If the angles of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it are complementary, then the height of the tower is
(a) $a b$
(b) $\sqrt{a b}$
(c) $\frac{a}{b}$
(d) $\sqrt{\frac{a}{b}}$
Let be the height of tower.
Given that: angle of elevation of top of the tower are and .
Distance and
Here, we have to find the height of tower.
So we use trigonometric ratios.
In a triangle,
$\Rightarrow \tan C=\frac{A B}{B C}$
$\Rightarrow \tan \left(90^{\circ}-\theta\right)=\frac{h}{h}$
$\Rightarrow \cot \theta=\frac{h}{b}$
Again in a triangle ABD,
$\tan D=\frac{A B}{B D}$
$\Rightarrow \tan \theta=\frac{h}{a}$
$\Rightarrow \frac{1}{\cot \theta}=\frac{h}{a}$
$\Rightarrow \frac{b}{h}=\frac{h}{a}$
$\Rightarrow h^{2}=a b$
$\Rightarrow h=\sqrt{a b}$
Put $\cot \theta=\frac{h}{b}$
Hence the correct option is $b$.