If the angles of elevation of the top of a tower form tow points at distances a and

Solution:

(b) $\sqrt{a b}$

Let $A B$ be the tower and $C$ and $D$ be the points of observation on $A C$.

Let:

$\angle A C B=\theta, \angle A D B=90-\theta$ and $A B=h \mathrm{~m}$

Thus, we have:

$A C=a, A D=b$ and $C D=a-b$

Now, in the right ∆ABC, we have:

$\tan \theta=\frac{A B}{A C} \Rightarrow \frac{h}{a}=\tan \theta$    .......(i)

In the right ∆ABD, we have:

$\tan (90-\theta)=\frac{A B}{A D} \Rightarrow \cot \theta=\frac{h}{b} \quad$...(ii)

On multiplying (i) and (ii), we have:

$\tan \theta \times \cot \theta=\frac{h}{a} \times \frac{h}{b}$

$\Rightarrow \frac{h}{a} \times \frac{h}{b}=1 \quad\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$

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

$\Rightarrow h=\sqrt{a b} \mathrm{~m}$

Hence, the height of the tower is $\sqrt{a b} \mathrm{~m}$.