In the given diagram, O is the centre of the circle and the tangent DE touches the circle at B. If ∠ADB = 32°. Find the values of x and y.

Given,

∠ADB = 32°

From figure,

∠ABC = 90° (Angle in a semicircle is a right angle)

We know that,

Angle between a tangent and a chord through point of contact is equal to an angle in the alternate segment.

∠BCA = ∠ABE = y

∠DBC = ∠BAC = x

Since, DE is a straight line, thus

⇒ ∠DBC + ∠ABC + ∠ABE = 180°

⇒ x + 90° + y = 180°

⇒ x + y = 180° - 90°

⇒ x + y = 90° ....(1)

In triangle ADB,

⇒ ∠ADB + ∠BAD + ∠ABD = 180°

⇒ 32° + x + (∠DBC + ∠ABC) = 180°

⇒ 32° + x + x + 90° = 180°

⇒ 2x + 122° = 180°

⇒ 2x = 180° - 122°

⇒ 2x = 58°

⇒ x = $\frac{58°}{2}$ = 29°.

Substituting value of x in equation (1), we get :

⇒ 29° + y = 90°

⇒ y = 90° - 29°

⇒ y = 61°.

Hence, x = 29° and y = 61°.