https://file.aiquickdraw.com/mathbot/file/17390365142184h34nb1f.jpg

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**!

In the diagram, PQ is the diameter of a circle, center O. ∠ABR = 46°. (a) Explain why ∠PRQ = 90°. (b) Explain why ∠AOR = 92°. (c) Find ∠OAR. Give a reason for your answer.

Explore a geometry problem involving the circle theorems, including the angle in a semicircle, cyclic quadrilaterals, and isosceles triangles. This problem requires knowledge of key theorems related to circle geometry and angle properties. Suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation