In the given figure, O is the centre of the circle and ∠AOB = 110°. Calculate:

(i) ∠ACO

(ii) ∠CAO.

(i) We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

⇒ ∠AOB = 2∠ACO

⇒ 110° = 2∠ACO

⇒ ∠ACO = $\dfrac{110^{\circ}}{2}$

⇒ ∠ACO = 55°

Hence, ∠ACO = 55°.

(ii) From figure,

⇒ ∠COA + ∠AOB = 180° [Linear pair]

⇒ ∠COA + 110° = 180°

⇒ ∠COA = 180° - 110°

⇒ ∠COA = 70°

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

In ΔABC,

⇒ ∠COA + ∠ACO + ∠CAO = 180°

⇒ 70° + 55° + ∠CAO = 180°

⇒ 125° + ∠CAO = 180°

⇒ ∠CAO = 180° - 125°

⇒ ∠CAO = 55°.

Hence, ∠CAO = 55°.