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Let's go through these two questions step by step.

### Question 22:
You are given a circle with center \( O \), and the points \( A \), \( B \), \( C \), and \( D \) lie on the circle. You also know that:
- \( AOC \) is a straight line.
- \( \angle BÂC = 42^\circ \).

You are asked to find:
1. \( \angle B\hat{D}C \)
2. \( \angle A\hat{B}C \)
3. \( \angle A\hat{C}B \)

#### Solution to (a): \( \angle B\hat{D}C \)
In a cyclic quadrilateral, the opposite angles add up to \( 180^\circ \). Since \( \angle B\hat{A}C = 42^\circ \), the angle at \( D \) opposite to \( \angle B\hat{A}C \) is:
\[
\angle B\hat{D}C = 180^\circ - 42^\circ = 138^\circ
\]

#### Solution to (b): \( \angle A\hat{B}C \)
\( \triangle ABC \) is an inscribed triangle in the circle. Using the property that the angle subtended by an arc at the center is twice the angle subtended on the circle:
\[
\angle A\hat{B}C = \frac{1}{2} \times \angle AOC = \frac{1}{2} \times 180^\circ = 90^\circ
\]

#### Solution to (c): \( \angle A\hat{C}B \)
Since \( \triangle ABC \) is isosceles (because \( A \), \( B \), and \( C \) lie on the circle), the sum of the angles in a triangle is \( 180^\circ \):
\[
\angle A\hat{C}B = 180^\circ - \angle B\hat{A}C - \angle A\hat{B}C = 180^\circ - 42^\circ - 90^\circ = 48^\circ
\]

---

### Question 23:
You are given another circle with center \( O \) and points \( A \), \( B \), \( C \), and \( D \) on the circle. You are told that:
- \( AOD \) is a straight line.
- \( BO \) is parallel to \( CD \).
- \( \angle C\hat{D}A = 36^\circ \).

You are asked to find:
1. \( \angle B\hat{O}A \)
2. \( \angle B\hat{C}A \)
3. \( \angle D\hat{C}B \)
4. \( \angle O\hat{B}C \)

#### Solution to (a): \( \angle B\hat{O}A \)
Since \( AOD \) is a straight line, the angle subtended by \( AB \) at the center will be half of \( \angle AOD \):
\[
\angle B\hat{O}A = \frac{1}{2} \times \angle AOD = \frac{1}{2} \times 180^\circ = 90^\circ
\]

#### Solution to (b): \( \angle B\hat{C}A \)
\( \angle B\hat{C}A \) is the same as \( \angle C\hat{D}A \) because of the parallel lines \( BO \parallel CD \), so:
\[
\angle B\hat{C}A = 36^\circ
\]

#### Solution to (c): \( \angle D\hat{C}B \)
Using the property of cyclic quadrilaterals that opposite angles sum to \( 180^\circ \), we have:
\[
\angle D\hat{C}B = 180^\circ - \angle B\hat{C}A = 180^\circ - 36^\circ = 144^\circ
\]

#### Solution to (d): \( \angle O\hat{B}C \)
Since \( BO \parallel CD \) and \( \angle D\hat{C}B = 144^\circ \), we can conclude that the angle subtended at \( B \) by \( O \) is the remaining angle:
\[
\angle O\hat{B}C = 180^\circ - 144^\circ = 36^\circ
\]

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Would you like more detailed explanations or have any further questions? Here are some related questions to deepen understanding:

1. How would you find angles in different cyclic quadrilateral setups?
2. Can the properties of parallel lines help in non-cyclic geometrical problems?
3. What is the relationship between inscribed angles and central angles in general?
4. How would the problem change if the line \( AOC \) were not straight?
5. How does the angle-sum property help in circle theorems?

**Tip:** Remember that in cyclic quadrilaterals, opposite angles sum to \( 180^\circ \), which is a key property for solving many angle-related problems.

The diagram shows a circle, centre O, passing through A, B, C, and D. AOC is a straight line and ∠BÂC = 42°. Find: (a) ∠B̂DC, (b) ∠ÂBC, (c) ∠ÂCB. The second question presents a similar circle and asks to find angles based on given data.

Explore how to solve geometry problems involving cyclic quadrilaterals, parallel lines, and angles in a circle. Learn key theorems such as the Cyclic Quadrilateral Theorem and the Inscribed Angle Theorem. Suitable for students in Grades 10-12.

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