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Question

In the given figure, O is the centre of the circle. If A is any point on minor arc BC, then $$\angle BAC-\angle OBC=$$ _________.
725955_4e7aa35ce64e4ed88e484daa3754a527.png


A
45o
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B
135o
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C
150o
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D
90o
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Solution

The correct option is C $$90^o$$
$$\angle BAC$$ and $$\angle BOC$$ are supplementary.
$$\therefore \angle BOC+\angle BAC=180^o\dots eqn (1)$$
 and $$\angle OBC=\angle OCB\,(\because OB=OC) $$
$$\Rightarrow \angle OBC=90^o-\angle BOC $$
Putting the above value of $$\angle BOC$$ in equation (1), we get
$$\angle 90^o-\angle OBC+\angle BAC=180^o$$
$$\therefore \angle BAC-\angle OBC=90^o$$

Mathematics

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