In the figure (i) given below, AB is diameter of a circle. If DC is parallel to AB and ∠CAB = 25°, find (i) ∠ADC (ii) ∠DAC.

(i) Join AD.

From figure,

∠BDC = ∠BAC = 25°. (∵ angles in same segment are equal.)

∠ADB = 90° (∵ angle in semicircle = 90°.)

∠ADC = ∠ADB + ∠BDC = 90° + 25° = 115°.

Hence, the value of ∠ADC = 115°.

(ii) ∠ACD = ∠CAB = 25° (∵ alternate angles are equal)

Since sum of angles in a triangle = 180°.

In △ADC,

⇒ ∠ADC + ∠DAC + ∠ACD = 180°
⇒ 115° + ∠DAC + 25° = 180°
⇒ ∠DAC + 140° = 180°
⇒ ∠DAC = 180° - 140° = 40°.

Hence, the value of ∠DAC = 40°.