A pole 50 m high stands on a building 250 m high. To an observerat a height of 300 m, the building and the pole subtend equal angles. The distance of the observer from the top of the pole will be

Given : Let O be the observer and XY is the pole = 50 m and YZ is the building = 250 m. $$\angle$$ XOY = $$\angle$$ YOZ = $$\theta$$

To find : XO = $$x$$ = ?

Solution : In right $$\triangle$$ XOY,

=> $$tan(\angle XOY)=\frac{YX}{XO}$$

=> $$tan\theta=\frac{50}{x}$$ -------------(i)

In right $$\triangle$$ XOZ,

=> $$tan(\angle XOZ)=\frac{ZX}{XO}$$

=> $$tan(2\theta)=\frac{300}{x}$$

=> $$\frac{2tan\theta}{1-tan^2\theta}=\frac{300}{x}$$

Substituting value from equation (i), we get :

=> $$\frac{2\times50/x}{1-(50/x)^2}=\frac{300}{x}$$

=> $$1-(\frac{50}{x})^2=\frac{1}{3}$$

=> $$(\frac{50}{x})^2=1-\frac{1}{3}=\frac{2}{3}$$

=> $$\frac{50}{x}=\sqrt{\frac{2}{3}}$$

=> $$x=\frac{50\sqrt3}{\sqrt2}$$

=> $$x=25\sqrt6$$ m

=> Ans - (D)