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?

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.