In the given figure, AB = BC = DC and ∠AOB = 50°. Find :

(i) ∠AOC

(ii) ∠AOD

(iii) ∠BOD

(iv) ∠OAC

(v) ∠ODA

(i) We know that,

Equal chords subtend equal angles at the center.

Since, AB = BC = DC

∴ ∠BOC = ∠AOB = 50°.

From figure,

⇒ ∠AOC = ∠BOC + ∠AOB = 50° + 50° = 100°.

Hence, ∠AOC = 100°.

(ii) We know that,

Equal chords subtend equal angles at the center.

Since, AB = BC = DC

∴ ∠DOC = ∠AOB = 50°.

From figure,

⇒ ∠AOD = ∠AOC + ∠DOC = 100° + 50° = 150°.

Hence, ∠AOD = 150°.

(iii) From figure,

⇒ ∠BOD = ∠BOC + ∠DOC = 50° + 50° = 100°.

Hence, ∠BOD = 100°.

(iv) Join AC.

In △ AOC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OAC = ∠OCA = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ y + y + 100° = 180°

⇒ 2y + 100° = 180°

⇒ 2y = 180° - 100°

⇒ 2y = 80°

⇒ y = $\dfrac{80°}{2}$

⇒ y = 40°

⇒ ∠OAC = 40°.

Hence, ∠OAC = 40°.

(v) Join AD.

In △ OAD,

⇒ OA = OD (Radius of same circle)

⇒ ∠ODA = ∠OAD = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAD + ∠ODA + ∠AOD = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°

⇒ ∠ODA = 15°.

Hence, ∠OAD = 15°.