Find the angle of elevation of the top of a $$250 \sqrt{3}$$​ m high tower, from a point which is 250 m away from its foot.

Height of the tower = 250√3 m

Distance from the base of the tower to the observation point = 250 m

The angle of elevation $$\theta$$​ can be found using the tangent function, which relates the height of the tower to the distance from the observation point:

​$$\tan(\theta) = \frac{\text{Height of the tower}}{\text{Distance from the base of the tower}}$$

value of $$\tan 60^0 = \sqrt{3}$$​​