Q14 of 141 Page 464

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

Given: ∠ACB = 40°


We know that ,


∠ AOB = 2×∠ACB


∠ AOB = 2×40 = 80°


∴ ∠AOB = 80°


In ΔAOB


OA = OB (radius)


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


Let ∠OAB = ∠OBA = x


By angle sum property


∠AOB + ∠OAB + ∠OBA = 180°


80 + x + x = 180°


80 + 2x = 180°


2x = 180° – 80° = 100°


x = = 50°


∴


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