Question

If the angles of elevation of a tower from two points at a distance a and b (where, b>a) from its foot and lying on the same side are ${60}^{\circ }$ and ${30}^{\circ }$, then height of the tower is

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

Answer 1

The correct option is C

$\sqrt{ab}$

Let AB be the tower.
Such that

CB=a and BD=b
In $\Delta ABC,tan{60}^{\circ }=\frac{AB}{BC}=\frac{AB}{a}$
$\Rightarrow$ $AB=\theta \sqrt{3}$ ....(i)\
In $\Delta ABC$
$tan{30}^{\circ }=\frac{AB}{BC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{b}$
$\Rightarrow$ $AB=\frac{b}{\sqrt{3}}$ ....(ii)
On multiplying Eqs. (i) and (ii), we get
$A{B}^2=ab\Rightarrow AB=\sqrt{ab}$