A boy, 1.6 m tall, is 20 m away from a tower and observes the angle of elevation of the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.

(i) From figure,

When angle of elevation is 45°, then CE is the tower.

In △ADE,

$\Rightarrow \text{tan 45°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow 1 = \dfrac{DE}{AD} \\[1em] \Rightarrow AD = DE \\[1em] \Rightarrow DE = 20 \text{ meters}.$

From figure,

CD = AB.

CE = CD + DE = 1.6 + 20 = 21.6 meters.

Hence, height of tower when angle of elevation is 45° is 21.6 meters.

(ii) From figure,

When angle of elevation is 60°.

In △ADF,

$\Rightarrow \text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DF}{AD} \\[1em] \Rightarrow DF = AD\sqrt{3} \\[1em] \Rightarrow DF = 20 \times 1.732 = 34.64 \text{ meters}.$

From figure,

CD = AB.

CF = CD + DF = 1.6 + 34.64 = 36.24 meters.

Hence, height of tower when angle of elevation is 60° is 36.24 meters.