(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°