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 \[6400+{{x}^{2}}=200x\] \[\Rightarrow \,{{x}^{2}}-200\,x+6400\,=0\] \[\Rightarrow \,x=40\]You need to login to perform this action.
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