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Question

In the figure, BC is a chord of the circle with centre O and A is a point on the minor arc BC. Then, $$\angle BAC -  \angle OBC$$ is equal to

614192_017293255be24d5bb3bd1e57a03385a0.png


A
30o
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B
60o
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C
80o
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D
90o
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Solution

The correct option is D $$90^o$$
Take any point M on the major arc of BC of a circle.
Join AM, BM, and CM
Now, we have a cyclic quadrilateral BACM
As sum of opposite angles of a quadrilateral is $$180^{o}$$, 
$$\therefore\ \angle BAC+\angle BMC=180^{o}$$ ........ (1)
As we know, the angle subtended by a chord at the centre is double of the angle subtended by a chord on the circle
$$\therefore\ \angle BOC = 2\angle BMC$$ ........ (2)
From (1) and (2),
$$\angle BAC+\dfrac{1}{2} \angle BOC=180^{o}$$ ........ (3)
Now, $$BO=OC$$ ........ (radii of circle)
$$\therefore\ \angle OBC=\angle OCB$$
In $$\triangle OBC$$,
$$\angle OBC+\angle OCB+\angle BOC=180^{o}$$
$$2\angle OBC+\angle BOC=180^{o}$$
$$\implies\ \angle BOC=180^{o}-2\angle OBC$$ ........ (4)
From (3) and (4),
$$\angle BAC+\dfrac{1}{2}(180^{o}-2\angle OBC)=180^{o}$$
$$\implies\ \angle BAC+90^{o}-\angle OBC=180^{o}$$
$$\therefore\ \angle BAC-\angle OBC=90^{o}$$
Hence, option D is correct.

Mathematics

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