In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 50°.

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°

 

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