Question

A flag staff of $5$ mts high stands on a building of $25$ mt high. At an observer at a height of $30$ mt the flag staff and the building subtend equal angles. The distance of the observer from the top of the flag staff is

• A. $5\sqrt{3}$
• B. $5\sqrt{2}$
• C. $5\sqrt{6}$
• D. $\frac{5\sqrt{3}}{\sqrt{2}}$

Answer 1

Answer: (D)

$\frac{5\sqrt{3}}{\sqrt{2}}$

As per the data we have drawn the diagram.

Consider OB is the bisector of angle COA of triangle COA.

$\frac{OC}{OA}=\frac{BC}{AB}$

$\frac{x}{\sqrt{x^2+{30}^2}}=\frac{5}{25}$

$x=\frac{\sqrt{x^2+{30}^2}}{5}$

Squaring on both the sides we get,

$x^2=\frac{x^2+900}{25}$

$25x^2-x^2=900$

$24x^2=900$

$x^2=\frac{900}{24}$

$x=\frac{30}{\sqrt{24}}$

$x=\frac{30}{2\sqrt{6}}$

$x=\frac{15}{\sqrt{2\times 3}}$

$x=\frac{\sqrt{3}\sqrt{3}\times 5}{\sqrt{2\times 3}}$

$x=\frac{5\sqrt{3}}{\sqrt{2}}$ m