From the top of a cliff, 60 meters high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60°. Find the height of the tower.

Let CD be the cliff so CD = 60 meters and AB be the tower.

Since, alternate angles are equal,

∴ ∠ECA = ∠CAF = 30° and ∠ECB = ∠CBD = 60°.

Let AF = BD = a meters.

In △BCD,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{BD}\\[1em] \Rightarrow CD = \sqrt{3}BD \\[1em] \Rightarrow CD = \sqrt{3}a \\[1em] \Rightarrow 60 = \sqrt{3}a \\[1em] \Rightarrow a = \dfrac{60}{\sqrt{3}} \\[1em] \Rightarrow a = \dfrac{60}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow a = \dfrac{60\sqrt{3}}{3} \\[1em] \Rightarrow a = 20\sqrt{3} \text{ meters}.$

In △AFC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CF}{AF}\\[1em] \Rightarrow CF = \dfrac{AF}{\sqrt{3}} \\[1em] \Rightarrow CF = \dfrac{a}{\sqrt{3}} \\[1em] \Rightarrow CF = \dfrac{20\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow CF = 20 \text{ meters}.$

From figure,

⇒ AB = DF = CD - CF

⇒ AB = CD - CF

⇒ AB = 60 - 20 = 40 meters.

Hence, the height of tower = 40 meters.