In the diagram, XT and ZT are tangents to a circle with centre O at X and Z. If angle OZX = 25 degrees, calculate angle OXZ, angle XOZ and angle XWZ.

Step 1: Identify given information and make an assumption.
Given that $XT$ and $ZT$ are tangents to a circle with center $O$ at points $X$ and $Z$ respectively.
The diagram shows $\angle OZX = 25^\circ$. The text states $\angle OZY = 25^\circ$. We will assume that $\angle OZY$ in the text refers to $\angle OZX$ as depicted in the diagram.

i) Calculate $\angle OXZ$
$OX$ and $OZ$ are radii of the same circle, so $OX = OZ$.
Therefore, $\triangle OXZ$ is an isosceles triangle.
In an isosceles triangle, the angles opposite the equal sides are equal.

$$\angle OXZ = \angle OZX$$

From the diagram, $\angle OZX = 25^\circ$.

$$\angle OXZ = 25^\circ$$

Reason: Base angles of an isosceles triangle are equal.

ii) Calculate $\angle XOZ$
The sum of angles in a triangle is $180^\circ$. In $\triangle OXZ$:

$$\angle XOZ + \angle OXZ + \angle OZX = 180^\circ$$

Substitute the known values:

$$\angle XOZ + 25^\circ + 25^\circ = 180^\circ$$ $$\angle XOZ + 50^\circ = 180^\circ$$ $$\angle XOZ = 180^\circ - 50^\circ$$ $$\angle XOZ = 130^\circ$$

Reason: Sum of angles in a triangle is $180^\circ$.

iii) Calculate $\angle XWZ$
$\angle XWZ$ is the angle subtended by arc $XZ$ at the circumference.
$\angle XOZ$ is the angle subtended by the same arc $XZ$ at the center.
The angle at the center is twice the angle at the circumference subtended by the same arc.

$$\angle XWZ = \frac{1}{2} \angle XOZ$$

Substitute the value of $\angle XOZ$:

$$\angle XWZ = \frac{1}{2} \times 130^\circ$$ $$\angle XWZ = 65^\circ$$

Reason: Angle at the center is twice the angle at the circumference.

iv) Calculate $\angle XTZ$
$OX$ is a radius and $XT$ is a tangent at point $X$. A radius is perpendicular to the tangent at the point of tangency.

$$\angle OXT = 90^\circ$$

Similarly, $OZ$ is a radius and $ZT$ is a tangent at point $Z$.

$$\angle OZT = 90^\circ$$

Consider the quadrilateral $OXTZ$. The sum of interior angles in a quadrilateral is $360^\circ$.

$$\angle XTZ + \angle OXT + \angle OZT + \angle XOZ = 360^\circ$$

Substitute the known values:

$$\angle XTZ + 90^\circ + 90^\circ + 130^\circ = 360^\circ$$ $$\angle XTZ + 310^\circ = 360^\circ$$ $$\angle XTZ = 360^\circ - 310^\circ$$ $$\angle XTZ = 50^\circ$$

Reason: Radius is perpendicular to tangent at the point of contact. Sum of angles in a quadrilateral is $360^\circ$.

Final Answers:

i) $\boxed{\text{25}^\circ}$

ii) $\boxed{\text{130}^\circ}$

iii) $\boxed{\text{65}^\circ}$

iv) $\boxed{\text{50}^\circ}$