An observer who is 1.6 m tall is 45 m away from a pole. The angle of elevation of the top of the pole from his eyes is 30º. The height (in m) of the pole is closest to:
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 27.6
Let height of observer = AB
√3 unit = 45 m
⇒ 1 unit = 45/√3 = 25.98 m
So,
ED = 25.98 m
So,
Height of tower = ED + CD
= 25.98 + 1.6
= 27.58 m