question_answer

A rectangle ABCD is inscribed in a circle with centre O. If AC is the diagonal and \[\angle \,\text{BAC}\,\text{= 30}{}^\circ \], then radius of the circle will be equal to

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 \[\Delta \,BMO,\]                 \[\frac{BM}{BO}=\sin {{30}^{o}}=\frac{1}{2}\] \[\therefore \]  \[BO=2B\,M\]                 \[=2.\frac{1}{2}BC=BC\]