Question

A tower stands vertically on the ground. The angle of elevation of the top of
the tower from a point on the ground, which is 90 m away from the foot of
the tower is 30°. Find the height of the tower (Take V3 = 1.73)​

Answer 1

Given:

• The angle of elevation of the top of the tower is 30°.
• The distance between the foot of the tower and the point of elevation is 90 m.
• √3 = 1.73

To find: The height of the tower.

Answer:

(Diagram for reference attached below.)

In triangle ABC,

$\sf tan\ {\theta}^{\circ}\ \ =\ \dfrac{Opposite}{Base}\\\\\\tan\ {30}^{\circ}\ =\ \dfrac{AB}{BC}\\\\\\\sf \bf (tan\ {30}^{\circ}\ =\ \dfrac{1}{\sqrt{3}})\\\\\\\sf \dfrac{1}{\sqrt{3}}\ =\ \dfrac{AB}{90} \\\\\\\dfrac{90}{\sqrt{3}}\ =\ AB\\ \\\\\\On\ rationalizing,\\\\\\\sf \dfrac{90\sqrt{3}}{3}\ =\ \sf AB\\\\\\30\sqrt{3}\ =\ AB\\ \\\\30\ \times\ 1.73\ =\ AB\\\\51.9\ =\ AB$

Therefore, the height of the tower is 51.9 m.