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From a point P on a level ground, the angle of elevation of the top of the tower is 30°. If the tower is 100 m high, the distance of the point P from the foot of the tower is :
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- 100/ √3 m
- 100 √3 m
- 50√3 m
- 50/√3 m
Correct Option: B
Let us draw the figure from the given question.
Let, OQ be the tower.
Then, OQ = 100m and ∠OPQ = 30°.
| In Δ OPQ, tan30° | = | OQ | ⇒ | 1 | = | 100 |
| OP | √3 | OP |
| ⇒ | OP | = | 100 √3 m. |
Hence , the distance of the point P from the foot of the tower is 100 √3 m.
