Question

The angle of elevation of the top of a tower from a point on the same level as the foot of the tower is 30° On advancing 150 m towards the foot of the tower, the angle of elevation becomes 60°. Show that the height of the tower is 129.9 metres.

Answer 1

Let $AB$ be the tower.
We have:
$CD\hspace{0.17em}=150$ m, ∠$ACB={30}^o$ and ∠$ADB={60}^o$
Let:
$AB=h$ m and $BD=x$ m

In the right ∆$ABD$, we have:
$\frac{AB}{AD}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{h}{x}=\sqrt{3}$
⇒ $x=\frac{h}{\sqrt{3}}$
Now, in the right ∆$ACB$, we have:
$\frac{AB}{BC}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{h}{x+150}=\frac{1}{\sqrt{3}}$
⇒ $\sqrt{3}h=x+150$

On putting $x=\frac{h}{\sqrt{3}}$ in the above equation, we get:
$\sqrt{3}h=\frac{h}{\sqrt{3}}+150$
⇒ $3h=h+150\sqrt{3}$
⇒ $2h=150\sqrt{3}$
⇒ $h=\frac{150\sqrt{3}}{2}=75\sqrt{3}=75\times 1.732=129.9$ m
Hence, the height of the tower is 129.9 m.