question_answer

The vertical poles, \[AB\] of length \[2\,\,m\] and \[CD\] of length \[20\,\,m\] are erected with base \[B\] and \[D\] respectively. It is given that distance between the poles is more than \[20\,\,m\] and \[(\angle ACB)=\frac{2}{77}\], then distance between the poles is

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\[72\,\,m.\]