A) \[68\,\,m\]
B) \[24\,\,m\]
C) \[72\,\,m\]
D) None of these
Correct Answer: C
Solution :
We have,\[\angle ACB=\theta \], we have \[\tan \theta =\frac{2}{77},\,\,AB=2,\,\,C=20\] Let \[\angle BCD=\alpha \] Then, \[\tan \alpha =\frac{BD}{CD}=\frac{BD}{20}\] and \[\tan (\theta +\alpha )=\frac{AE}{CE}=\frac{BD}{18}\] \[\Rightarrow \] \[\frac{BD}{18}=\tan (\theta +\alpha )\] \[\Rightarrow \] \[\frac{BD}{18}=\frac{\tan \theta +\tan \alpha }{1-\tan \theta \tan \alpha }\] \[\Rightarrow \] \[\frac{BD}{18}=\frac{\frac{2}{77}+\frac{BD}{20}}{1-\frac{2}{77}\cdot \frac{BD}{20}}\] \[\Rightarrow \] \[B{{D}^{2}}-77BD+360=0\] \[\Rightarrow \] \[(BD-5)(BD-72)=0\] \[\Rightarrow \] \[BD=5,\,\,72\] But \[BD>20\] \[\therefore \] \[BD=72\,\,cm\] Hence, the distance between poles is\[72\,\,m.\]You need to login to perform this action.
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