question_answer

The angle of elevation of the top of a tower at distance of 500 meters from the foot is \[30{}^\circ \] The height of the tower is

A) \[250\sqrt{3}\,\,meters\]

B) \[\frac{500}{\sqrt{3}}\,\,meters\]

C) \[500\sqrt{3}\,\,meters\]

D) \[250\,\,meters\]

Correct Answer: C

Solution :

 From \[\Delta \,POT,\,\frac{OT}{PO}={{30}^{o}}\] or           \[h=500\sqrt{3}\]