Q3 of 141 Page 464

In the given figure, O is the centre of a circle. If ∠OAB = 40°, then ∠ACB = ?

Given: ∠OAB = 40°


Consider ΔAOB


Here,


OA = OB (radius)


∠OBA = ∠OAB = 40° (angles opposite to equal sides are equal)


By angle sum property


∠OBA + ∠OAB + ∠AOB = 180°


40° + 40° + ∠AOB = 180°


∠AOB = 180° — 40° — 40° = 100°


We know that,


∠AOB = 2× ∠ACB


∠AOB = ∠ACB


×100° = ∠ACB


∠ACB = 50°


∴∠ACB = 50°

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