Question

A , B , C are three collinear points on the ground such that B lies between A and C and AB=10 m. If the angles of the elevation of the top of a vertical tower standing at C are respectively ${30}^0$ and ${60}^0$ as seen from A and B, then the height of the tower is

• A. $5\sqrt{3}m$
• B. $5m$
• C. $\frac{10\sqrt{3}}{3}$
• D. $\frac{20\sqrt{3}}{3}$

Answer 1

Answer: (A)

$5\sqrt{3}m$

Let A,B,C be the points on ground and PC be the vertical tower.

Let $BC=xm\therefore AC=(x+10)m$

$\text{In}\Delta PCB,\angle PCB=90°,\angle PBC=60°\therefore \tan 60°=\frac{PC}{BC}\sqrt{3}=\frac{PC}{x}\therefore PC=x\sqrt{3}m\to \left(1\right)$

$\text{In}\Delta PCA,\angle PCA=90°,\angle PAC=30°\tan 30°=\frac{PC}{AC}=\frac{PC}{AB+BC}\Rightarrow \frac{1}{\sqrt{3}}=\frac{PC}{10+x}\Rightarrow PC=\frac{(10+x)}{\sqrt{3}}m\to \left(2\right)$

From $\left(1\right)\&\left(2\right)$

$x\sqrt{3}=\frac{10+x}{\sqrt{3}}3x=10+x2x=10\Rightarrow x=5m$

$\therefore$ Height of tower $\left(PC\right)=5\sqrt{3}m$