Question 131

An observer who is 1.62 m tall is 45 m away from a pole. The angle of elevation of the top of the pole from his eyes is $$30^\circ$$. The height (in m) ofthe pole is closest to:

Solution

Given that observer 1.62 m away from 45 m angle $$30^\circ$$

AB= 45 m

then in the diagram

from the above figure $$\tan 30^\circ = \frac{h-1.62} {45} $$

$$\Rightarrow h-1.62 = 45 \times \frac{1} {\sqrt{3}} $$

$$\Rightarrow h= 45 \times \frac{1} {\sqrt{3}} + 1.62 $$

$$\Rightarrow h= 45 \times 0.577 +1.62 $$

$$\Rightarrow h= 27.58 $$ m

$$\Rightarrow h= 27.6 $$ m Ans

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