In the given figure, AOB is a diameter and DC is parallel to AB. If ∠CAB = x°; find (in terms of x) the values of :

(i) ∠COB,

(ii) ∠DOC,

(iii) ∠DAC,

(iv) ∠ADC.

(i) We know that,

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

⇒ ∠COB = 2∠CAB = 2x

Hence, ∠COB = 2x.

(ii) As DC || OB

⇒ ∠OCD = ∠COB = 2x [Alternate angles]

In △OCD,

OC = OD [Radius of same circle]

∠ODC = ∠OCD = 2x [Angles opposite to equal sides are equal]

⇒ ∠ODC + ∠OCD + ∠DOC = 180°

⇒ 2x + 2x + ∠DOC = 180°

⇒ ∠DOC = 180° - 4x.

Hence, ∠DOC = 180° - 4x.

(iii) We know that,

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

⇒ ∠DOC = 2∠DAC

⇒ ∠DAC = $\dfrac{1}{2}$∠DOC = $\dfrac{1}{2}$ x (180° - 4x) = 90° - 2x.

Hence, ∠DAC = 90° - 2x.

(iv) DC || AO

∴ ∠ACD = ∠OAC = x (Alternate angles are equal)

In △ADC,

⇒ ∠ADC + ∠DAC + ∠ACD = 180° [Angle sum property of triangle]

⇒ ∠ADC + 90° - 2x° + x = 180°

⇒ ∠ADC + 90° - x = 180°

⇒ ∠ADC = 180° - 90° + x

⇒ ∠ADC = 90° + x

Hence, ∠ADC = 90° + x.