question_answer

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

A)  20m                            

B)  40m

C)  60 m                           

D)  80 m

Correct Answer: B

Solution :

            Let height of tower be x               In \[\Delta ABD\]             \[\tan (\theta +\alpha )\,=\frac{x}{40}\]             In \[\Delta ABC,\]             \[\tan \theta =\frac{x}{160}\]             \[\therefore \,\,\tan \alpha =\tan \,((\theta +\alpha )\,-\theta )\]             \[=\frac{\tan (\theta +\alpha )\,-\tan \theta }{1+\tan (\theta +\alpha ).\,\tan \theta }\] \[\frac{=\frac{x}{40}-\frac{x}{160}}{1+\frac{{{x}^{2}}}{(40)(160)}}=\frac{3}{5}\] \[\left( As,\,\,\tan \alpha =\frac{3}{5} \right)\] \[\therefore \] On solving \[6400+{{x}^{2}}=200x\] \[\Rightarrow \,{{x}^{2}}-200\,x+6400\,=0\] \[\Rightarrow \,x=40\]