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 = $$x$$ =Â ?
Solution :Â 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)
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