Question:
In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 50°. Then, ∠AOC = ?
(a) 50°
(b) 70°
(c) 20°
(d) 60°
Solution:
(d) 60°
We have:
OA = OB (Radii of a circle)
⇒ ∠OBA= ∠OAB = 20°
In ΔOAB, we have:
∠OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)
⇒ 20° + 20° + ∠AOB = 180°
⇒ ∠AOB = (180° - 40°) = 140°
Again, we have:
OB = OC (Radii of a circle)
⇒ ∠OBC = ∠OCB = 50°
In ΔOCB, we have:
∠OCB + ∠OBC + ∠COB = 180° (Angle sum property of a triangle)
⇒ 50° + 50° + ∠COB = 180°
⇒ ∠COB = (180° - 100°) = 80°
Since ∠AOB = 140°, we have:
∠AOC + ∠COB = 140°
⇒∠AOC + 80° = 140°
⇒ ∠AOC = (180° - 80°) = 60°