In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, q and r in terms of x.

We know that,

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

⇒ ∠AOB = 2∠ACB

⇒ x = 2q

⇒ q = $\dfrac{x}{2}$.

Angles in the same segment are equal.

∴ ∠ADB = ∠ACB = q = $\dfrac{x}{2}$

From figure,

∠ADC = 90° [Angle in semi-circle is a right angle.]

∠BDC = ∠ADC - ∠ADB = 90° - $\dfrac{x}{2}$.

∴ r = 90° - $\dfrac{x}{2}$.

In △EBC,

⇒ ∠EBC + ∠CEB + ∠ECB = 180° [Angle sum property of triangle]

⇒ ∠EBC + 90° + q = 180°

⇒ ∠EBC = 90° - q

From figure,

∠DBC = ∠EBC = 90° - q = 90° - $\dfrac{x}{2}$.

Angles in the same segment are equal.

∴ ∠DAC = ∠DBC

⇒ p = 90° - $\dfrac{x}{2}$.

Hence, p = 90° - $\dfrac{x}{2}$, q = $\dfrac{x}{2}$ and r = 90° - $\dfrac{x}{2}$.