The figure given below, shows a circle with centre O.

Given : ∠AOC = a and ∠ABC = b.

(i) Find the relationship between a and b.

(ii) Find the measure of angle OAB, if OABC is a parallelogram.

(i) We know that,

Angle at the centre is double the angle at the circumference subtended by the same chord

∴ ∠ABC = $\dfrac{1}{2}$ Reflex ∠COA

⇒ b = $\dfrac{1}{2}$ (360° - a)

⇒ 2b = 360° - a

⇒ a + 2b = 360° …..(1)

Hence, relationship between a and b is given by the equation : a + 2b = 360°.

(ii) From equation 1, we get :

⇒ a + 2b = 360°

⇒ a = 360° - 2b

As OABC is a parallelogram, the opposite angles are equal.

So, a = b

⇒ 360° - 2b = b

⇒ 3b = 360°

⇒ b = $\dfrac{360°}{3}$ = 120°

Let ∠OAB = x and ∠OCB = x.

⇒ ∠OAB + ∠OCB + ∠AOC + ∠ABC = 360°

⇒ x + x + a + b = 360°

⇒ 2x + 120° + 120° = 360°

⇒ 2x + 240° = 360°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

Hence, ∠OAB = 60°.