In the given figure, AB represents a pole and CD represents a 60 m high tower, both of which are standing on the same horizontal plane. From the top of the tower, the angles of depression of the top and the foot of the pole are 30° and 60° respectively. Calculate:

(i) the horizontal distance between the pole and the tower,

(ii) the height of the pole.

(i) Given,

Height of the tower CD = 60 m

Angle of depression to the foot of pole (A) = 60°

Angle of depression to the top of pole (B) = 30°

Let the horizontal distance between the foot of the pole and the foot of the tower be x.

Considering right angled △ACD, we get :

$\Rightarrow \tan 60^{\circ} = \dfrac{\text{perpendicular}}{\text{base}} = \dfrac{CD}{AC} \\[1em] \Rightarrow \sqrt3 = \dfrac{60}{AC} \\[1em] \Rightarrow AC = \dfrac{60}{\sqrt3}\\[1em] \Rightarrow AC = \dfrac{60\times \sqrt3}{\sqrt3 \times \sqrt3}\\[1em] \Rightarrow AC = 20\sqrt3 \\[1em] \Rightarrow AC = 34.64 \text{ m.}$

Hence, horizontal distance between pole and tower = 34.64 m.

(ii) In △DEB,

$\Rightarrow \tan 30^{\circ} = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{DE}{BE} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{x}{20\sqrt3} \\[1em] \Rightarrow 1 = \dfrac{x}{20}\\[1em] \Rightarrow x = 20 \text{ m.}$

Height of the pole AB = CE = 60 - DE

= 60 - x

= 60 - 20

= 40 m.

Hence, height of the pole = 40 m.