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: $\angle O\hat{B}C = 180^\circ - 144^\circ = 36^\circ$

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.