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, \[BP=\frac{2ca}{c+a}\cos \frac{B}{2}\] \[\Rightarrow \,\,\,\,\,\,\,\,\frac{1}{a}+\frac{1}{c}=\frac{1}{40}\]You need to login to perform this action.
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