question_answer

If O is the circumcentre of a \[\Delta \,ABC\] lying inside the triangle, then \[\angle OBC+\angle BAC\] is equal to

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

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

C) \[110{}^\circ \]            

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

Correct Answer: B

Solution :

Let \[\angle A=x{}^\circ \]

\[\therefore \]      \[\angle BOC=2x{}^\circ \]

and       \[\angle BOE=x{}^\circ \]

Again, let \[\angle OBE=y{}^\circ \]

and we have \[\angle OEB=90{}^\circ \]

In \[\Delta BOE,\]

\[x{}^\circ +y{}^\circ +90{}^\circ =180{}^\circ \]

\[\Rightarrow \]               \[x{}^\circ +y{}^\circ =90{}^\circ \]

\[\therefore \]      \[\angle OBC+\angle BAC=90{}^\circ \]