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From a point P on a level ground, the angle of elevation of the top tower is 30º. If the tower is 100 m high, the distance of point P from the foot of the tower is:
Answer & Solution
Correct Answer:
Option
C
Let AB be the tower.
Then, ∠APB = 30° and AB = 100m
$$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
⇒ AP = AB × √3 = 100 × √3
⇒ AP = 100 × 1.73 = 173m
Then, ∠APB = 30° and AB = 100m
$$\frac{{AB}}{{AP}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
⇒ AP = AB × √3 = 100 × √3
⇒ AP = 100 × 1.73 = 173m
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