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 \[\Delta BOE,\] |
\[x{}^\circ +y{}^\circ +90{}^\circ =180{}^\circ \] |
\[\Rightarrow \] \[x{}^\circ +y{}^\circ =90{}^\circ \] |
\[\therefore \] \[\angle OBC+\angle BAC=90{}^\circ \] |
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