Question

If the angles of the 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 ${30}^0$ and ${60}^0$, then the height of the tower is

• A. $\sqrt{a+b}$
• B. $\sqrt{ab}$
• C. $\sqrt{a-b}$
• D. $\sqrt{\frac{a}{b}}$

Answer 1

The correct option is B $\sqrt{ab}$

Let AB be the height of the tower and B be its foot

$AB=h$

$BD=b$

$BC=a$

$\angle ADC={60}^o$

$\angle ACB={30}^o$

In $△ABC$

$\tan 30=\frac{AB}{BC}$

$\frac{1}{\sqrt{3}}=\frac{h}{a}$

$h=\frac{a}{\sqrt{3}}.....\left(1\right)$

In $△ADB$

$\tan 60=\frac{h}{b}$

$\sqrt{3}=\frac{h}{}$

$h=b\sqrt{3}.....\left(2\right)$

Multiplying (1) and (2)

$h^2=\frac{a}{\sqrt{3}}\times b\sqrt{3}$

$h^2=ab$

$h=\sqrt{ab}$