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Let's solve the problem step by step.

### Given:
- \( \angle AOB = 132^\circ \)
- \( \angle ACD = 59^\circ \)
- \( AOC \) is a straight line (so \( \angle AOC = 180^\circ \))

### Required:
- (a) \( \angle ATB \)
- (b) \( \angle BDA \)
- (c) \( \angle BDC \)
- (d) \( \angle OBD \)

### Step-by-Step Solution:

#### (a) Finding \( \angle ATB \)
- \( TA \) and \( TB \) are tangents to the circle at points \( A \) and \( B \).
- The angle between the tangents \( \angle ATB \) is given by \( 180^\circ - \angle AOB \).

So,
\[
\angle ATB = 180^\circ - 132^\circ = 48^\circ
\]

#### (b) Finding \( \angle BDA \)
- \( \angle BDA \) is an exterior angle to the triangle \( OBD \) at \( D \).
- We know \( \angle ACD = 59^\circ \) and \( \angle AOB = 132^\circ \).
- The angle at the center is twice the angle subtended on the circumference by the same arc.

Therefore, 
\[
\angle BDA = \frac{1}{2} \times \angle BOC = \frac{1}{2} \times 132^\circ = 66^\circ
\]

#### (c) Finding \( \angle BDC \)
- \( \angle BDC \) can be found using the exterior angle theorem or by considering the angles within quadrilateral \( BDC \) with the circle.
- The sum of the interior angles of quadrilateral \( BDCO \) is \( 360^\circ \).
  
From quadrilateral \( BDC \), we know:
\[
\angle BDA = 66^\circ
\]
So:
\[
\angle BDC = 180^\circ - \angle BDA - \angle BCD 
\]
Since \( \angle BCD \) is given as \( 59^\circ \), we have:
\[
\angle BDC = 180^\circ - 66^\circ - 59^\circ = 55^\circ
\]

#### (d) Finding \( \angle OBD \)
- \( \angle OBD \) can be found using the angles in \( \triangle OBD \).
  
We know:
\[
\angle OBD = 180^\circ - \angle BDA - \angle BDC 
\]
So, substituting:
\[
\angle OBD = 180^\circ - 66^\circ - 55^\circ = 59^\circ
\]

Thus, the final answers are:
- (a) \( \angle ATB = 48^\circ \)
- (b) \( \angle BDA = 66^\circ \)
- (c) \( \angle BDC = 55^\circ \)
- (d) \( \angle OBD = 59^\circ \)

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

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**Relative Questions:**
1. How can you find the angle at the center of a circle from the subtended angle at the circumference?
2. What is the exterior angle theorem, and how is it applied in circle theorems?
3. Why is the angle subtended by a tangent and a radius always 90 degrees?
4. How does knowing that \( AOC \) is a straight line help in solving the problem?
5. How can you determine the interior angles of a cyclic quadrilateral?

**Tip:** Always remember that the angle subtended at the center of a circle by an arc is twice the angle subtended at any point on the remaining circumference.

In the diagram, the points A, B, C, and D lie on the circle with center O. TA and TB are tangents touching the circle at A and B, respectively. Given angles: ∠AOB = 132°, ∠ACD = 59°, and AOC is a straight line. Find (a) ∠ATB, (b) ∠BDA, (c) ∠BDC, and (d) ∠OBD.

Learn how to solve circle theorem problems involving tangents, chords, and cyclic quadrilaterals. Detailed solutions to find angles ∠ATB, ∠BDA, ∠BDC, and ∠OBD with relevant theorems and formulas. Suitable for high school students in Grades 10-12.

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