Question

The sum of 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 is $90^{\circ}$. Then, the height of the tower is

• A. $a^{2} b$
• B. $a b^{2}$
• C. $\sqrt{a b}$
• D. $ab$

Answer 1

Answer: (C)

Given, $\theta+\phi=90^{\circ}$
Let the height of a tower $=h$
Then, In $\Delta A C P$
$\tan \theta=\frac{h}{a}$ ...(i)
In $\Delta A B P$
$\tan \phi=\frac{h}{b}$ ...(ii)

$\because \tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}$
$\Rightarrow \tan 90^{\circ}=\frac{\frac{h}{a}+\frac{h}{b}}{1-\frac{h}{a} \cdot \frac{h}{b}}=\infty$
$\Rightarrow 1-\frac{h^{2}}{a b}=0$
$\Rightarrow h^{2}=ab$
$\Rightarrow h=\sqrt{ab}$
Hence, the required height of a tower is $\sqrt{a b}$.