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

(i) ∠ABC

(ii) ∠BCO

(iii) ∠OAB

(iv) ∠BCA

(i) From figure,

OA = OC [Radius of same circle]

In Δ AOC,

∠OCA = ∠OAC = 50° (As angles opposite to equal sides in a triangle are equal)

We know that,

The sum of the three interior angles of any triangle is always 180°.

∠AOC + ∠OCA + ∠OAC = 180°

∠AOC + 50° + 50° = 180°

∠AOC = 180° − 50° − 50° = 80°.

We know that,

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

∠ABC = $\dfrac{1}{2}$ ∠AOC

∠ABC = $\dfrac{1}{2}$ 80°

∠ABC = 40°.

Hence, ∠ABC = 40°.

(ii) From figure,

∠BOC = ∠AOB − ∠AOC = 140° − 80° = 60°.

OB = OC (Radius of same circle)

Let ∠OBC = ∠OCB = x (As angles opposite to equal sides are equal)

We know that,

The sum of the three interior angles of any triangle is always 180°.

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120^{\circ}}{2}$

⇒ x = 60°.

Hence, ∠BCO = 60°.

(iii) In ∆AOB, we have

OA = OB (radius of same 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 = $\dfrac{40^{\circ}}{2}$

⇒ ∠OAB = 20°.

Hence, ∠OAB = 20°.

(iv) We know that,

⇒ ∠BCO = 60°

⇒ ∠OCA = 50°

From figure,

⇒ ∠BCA = ∠BCO + ∠OCA

⇒ ∠BCA = 60° + 50°

⇒ ∠BCA = 110°.

Hence, ∠BCA = 110°.