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

  $a b^2$
• Correct answer
  C

  $\sqrt{a b}$
• D

  $a b$

Solution

Given, $\theta+\phi=90^{\circ}$ Let the height of a tower $=h$ Then, In $\triangle A C P$ $\tan \theta=\frac{h}{a}$ ...(i) In $\triangle A B P$ $\tan \phi=\frac{h}{b}$ ...(ii)
$\because \quad \tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \cdot \tan \phi}$ $\tan 90^{\circ}=\frac{\frac{h}{a}+\frac{h}{b}}{1-\frac{h}{a} \cdot \frac{h}{b}}=\infty$ $\Rightarrow \quad 1-\frac{h^2}{a b}=0$ $\Rightarrow \quad h^2=a b$ $\Rightarrow \quad h=\sqrt{a b}$ Hence, the required height of a tower is $\sqrt{a b}$.

Hence, the correct answer is option: 'C'.