A vertical pole and a vertical tower are on the same level ground in such a way that 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 bottom of the tower is 30°. Find :

(i) the height of the tower, if the height of the pole is 20 m;

(ii) the height of the pole, if the height of the tower is 75 m

(i) Let AB be the pole and CD be the tower.

Given,

Length of pole (AB) = 20 m.

From figure,

CE = AB = 20 m.

In △AEC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CE}{AE} \\[1em] \Rightarrow AE = \sqrt{3} CE \\[1em] \Rightarrow AE = 20\sqrt{3} \text{ m}.$

In △AED,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DE}{AE} \\[1em] \Rightarrow DE = AE\sqrt{3} \\[1em] \Rightarrow DE = 20\sqrt{3} \times \sqrt{3} = 20 \times 3 = 60 \text{ m}.$

From figure,

CD = DE + CE = 60 + 20 = 80 m.

Hence, the height of the tower = 80 m.

(ii) Given,

Length of tower (CD) = 75 m

From figure,

AE = BC = y (let)

In △AEC,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CE}{AE} \\[1em] \Rightarrow CE = \dfrac{AE}{\sqrt{3}} \\[1em] \Rightarrow CE = \dfrac{y}{\sqrt{3}} \text{ m}.$

In △AED,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{DE}{AE} \\[1em] \Rightarrow DE = AE\sqrt{3} \\[1em] \Rightarrow DE = y\sqrt{3} \text{ m}.$

From figure,

⇒ CD = CE + DE

$\Rightarrow \dfrac{y}{\sqrt{3}} + \sqrt{3}y = 75 \\[1em] \Rightarrow \dfrac{y + 3y}{\sqrt{3}} = 75 \\[1em] \Rightarrow \dfrac{4y}{\sqrt{3}} = 75 \\[1em] \Rightarrow y = \dfrac{75\sqrt{3}}{4} \text{ m}.$

We know that,

CE = $\dfrac{y}{\sqrt{3}} = \dfrac{\dfrac{75\sqrt{3}}{4}}{\sqrt{3}} = \dfrac{75}{4}$ = 18.75 meters.

Hence, height of pole = 18.75 meters.