In the figure (i) given below, O is the center of the circle. If ∠AOC = 150°, find

(i) ∠ABC

(ii) ∠ADC.

(i) Arc AC subtends ∠AOC on center and ∠ABC at point B of the circle.

∴ ∠AOC = 2∠ABC (∵ as angle subtended on center of circle is double the angle subtended on any other part.)

⇒ 150° = 2∠ABC

⇒ ∠ABC = $\dfrac{150°}{2}$

⇒ ∠ABC = 75°.

Hence, the value of ∠ABC = 75°.

(ii) From figure,

ABCD is a cyclic quadrilateral.

Since opposite angles sum is 180° in cyclic quadrilateral

⇒ ∠ABC + ∠ADC = 180°
⇒ 75° + x° = 180°
⇒ x° = 180° - 75°
⇒ x° = 105°.

Hence, the value of ∠ADC = 105°.