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

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(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.
$$\begin{aligned}\Rightarrow ∠CYX = \frac{1}{2}∠COX \\[1em] \Rightarrow ∠CYX = \frac{58°}{2} \\[1em] \Rightarrow ∠CYX = 29°\end{aligned}$$
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,
$$\begin{aligned}\Rightarrow ∠XZY + ∠ZXY + ∠ZYX = 180° \\[1em] \Rightarrow ∠ZYX + 2∠XZY = 180° \\[1em] \Rightarrow 2∠XZY = 180° - ∠ZYX \\[1em] \Rightarrow ∠XZY = \frac{180° - ∠ZYX}{2} \\[1em] = \frac{180° - 66°}{2} \\[1em] = \frac{114°}{2} \\[1em] = 57° .\end{aligned}$$
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°.