https://file.aiquickdraw.com/mathbot/file/1729002850149oubg5xkf.jpg

Let's analyze and solve each question from the image one by one.

### Question 1
In the first diagram, we have a circle with points \(P\), \(Q\), and \(O\), where \(O\) is the center, and the angle at \(O\) is \(120^\circ\). You are asked to find:
1. \( \angle P \)
2. \( \angle Q \)

#### Solution:
- In a circle, the angle subtended at the center by a chord (in this case, \(POQ\)) is twice the angle subtended on the circumference by the same chord. This is known as the *angle at the center theorem*.

  Given:
  - \( \angle POQ = 120^\circ \)
  
  By the theorem:
  - \( \angle P = \angle Q = \frac{1}{2} \times \angle POQ = \frac{1}{2} \times 120^\circ = 60^\circ \)

Thus, \( \angle P = 60^\circ \) and \( \angle Q = 60^\circ \).

---

### Question 2
In this diagram, \(PR\) is the diameter of the circle, and you are asked to:
1. Find \( \angle PRQ \)
2. Find \( \angle PQS \)
3. Find \( \angle POS \)

#### Solution:
- **a. \( \angle PRQ \):** Since \(PR\) is the diameter, by the *angle in a semicircle theorem*, the angle subtended by a diameter on the circle is always \(90^\circ\).
  
  Therefore, \( \angle PRQ = 90^\circ \).

- **b. \( \angle PQS \):** From the diagram, \( \angle PQS = 40^\circ \) (directly labeled).

- **c. \( \angle POS \):** This angle can be calculated by using the sum of angles around point \(O\). We know from the diagram that:
  \[
  \angle POS = 180^\circ - 60^\circ = 120^\circ
  \]
  Therefore, \( \angle POS = 120^\circ \).

---

### Question 3
In the third diagram, the circle has radius \(7\) cm and point \(B\) lies outside the circle. The line segments \(AB\) and \(BC\) are tangents to the circle. You are asked to find:
1. The length of \(AB\)
2. The area of quadrilateral \(OABE\)
3. The length of \(AC\)

#### Solution:
- **a. Length of \(AB\):**
  Given:
  - \(OB = 25\) cm (distance from the center of the circle to point \(B\)).
  - The radius of the circle is \(7\) cm.

  Using the Pythagorean theorem for the right triangle \(OAB\), where \(OB\) is the hypotenuse:
  \[
  OB^2 = OA^2 + AB^2
  \]
  Substituting the values:
  \[
  25^2 = 7^2 + AB^2 \implies 625 = 49 + AB^2 \implies AB^2 = 576 \implies AB = \sqrt{576} = 24 \, \text{cm}
  \]

- **b. Area of quadrilateral \(OABE\):**
  Quadrilateral \(OABE\) consists of a right triangle \(OAB\) and a rectangle \(OABE\). The area of the triangle \(OAB\) can be calculated as:
  \[
  \text{Area of } OAB = \frac{1}{2} \times OA \times AB = \frac{1}{2} \times 7 \times 24 = 84 \, \text{cm}^2
  \]

- **c. Length of \(AC\):**
  Since \(AB = AC\) (tangents drawn from an external point to a circle are equal), the length of \(AC = 24\) cm.

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Would you like more detailed steps or clarifications on any of the solutions? Here are some additional related questions:

1. What is the relationship between angles subtended by the same arc at different points?
2. How would you calculate the length of a tangent from a different point outside the circle?
3. Can you derive the general formula for the area of a sector of a circle?
4. How do we know that \( \angle PRQ \) is \(90^\circ\) from the diagram?
5. How can we use the Pythagorean theorem in problems involving tangents?

**Tip:** For problems involving circles, always remember the fundamental properties of tangents, chords, and angles subtended by arcs or diameters.

The user provided an image containing three geometry problems related to circles. The problems ask for the calculation of angles and lengths in different circle-related diagrams.

This page explains solutions to circle geometry problems involving calculating angles subtended by arcs, finding lengths of tangents using the Pythagorean theorem, and applying properties of angles in a semicircle. Suitable for students in Grade 10-11.

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