Q5 of 141 Page 420

In the given figure, is the center of the circle. If find

∠AOB = 2 × ∠ACB


⇒ ∠AOB = 2 × 50°


⇒ ∠AOB = 100°


OA =OB [Radius of the circle]


∴ ∠OAB = ∠OBA _________________ (i)


In triangle AOB,


∠OAB + ∠OBA + ∠AOB = 180°[Sum of angles of triangle]


⇒ 2 ∠OAB + 100° = 180°[From equation (i)]


⇒ 2 ∠OAB = 80°


⇒ ∠OAB = 40°


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