle of elevation of the top of a tower is $30^\circ$. If the tower is 100 m high, the distance of point P from the foot of the tower is:
- 149 m
- 156 m
- 173 m
- 200 m
Correct answer: 173 m
Solution
In the right triangle, $\tan 30^\circ = \frac{\text{height}}{\text{distance}} = \frac{100}{d}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get $d = 100\sqrt{3} \approx 173$ m.
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