Question

The angle of elevation of the top of a tower from a point on the ground ${30}^{\circ }$ after walking $200$ m towards the tower the angle of elevation becomes ${60}^{\circ }$. The height of the tower is

• A. $100\sqrt{3}$ m
• B. $200\sqrt{3}$ m
• C. $100$ m
• D. $200$ m

Answer 1

The correct option is A $100\sqrt{3}$ m

Let $h$ be the height of the tower

In $\Delta ABC$

$tan{60}^0=\frac{AB}{BC}$

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

$=>x=\frac{h}{\sqrt{3}}$ .......(i)

In $\Delta ABD$

$tan{30}^0=\frac{AB}{BD}$

$=>\frac{1}{\sqrt{3}}=\frac{h}{x+200}$

$=>x+200=\sqrt{3}h$

$=>x=\sqrt{3}h-200$ .......(ii)

From (i) and (ii) we get,

$\frac{h}{\sqrt{3}}=\sqrt{3}h-200$

$=>h=3h-200\sqrt{3}$

$=>2h=200\sqrt{3}$

$h=>100\sqrt{3}$m

$=>h=100x1.732$

$=>h=173.2$ metre

∴ Height of tower =$173.2$ m.