In the given diagram, O is the center of circle circumscribing the △ABC, CD is perpendicular to chord AB. ∠OAC = 32°. Find each of the unknown angles, x, y and z.

In △OAC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OCA = ∠OAC = 32° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠OCA + ∠OAC + ∠AOC = 180°

⇒ 32° + 32° + ∠AOC = 180°

⇒ ∠AOC + 64° = 180°

⇒ ∠AOC = 180° - 64°

⇒ ∠AOC (x°) = 116°.

We know that,

Angle subtended by an arc at the center of the circle is twice the angle subtended at the remaining part of the circumference.

⇒ ∠AOC = 2∠ABC

⇒ x° = 2y°

⇒ y° = $\dfrac{x°}{2} = \dfrac{116°}{2}$ = 58°.

In △ BDC,

⇒ ∠BDC + ∠BCD + ∠CBD = 180°

⇒ 90° + z° + y° = 180°

⇒ z° = 180° - 90° - y° = 90° - 58° = 32°.

Hence, x° = 116°, y° = 58° and z° = 32°.