Question

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower.

Answer 1

Let $AB$ be the tower standing vertically on the ground and O be the position of the observer.
We now have:
$OA=20m,$∠$OAB\hspace{0.17em}={90}^o$ and ∠$AOB$ $={60}^o$
Let:
$AB=h$ m

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

⇒ $\frac{h}{20}=\sqrt{3}$

⇒ $h=20\sqrt{3}$ = $(20\times 1.732)=34.64$

Hence, the height of the pole is 34.64 m.