X, Y, Z and C are the points on the circumference of a circle with centre O. AB is a tangent to the circle at X and ZY = XY. Given ∠OBX = 32° and ∠AXZ = 66°. Find:

(a) ∠BOX

(b) ∠CYX

(c) ∠ZYX

(d) ∠OXY

(a) Given,

In ΔBOX, OX ⟂ BX (radius ⟂ tangent at its point of contact)

⇒ ∠OXB = 90°.

By angle‑sum property of triangle,

⇒ ∠BOX + ∠OBX + ∠OXB = 180°

⇒ ∠BOX + 32° + 90° = 180°

⇒ ∠BOX + 122° = 180°

⇒ ∠BOX = 180° - 122°

⇒ ∠BOX = 58°.

Hence, ∠BOX = 58°.

(b) From figure,

∠COX = ∠BOX = 58°

We know that,

The angle which, an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

$\Rightarrow ∠CYX = \dfrac{1}{2}∠COX \\[1em] \Rightarrow ∠CYX = \dfrac{58°}{2} \\[1em] \Rightarrow ∠CYX = 29°$

Hence, ∠CYX = 29°.

(c) We know that,

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

∠ZYX = ∠AXZ = 66°

Hence, ∠ZYX = 66°.

(d) From figure,

In isosceles ΔZXY,

ZY = XY

We know that,

The angles opposite to equal side of a triangle are equal.

∠ZXY = ∠XZY

By angle sum property in ΔXYZ,

$\Rightarrow ∠XZY + ∠ZXY + ∠ZYX = 180° \\[1em] \Rightarrow ∠ZYX + 2∠XZY = 180° \\[1em] \Rightarrow 2∠XZY = 180° - ∠ZYX \\[1em] \Rightarrow ∠XZY = \dfrac{180° - ∠ZYX}{2} \\[1em] = \dfrac{180° - 66°}{2} \\[1em] = \dfrac{114°}{2} \\[1em] = 57° .$

We know that,

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

∠YXB = ∠XZY = 57°.

Also,

∠ZXY = ∠XZY = 57°.

From figure,

⇒ ∠OXY = ∠OXB - ∠YXB

⇒ ∠OXY = 90° - 57° = 33°.

Hence, ∠OXY = 33°.