In the figure (i) given below, AB is a chord of the circle with centre O, BT is tangent to the circle. If ∠OAB = 32°, find the values of x and y.

In △OAB,

OA = OB (∵ both are radius of the common circle.)

So, △OAB is a isosceles triangle with,

∠OBA = ∠OAB = 32°.

Since sum of angles in a triangle = 180°.

In △OAB,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°
⇒ 32° + 32° + ∠AOB = 180°
⇒ 64° + ∠AOB = 180°
⇒ ∠AOB = 180° - 64°
⇒ ∠AOB = 116°.

Arc AB subtends ∠AOB at centre and ∠ACB at remaining part of circle.

∴ ∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of the circle.)

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

From figure,

∠ABT = ∠ACB = 58° (∵ angles in alternate segments are equal.)

∴ x = 58°.

Hence, the value of x = 58° and y = 58°.