Question

A man standing at a point P is watching the top of a tower, which makes an angle of elevation of ${30}^{\circ }$.The man walks some distance towards the tower and then his angle of elevation of the top of the tower is ${60}^{\circ }$. If the height of the tower is 30 m, then the distance he moves is:

• A. $22m$
• B. $22\sqrt{3}m$
• C. $20m$
• D. $20\sqrt{3}m$

Answer 1

The correct option is D $20\sqrt{3}m$

In $△APC$, In $△PBC$,

$\tan {30}^o=\frac{30}{x+y}$ $\tan {60}^o=\frac{30}{y}$

$\frac{1}{\sqrt{3}}=\frac{30}{x+y}$ $\sqrt{3}=\frac{30}{y}$

$\therefore x+y=30\sqrt{3}$ $\therefore y=\frac{30}{\sqrt{3}}$

$\therefore x+\frac{30}{\sqrt{3}}=30\sqrt{3}$

$\therefore x=30\sqrt{3}-\frac{30}{\sqrt{3}}=\frac{30\times 3-30}{\sqrt{3}}=\frac{90-30}{\sqrt{3}}=\frac{60}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$\therefore x=\frac{60\sqrt{3}}{3}$ $\therefore x=20\sqrt{3}m$