JEE Main & Advanced
Sample Paper
JEE Main Sample Paper-24
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 \[6400+{{x}^{2}}=200x\] \[\Rightarrow \,{{x}^{2}}-200\,x+6400\,=0\] \[\Rightarrow \,x=40\]