Question

The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30°. The height of the tower is

(a) 30 m
(b) $10\sqrt{3}\mathrm{m}$
(c) 20 m
(d) $10\sqrt{2}\mathrm{m}$

Answer 1

(b) $10\sqrt{3}\mathrm{m}$
Let $AB$ be the tower and $O$ be the point of observation.
Also,
∠$AOB={30}^o$ and $OB=30$ m
Let:
$AB=h$ m

In ∆$AOB$, we have:
$\frac{AB}{OB}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{h}{30}=\frac{1}{\sqrt{3}}$
⇒ $h=\frac{30}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{30\sqrt{3}}{3}=10\sqrt{3}$ m

Hence, the height of the tower is $10\sqrt{3}$ m.