(i) In Figure (1), is the center of the circle. If and find (ii) In figure (2), and are three points on the circle with center such that and Find
(i) Join OB.
∠OAB = ∠OBA = 40°[Because OB = OA]
∠OCB = ∠OBC = 30°[Because OB = OC]
∠ABC = ∠OBA + ∠OBC
⇒ ∠ABC = 40°+ 30°
⇒ ∠ABC = 70°
∠AOC = 2 × ∠ABC
⇒ ∠AOC = 2 × ∠ABC
⇒ ∠AOC = 2 × 70°
⇒ ∠AOC = 140°
(ii)
∠BOC = 360° - (∠AOB + ∠AOC) [Sum of all angles at a point = 360°]
⇒ ∠BOC = 360° - (90° + 110°)
⇒ ∠BOC = 360° - 200°
⇒ ∠BOC = 160°
We know that ∠BOC = 2 × ∠BAC
⇒ ∠BAC = (1/2) × ∠BOC
⇒ ∠BAC = (1/2) × 160°
⇒ ∠BAC = 80°