A 20 m high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole, is 60° and the angle of elevation of the top of the pole, as seen from the foot of tower is 30°. Find :

(i) the height of the tower;

(ii) the horizontal distance between the pole and the tower.

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

In △ABD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = \sqrt{3}AB \\[1em] \Rightarrow BD = 1.732 \times 20 \\[1em] \Rightarrow BD = 34.64 \text{ m}.$

In △CBD,

$\text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{CD}{BD} \\[1em] \Rightarrow CD = \sqrt{3}BD \\[1em] \Rightarrow CD = 1.732 \times 34.64 \\[1em] \Rightarrow CD = 60 \text{ m}.$

Hence, height of tower = 60 meters.

(ii) From part (i),

BD = 34.64 m

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