question_answer

In \[\Delta ABC,\] \[\angle B=\frac{2\pi }{3}.\]A line through B meets AC internally at P. If PQ is perpendicular to AB and PR is perpendicular to BC such that \[PQ=PR=20\sqrt{3}\]units, then the value of \[\frac{1}{AB}+\frac{1}{BC}\] is

A) \[\frac{1}{20}\]                    

B)        \[\frac{1}{40}\]             

C)   \[\frac{1}{60}\]                     

D)       \[\frac{1}{80}\]

Correct Answer: B

Solution :

[b] \[\angle B=120{}^\circ \] \[PQ=PR=20\sqrt{3}\] So, BP is angular bisector of B. \[BP=\frac{PR}{\sin 60{}^\circ }=40\] Now, \[BP=\frac{2ca}{c+a}\cos \frac{B}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{1}{a}+\frac{1}{c}=\frac{1}{40}\]