Question

The angles of elevation of the top of a vertical tower from two points. 30 metros apart, and on the same straight line passing through the base of tower, are ${30}^0$ and ${60}^0$ respectively. The height
of the tower is

• A. $10$ m
• B. $15$ m
• C. $15\sqrt{3}$
• D. $30$ m

Answer 1

The correct option is C $15\sqrt{3}$

Let AB is tower, the angles of elevation of the top of a vertical tower AB from two points D and C are ${30}^0$ and ${60}^0$ respectively.
Let the height of tower be h metres
and $BC=xm$
In triangle $ABC,\frac{AB}{BC}=tan{60}^0$
$\therefore \frac{h}{x}=\sqrt{3}$...........(i)
In triangle ABD,$\frac{AB}{BD}=tan{30}^0$
$\therefore \frac{h}{(x+30)}=\frac{1}{\sqrt{3}}$
or $\sqrt{3}h=x+30$
Put value of x from equation Ci) in equation (ii), we get
$\sqrt{3}h=\frac{h}{\sqrt{3}}+30$
or $\sqrt{3}h-\frac{h}{\sqrt{3}}=30$
or $\frac{2h}{\sqrt{3}}=30$
$\therefore h=\frac{30\sqrt{3}}{2}=15\sqrt{3}$
So, height of the tower is $15\sqrt{3}$metre.