Let's go through each part of the question step by step.

Given Information:

• $PQ$ is the diameter of the circle with center $O$.
• $\angle ABR = 46^\circ$.

(a) Explain why $\angle PRQ = 90^\circ$.

Since $PQ$ is the diameter, the angle subtended by the diameter on the circumference is always 90° (by the angle in a semicircle theorem).

Thus, $\angle PRQ = 90^\circ$.

(b) Explain why $\angle AOR = 92^\circ$.

Since $\angle ABR = 46^\circ$, we use the property that the opposite angles in a cyclic quadrilateral sum to 180°.

Thus, $\angle ABR + \angle AOR = 180^\circ$ $46^\circ + \angle AOR = 180^\circ$ $\angle AOR = 180^\circ - 88^\circ = 92^\circ$

(c) Find $\angle OAR$ and give a reason for your answer.

Since $O$ is the center of the circle, $AO$ and $OR$ are radii of the circle, meaning that $\triangle AOR$ is an isosceles triangle.

In $\triangle AOR$, we know: $\angle AOR = 92^\circ$

The sum of angles in a triangle is $180^\circ$, so: $\angle OAR + \angle ORA + \angle AOR = 180^\circ$ $\angle OAR + \angle ORA + 92^\circ = 180^\circ$

Since $\angle OAR = \angle ORA$ (because the triangle is isosceles): $2\angle OAR = 88^\circ$ $\angle OAR = 44^\circ$

Final Answers:

• (a): $\angle PRQ = 90^\circ$ (Angle in a semicircle).
• (b): $\angle AOR = 92^\circ$ (Opposite angles in a cyclic quadrilateral sum to $180^\circ$).
• (c): $\angle OAR = 44^\circ$ (Using triangle sum theorem in an isosceles triangle).

Would you like a detailed explanation of any step? Here are five related questions to deepen your understanding:

1. What is the angle in a semicircle theorem, and why does it always hold?
2. What is a cyclic quadrilateral, and how do its opposite angles behave?
3. How can you prove that the sum of angles in a triangle is always $180^\circ$?
4. How does the isosceles triangle property help in angle calculations?
5. What other theorems can be applied when working with circle geometry?

Tip:

Whenever you see a diameter in a circle and a triangle is formed on the circumference, always check for a right angle!