question_answer

The upper \[\frac{3}{4}\]th portion of a vertical pole subtends an angle \[{{\tan }^{-1}}\frac{3}{5}\] at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is

A) 80 m

B) 20 m

C) 40 m

D) 60 m

Correct Answer: C

Solution :

[c] \[\theta =\alpha +\beta ,\beta ={{\tan }^{-1}}\left( \frac{3}{5} \right)\] or         \[\beta =\theta -\alpha \] \[\Rightarrow \tan \beta =\frac{\tan \theta -\tan \alpha }{1+\tan \theta .\tan \alpha }\] or \[\frac{3}{5}=\frac{\frac{h}{40}-\frac{h}{160}}{1+\frac{h}{40}.\frac{h}{160}}\] \[\Rightarrow {{h}^{2}}-200h+6400=0\Rightarrow h=40\] or 160 metre \[\therefore \] possible height = 40 metre