In the given figure, O is the centre of the circle and AD = AB. If ∠ABC = 32°, find the angles $x, y, z, v$ and $w$.

Given,

O is the centre of the circle, AD = AB and ∠ABC = 32°.

From the figure, BC passes through the centre O and hence BC is a diameter of the circle.

We know that,

Angle in a semi-circle is a right angle.

∴ ∠BAC = 90°

In ∆ABC,

⇒ ∠ACB + ∠BAC + ∠ABC = 180° [Angle sum property of triangle]

$$\begin{aligned}\Rightarrow x + 90^{\circ} + 32^{\circ} = 180^{\circ} \\[1em] \Rightarrow x = 180^{\circ} - 122^{\circ} \\[1em] \Rightarrow x = 58^{\circ}.\end{aligned}$$

We know that,

The angle between a chord and a tangent at the point of contact is equal to the angle in the alternate segment.

$$\begin{aligned}\Rightarrow y = \angle ACB \\[1em] \Rightarrow y = x \\[1em] \Rightarrow y = 58^{\circ}.\end{aligned}$$

We know that,

Angles in the same segment of a circle are equal.

Here, ∠ADB and ∠ACB are the angles in the same segment subtended by the chord AB.

$$\begin{aligned}\Rightarrow z = \angle ADB = \angle ACB \\[1em] \Rightarrow z = x \\[1em] \Rightarrow z = 58°.\end{aligned}$$

In ∆ADB,

⇒ AD = AB [Given]

⇒ ∠ABD = ∠ADB [Angles opposite to equal sides are equal]

⇒ $\angle ABD = z = 58°$

By angle sum property of triangle,

$$\begin{aligned}\Rightarrow v + \angle ADB + \angle ABD = 180° \\[1em] \Rightarrow v + 58° + 58° = 180° \\[1em] \Rightarrow v = 180° - 116° \\[1em] \Rightarrow v = 64°.\end{aligned}$$

In cyclic quadrilateral ADEB,

We know that,

The sum of the opposite angles of a cyclic quadrilateral is 180°.

$$\begin{aligned}\Rightarrow v + w = 180° \\[1em] \Rightarrow 64° + w = 180° \\[1em] \Rightarrow w = 180° - 64° \\[1em] \Rightarrow w = 116°.\end{aligned}$$

Hence, $\bm{x = 58°}, \bm{y = 58°}, \bm{z = 58°}, \bm{v = 64°}$ and $\bm{w} = 116°$.