The angle of elevation of the top of a tower from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.

Let AB be the tower and C be the point at a distance of 160 m from foot of tower.

Let,

AB = h meters.

From figure,

In △ABC,

$\Rightarrow \text{tan θ} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan 60°} = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h}{160} \\[1em] \Rightarrow h = 160\sqrt{3} \\[1em] \Rightarrow h = 160 \times 1.732 = 277.12 \text{ m}.$

Hence, the height of the tower is 277.12 m.