A) $ 60{}^\circ $
B) $ 30{}^\circ $
C) $ 40{}^\circ $
D) $ 50{}^\circ $
Correct Answer: C
$ \therefore $ $ \angle BOC=2\angle BAC $ [since, angle subtended at the centre is twice the angle subtended at the point A]
$ \therefore $ $ \angle BOC=2\times 50{}^\circ =100{}^\circ $ Now, $ \angle OBC=\angle OCB $ [ $ \because $ angle along same side OB = OC are radius] Now, in $ \Delta BOC $
$ \Rightarrow $ $ \angle OBC+\angle OCB+\angle BOC=180{}^\circ $ [angle sum property]
$ \Rightarrow $ $ 2\angle OBC=180{}^\circ -100{}^\circ $
$ \Rightarrow $ $ 2\angle OBC=80{}^\circ $
$ \Rightarrow $ $ \angle OBC=40{}^\circ $