question_answer

If 0 is the circumcentre of a triangle ABC lying inside the triangle, then \[\angle OBC+\angle BAC\] is equal to                [SSC CGL Tier II, 2015]

A) \[90{}^\circ \]

B) \[110{}^\circ \]

C) \[60{}^\circ \]

D) \[-120{}^\circ \]

Correct Answer: A

Solution :

[a] According to the question, \[\because \]       OB = OC = Radius of circle \[\therefore \]      \[\angle OBC=\angle OCB\] \[\because \]       O is the circumcentre of a\[\Delta ABC\] \[\therefore \]      \[\angle BOC=2\times \angle BAC\] In\[\Delta BOC,\]\[2\times \angle OBC+\angle BOC=180{}^\circ \]                                    \[(\therefore \angle OBC\angle OCB)\] \[\Rightarrow \]   \[2\times \angle OBC+2\times \angle BAC=180{}^\circ \] \[\therefore \]      \[\angle OBC+\angle BAC=90{}^\circ \]