A) \[\frac{\sqrt{3}}{2}BC\]
B) BC
C) \[\sqrt{3}\,\,BC\]
D) 2 BC
Correct Answer: B
Solution :
Join B and O. Then \[\angle BOC=2\,\angle BAC={{60}^{o}}\] Draw OM as perpendicular from O on BC. Then, \[BM=\frac{1}{2}BC\] and \[\angle BOM={{30}^{o}}\] From \[\Delta \,BMO,\] \[\frac{BM}{BO}=\sin {{30}^{o}}=\frac{1}{2}\] \[\therefore \] \[BO=2B\,M\] \[=2.\frac{1}{2}BC=BC\]You need to login to perform this action.
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