Question

A man finds that at a point due to South of a vertical tower the angle of elevation of the tower is $\frac{\pi }{3}$. He then walks due West $10\sqrt{6}$ in on the horizontal plane and find the angle of elevation of the tower to be $\frac{\pi }{6}$. Original distance of the men from the tower is

• A. $5\sqrt{3}m$
• B. $15\sqrt{3}m$
• C. $10\sqrt{3}m$
• D. None of these

Answer 1

The correct option is A $5\sqrt{3}m$

$\tan \left(\frac{\pi }{3}\right)=\frac{AO}{BO}=\frac{h}{x}⟹x=\frac{h}{\sqrt{3}}$

$\tan \left(\frac{\pi }{6}\right)=\frac{AO}{CO}=\frac{h}{y}⟹y=\sqrt{3}h$

$O{B}^2+B{C}^2=O{C}^2⟹x^2=(10\sqrt{6}{)}^2=y^2⟹\frac{h^2}{3}+600=3h^2⟹h=15$

$x=\frac{h}{\sqrt{3}}=\frac{15}{\sqrt{3}}=5\sqrt{3}$