In the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; find :

(i) ∠ACB,

(ii) ∠OBC,

(iii) ∠OAB,

(iv) ∠CBA.

Given, ∠AOB = 140° and ∠OAC = 50°

(i) We know that,

Angle at the center is double the angle at the circumference subtended by the same chord.

∠ACB = $\dfrac{1}{2}$Reflex ∠AOB

= $\dfrac{1}{2}$ (360° - 140°)

= $\dfrac{1}{2} \times 220°$ = 110°.

Hence, ∠ACB = 110°.

(ii) We know that,

The sum of angles in a quadrilateral is 360°

In quadrilateral OBCA,

∠OBC + ∠ACB + ∠OAC + ∠AOB = 360°

⇒ ∠OBC + 110° + 50° + 140° = 360°

⇒ ∠OBC + 300° = 360°

⇒ ∠OBC = 360° - 300° = 60°.

Hence, ∠OBC = 60°.

(iii) Join AB.

In ∆AOB, we have

OA = OB (radius of circle)

So, ∠OBA = ∠OAB (As angles opposite to equal sides are equal)

By angle sum property of a triangle we get,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OAB + 140° = 180°

⇒ 2∠OAB = 40°

⇒ ∠OAB = $\dfrac{40}{2}$ = 20°

Hence, ∠OAB = 20°.

(iv) We already found, ∠OBC = 60°.

⇒ ∠OBC = ∠CBA + ∠OBA

⇒ 60° = ∠CBA + 20°

⇒ ∠CBA = 60° - 20° = 40°

Hence, ∠CBA = 40°.