# If the angles of elevation of a tower from two points distant a and b (a>b)

Question:

If the angles of elevation of a tower from two points distant a and b (a>b) from its foot and in the same straight line from it are 30° and 60°, then the height  of the tower is

(a) $\sqrt{a+b}$

(b) $\sqrt{a b}$

(c) $\sqrt{a-b}$

(d) $\sqrt{\frac{a}{b}}$

Solution:

Let  be the height of tower

Given that: angle of elevation are and.

Distance  and

Here, we have to find the height of tower.

So we use trigonometric ratios.

In a triangle,

$\Rightarrow \tan C=\frac{A B}{B C}$

$\Rightarrow \tan 60^{\circ}=\frac{A B}{B C}$

$\Rightarrow \tan 60^{\circ}=\frac{h}{b}$

Again in a triangle ABD,

$\Rightarrow \tan D=\frac{A B}{B D}$

$\Rightarrow \tan 30^{\circ}=\frac{h}{a}$

$\Rightarrow \tan \left(90^{\circ}-60^{\circ}\right)=\frac{h}{a}$

$\Rightarrow \cot 60^{\circ}=\frac{h}{a}$

$\Rightarrow \frac{1}{\tan 60^{\circ}}=\frac{h}{a}$

$\Rightarrow \frac{b}{h}=\frac{h}{a} \quad$ Put $\tan 60^{\circ}=\frac{h}{b}$

$\Rightarrow h^{2}=a b$

$\Rightarrow h=\sqrt{a b}$

Hence the correct option is $b$.