A vertical pole and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 60° and the angle of depression of the foot of tower is 30°. Find the height of the tower if the height of the pole is 20 m.

Let AB be the pole and CD be the tower. Let length of tower (CD) be h metres.

Let distance between pole and tower (BD) be x meters.

From figure,

ABDE is a rectangle so,

DE = AB = 20 meters
AE = BD = x meters
CE = CD - DE = (h - 20) meters
∠EAD = ∠ADB = 30° (Alternate angles are equal)

Considering right angled △ABD we get,

$\Rightarrow \text{tan 30°} = \dfrac{AB}{BD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{20}{x} \\[1em] \Rightarrow x = 20\sqrt{3}$

Considering right angled △ACE we get,

$\Rightarrow \text{tan 60°} = \dfrac{CE}{AE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h - 20}{BD} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h - 20}{x} \\[1em] \Rightarrow \sqrt{3} = \dfrac{h - 20}{20\sqrt{3}} \\[1em] \Rightarrow 20\sqrt{3} \times \sqrt{3} = h - 20 \\[1em] \Rightarrow 60 = h - 20 \\[1em] \Rightarrow h = 80.$

Hence, the height of the tower is 80 meters.