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. ${30}^o$
• B. ${60}^o$
• C. ${80}^o$
• D. ${90}^o$

Answer 1

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),

$\angle BAC+\frac{1}{2}\angle BOC={180}^o$ ........ (3)

Now, $BO=OC$ ........ (radii of circle)

$\therefore \angle OBC=\angle OCB$

In $△OBC$,

$\angle OBC+\angle OCB+\angle BOC={180}^o$

$2\angle OBC+\angle BOC={180}^o$

$⟹\angle BOC={180}^o-2\angle OBC$ ........ (4)

From (3) and (4),

$\angle BAC+\frac{1}{2}({180}^o-2\angle OBC)={180}^o$

$⟹\angle BAC+{90}^o-\angle OBC={180}^o$

$\therefore \angle BAC-\angle OBC={90}^o$

Hence, option D is correct.