In the figure (ii) given below, C is a point on the minor arc AB of the circle with centre O. Given ∠ACB = p°, ∠AOB = q°, express q in terms of p. Calculate p if OACB is a parallelogram.

From figure,

⇒ ∠AOB + Reflex ∠AOB = 360°
⇒ q° + Reflex ∠AOB = 360°
⇒ Reflex ∠AOB = 360° - q°.

Arc AB subtends reflex ∠AOB at center and ∠ACB at the point C on the circle.

∴ Reflex ∠AOB = 2∠ACB

⇒ 360° - q° = 2 × p°
⇒ 360° - q° = 2p°
⇒ 2p° + q° = 360°
⇒ q° = 360° - 2p°
⇒ q° = 2(180° - p°)
⇒ q = 2(180 - p).

Given, OABC is a parallelogram, then

Opposite angles are equal.

∴ ∠AOB = ∠ACB

⇒ p° = q°
⇒ p° = 360° - 2p°
⇒ 3p° = 360°
⇒ p° = 120°.

Hence, q = 2(180 - p) and the value of p = 120.