Question

A tower is $5\sqrt{3}$ meters height. Find the angle of elevation of its top from a point $5$ meter away from its foot.

Answer 1

Solution:

$\tan \theta =\frac{opp}{adj}$

$\tan \theta =\frac{5\sqrt{3}}{5}$

$\tan \theta =\sqrt{3}$

$tan{60}^{\circ }=\sqrt{3}$

$\theta ={60}^{\circ }$