question_answer

A vertical pole (more than 100 m high) consists of two portions, the lower being one-third of the whole. If the upper portion subtends an angle \[{{\tan }^{-1}}\frac{1}{2}\] at a point in a horizontal plane through the foot of the pole and distance 40 ft from it, then the height of the pole is [AMU 1981]

A) 100 ft

B) 120 ft

C) 150 ft

D) None of these

Correct Answer: B

Solution :

Obviously, from figure \[\tan \alpha =\frac{h/3}{40}=\frac{h}{120}\]         .....(i) \[\tan \beta =\frac{h}{40}=\frac{3h}{120}\] .....(ii) Therefore \[\tan \theta =\tan (\beta -\alpha )\] \[\Rightarrow \] \[\frac{1}{2}=\frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3{{h}^{2}}}{14400}}\Rightarrow h=120,\,40\] But \[h=40\] cannot be taken according to the condition, therefore\[h=120ft\].