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 BAC -

OBC is equal to​

Answer 1

Answer:

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

,

∴ ∠BAC+∠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

∴ ∠BOC=2∠BMC ........ (2)

From (1) and (2),

∠BAC+

2

1

∠BOC=180

o

........ (3)

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

∴ ∠OBC=∠OCB

In △OBC,

∠OBC+∠OCB+∠BOC=180

o

2∠OBC+∠BOC=180

o

⟹ ∠BOC=180

o

−2∠OBC ........ (4)

From (3) and (4),

∠BAC+

2

1

(180

o

−2∠OBC)=180

o

⟹ ∠BAC+90

o

−∠OBC=180

o

∴ ∠BAC−∠OBC=90

o