In the given figure, O is the centre of the circle and ∠AOC = 160°. Prove that 3∠y − 2∠x = 140°.

We know that,

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

∠AOC = 2∠ABC

∠AOC = 2x

∠x = $\dfrac{1}{2}$ ∠AOC = $\dfrac{160^{\circ}}{2}$ = 80°.

In a cyclic quadrilateral, the sum of opposite angles is 180°.

In quadrilateral ABCD,

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠x + ∠y = 180°

⇒ 80° + ∠y = 180°

⇒ ∠y = 100°.

Substitute values in L.H.S of equation 3∠y − 2∠x = 140° :

= 3(100°) − 2(80°)

= 300° − 160°

= 140°.

As, L.H.S = R.H.S

Hence, proved that 3∠y − 2∠x = 140°.