In Fig. 10.6, if OAB = 40 �, then ACB is equal to:
In triangle AOB,
AO = OB = Radius
∴ ∠OAB = ∠OBA = 40° (∵ angles opposite to equal sides are equal)
Using the angle sum property of triangle, sum of all angles of a triangle is 180°,
∴ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ 40° + 40° + ∠AOB = 180°
⇒ ∠AOB = 180° - 40° - 40°
⇒ ∠AOB = 100°
By theorem “The angle subtended by an arc at the center of a circle is twice the angle subtended by it at remaining part of the circle”, we have:
∠AOB = 2 × ∠ACB
∠ACB = ∠AOB/2
= 100°/2
∴ ∠ACB = 50°