Question

The angle of elevation of a tower from a point on the ground is ${30}^o$. At a point on the horizontal line passing through the foot of the tower and $100$ meters nearer to it, the angle of elevation is found to be ${60}^o$, then the height of the tower is

• A. $50\sqrt{3}$ meters
• B. $\frac{50}{\sqrt{3}}$ meters
• C. $\frac{100}{\sqrt{3}}$ meters
• D. $100\sqrt{3}$ meters

Answer 1

The correct option is A $50\sqrt{3}$ meters

Let $CD=xm$

$BC=100m$

From $△ACD,$

$tan{60}^o=\frac{AD}{CD}$

$\Rightarrow \sqrt{3}=\frac{AD}{x}$

$\Rightarrow AD=x\sqrt{3}..........\left(i\right)$

From $△ABD$

$tan{30}^o=\frac{AD}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AD}{x+100}$

$\Rightarrow \sqrt{3}AD=x+100$

$\Rightarrow 3x=x+100$ $\left[By\right(i\left)\right]$

$\Rightarrow x=50$

Hence the height of the tower AD$=50\times \sqrt{3}m=50\sqrt{3}m$