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 mĀ angle $$30^\circ$$
AB= 45 m
then in the diagram
from the above figureĀ $$\tan 30^\circ = \frac{h-1.62} {45} $$
$$\Rightarrow h-1.62 = 45 \times \frac{1} {\sqrt{3}} $$
$$\Rightarrow h= 45 \timesĀ \frac{1} {\sqrt{3}} + 1.62 $$
$$\Rightarrow h= 45 \times 0.577 +1.62 $$
$$\Rightarrow h= 27.58 $$ mĀ
$$\Rightarrow h= 27.6Ā $$ m AnsĀ
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