Question

BC is the chord of circle with centre o ​

Answer 1

$\huge\mathfrak\red{Question}$

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 _____

$\huge\mathfrak\red{Answer}$

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 °

∴ ∠BAC+∠BMC=180° ____(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+ 1/2 ∠BOC=180°___ (3)

Now, BO=OC ___(radii of circle)

∴ ∠OBC=∠OCB

In △OBC

In △OBC,

∠OBC+∠OCB+∠BOC=180°

2∠OBC+∠BOC=180°

⟹ ∠BOC=180° −2∠OBC ____(4)

From (3) and (4)

BAC+ 1/2 (180° −2∠OBC)=180°

⟹ ∠BAC+90°−∠OBC=180°

∴ ∠BAC−∠OBC=90°

Hence,solved

hope it helps you