Question

From the top of a building 15 m high, the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find height of the tower, and the distance between the tower and the building.

Answer 1

Let $AB$ be the tower and $CD$ be the building.
We have:
$CD=15$ m, ∠$ADE={30}^o$ and ∠$ACB={60}^o$
Let $AB=h$ m and $BC=x$ m such that $ED=x$ m.

In the right ∆$ABC$, we have:
$\frac{h}{x}=\tan {60}^o=\sqrt{3}$
⇒ $x=\frac{h}{\sqrt{3}}$
Now, in the right ∆$AED$, we have:
$\frac{h-15}{x}=\tan {30}^o=\frac{1}{\sqrt{3}}$
On putting $x=\frac{h}{\sqrt{3}}$ in the above equation, we get:
$\frac{h-15}{{\displaystyle \frac{h}{\sqrt{3}}}}=\frac{1}{\sqrt{3}}$

⇒ $\frac{(h-15)\sqrt{3}}{h}=\frac{1}{\sqrt{3}}$
⇒ $3h-45=h$
⇒ $2h=45$
⇒ $h=22.5$ m

​Hence, the height of the tower is 22.5 m.
Now,
Distance between the tower and the building = $x=\frac{h}{\sqrt{3}}=\frac{22.5}{\sqrt{3}}=\frac{22.5}{1.732}=12.99$ m