In the figure (ii) given below, O is the center of the circle. If ∠AOB = 140° and ∠OAC = 50°, find

(i) ∠ACB

(ii) ∠OBC

(iii) ∠OAB

(iv) ∠CBA.

(i) From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ 140° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - 140°
⇒ Reflex ∠AOC = 220°.

Arc AB subtends Reflex ∠AOB at center and ∠ACB at point C of circle.

⇒ Reflex ∠AOB = 2 ∠ACB
⇒ 2∠ACB = 220°
⇒ ∠ACB = $\dfrac{220°}{2}$
⇒ ∠ACB = 110°.

Hence, the value of ∠ACB = 110°.

(ii) In Quadrilateral OABC,

⇒ ∠OAC + ∠ACB + ∠BOA + ∠OBC = 360°
⇒ 50° + 110° + 140° + ∠OBC = 360°
⇒ 300° + ∠OBC = 360°
⇒ ∠OBC = 360° - 300°
⇒ ∠OBC = 60°

Hence, the value of ∠OBC = 60°.

(iii) In △OAB,

OA = OB (Radius of the circle)

∠OAB = ∠OBA = x (∵ angles of equal sides in isosceles triangle are equal.)

Sum of angles in a triangle are equal,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°.
⇒ 140° + x + x = 180°
⇒ 140° + 2x = 180°
⇒ 2x = 40°
⇒ x = 20°.

Hence, the value of ∠OAB = 20°.

(iv) ∠CBA = ∠OBC - ∠OBA

⇒ ∠CBA = 60° - 20° ⇒ ∠CBA = 40°

Hence, the value of ∠CBA = 40°.