Question

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height 5 m. From a point on the plane the angles of elevation of the bottom and the top of the flagstaff are 30° and 60°. Find the height of the tower.

Answer 1

Let $OX$ be the horizontal line, $AC$ be the vertical tower and $BC$ be the vertical flagstaff such that $BC=5$ m.
Let:
$AC=h$ m and $OA=x$ m



In the right ∆$AOC$, we have:
$\frac{AC}{OA}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x}=\frac{1}{\sqrt{3}}$
⇒ $x=\sqrt{3}h$

Now, in the right ∆$AOB$, we have:
$\frac{AB}{OA}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{(h+5)}{x}=\sqrt{3}$

On putting $x=\sqrt{3}h$ in the above equation, we get:

⇒ $\frac{(h+5)}{\sqrt{3}h}=\sqrt{3}$
⇒ $h+5=3h$
⇒ $2h=5$
⇒ $h=\frac{5}{2}=2.5$ m
​Hence, the height of the tower is 2.5 m.