  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 A
30o  B
60o  C
80o  D
90o  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 CMNow, we have a cyclic quadrilateral BACMAs 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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