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In the circle below, DOB is a diameter.

Circle with quadrilateral ABCD, where all points are on the circumference and BD forms the diameter. Angle DAO is 28 degrees and andgle OCA is 24 degrees.

∠ADO =  blank                                    ∠ABC =  blank

∠OCA =  blank                                    ∠ACB =  blank

∠OBC =  blank                                    ∠CAB =  blank

24° 28° 34° 38° 52° 58° 114° 120°

To solve the angles in this circle, let's start by analyzing the given information:

- \( \angle DAO = 28^\circ \)
- \( \angle OCA = 24^\circ \)
- \( BD \) is a diameter, so \( \angle BOD = 180^\circ \) (because the angle subtended by a diameter in a circle is always a right angle).

We need to find the following angles:
1. \( \angle ADO \)
2. \( \angle ABC \)
3. \( \angle OCA \) (already given as \( 24^\circ \))
4. \( \angle ACB \)
5. \( \angle OBC \)
6. \( \angle CAB \)

### Step 1: \( \angle ADO \)
- In triangle \( \triangle ADO \), we know \( \angle DAO = 28^\circ \) and \( \angle AOD \) (because \( AO \) is a radius, and thus \( \angle AOD \) is 90° - \( \angle ODA \) ).

We need to find \( \angle ADO \). Since the angles in a triangle add up to \( 180^\circ \):

\[
\angle ADO + \angle DAO + \angle OAD = 180^\circ
\]

Given that \( \angle OAD = 28^\circ \), we can find \( \angle ADO \).

### Step 2: \( \angle ABC \)
- \( \angle ABC \) is subtended by the arc \( AC \). Using the property that the angle subtended by an arc at the center is twice the angle subtended at the circumference, we can find \( \angle ABC \).

### Step 3: \( \angle ACB \)
- Similarly, \( \angle ACB \) can be found using the properties of the angles in a circle, such as the fact that opposite angles of a cyclic quadrilateral sum to \( 180^\circ \).

### Step 4: \( \angle OBC \)
- This angle can be found by considering triangle \( OBC \) and using the known angles.

### Step 5: \( \angle CAB \)
- \( \angle CAB \) is the same as \( \angle ABC \) since they subtend the same arc.

Let me calculate these angles for you.Based on the calculations:

- \( \angle ADO = 90^\circ \)
- \( \angle ABC = 66^\circ \)
- \( \angle OCA = 24^\circ \) (as given)
- \( \angle ACB = 24^\circ \)
- \( \angle OBC = 66^\circ \)
- \( \angle CAB = 66^\circ \)

Would you like more details on any of these calculations, or do you have any other questions?

### Related Questions:
1. What is the relationship between the angles subtended by the same arc at the center and the circumference of the circle?
2. How does the angle subtended by a diameter relate to the angles at the circumference?
3. Can you determine if a quadrilateral is cyclic if one angle is \( 90^\circ \)?
4. How do you use the properties of cyclic quadrilaterals to determine unknown angles?
5. How does knowing that \( BD \) is a diameter help in solving the problem?

### Tip:
Remember that in a circle, the angle subtended by a chord at the center is twice the angle subtended at the circumference, which is a key property for solving such geometric problems.

Explore how to solve a circle geometry problem involving a diameter and angles at the circumference. Step-by-step solution provided, including angles subtended by a diameter and arcs. Suitable for high school students learning circle geometry.

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