Let's solve the given problem step by step.

Information from the diagram:

• $P\hat{B}C = 73^\circ$
• $A\hat{O}C = 92^\circ$
• $TA$ and $TB$ are tangents to the circle from $T$ to points $A$ and $B$.
• $O$ is the center of the circle.

Part (a) $A\hat{B}T$

Since $TA$ and $TB$ are tangents from the same point $T$, the angle between the tangents at point $B$ is equal to the angle subtended by the opposite arc $ACB$ at point $P$.

By the tangent-chord theorem, the angle $A\hat{B}T$ is equal to $P\hat{B}C$ (the angle subtended by chord $BC$ at $P$).

$A\hat{B}T = P\hat{B}C = 73^\circ$

Part (b) $B\hat{C}O$

The total angle $A\hat{O}C = 92^\circ$ is given, and since $O\hat{C}B$ and $B\hat{C}O$ are isosceles, the sum of these two equal angles must add up with $A\hat{O}C$ to make $180^\circ$ at point $C$ (since it is the center of the circle).

Therefore, the two angles $B\hat{C}O$ and $O\hat{C}B$ must be: $B\hat{C}O = O\hat{C}B = \frac{180^\circ - 92^\circ}{2} = 44^\circ$

Part (c) $A\hat{D}C$

Using the property of a cyclic quadrilateral, the opposite angles add up to $180^\circ$.

Therefore, in quadrilateral $AODC$, angle $A\hat{D}C$ is the exterior angle to angle $A\hat{O}C$. Hence: $A\hat{D}C = 180^\circ - A\hat{O}C = 180^\circ - 92^\circ = 88^\circ$

Final Answers:

• (a) $A\hat{B}T = 73^\circ$
• (b) $B\hat{C}O = 44^\circ$
• (c) $A\hat{D}C = 88^\circ$

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the tangent-chord theorem, and how does it apply in this context?
2. Why are angles subtended by the same arc equal in a circle?
3. What is a cyclic quadrilateral, and what are its properties?
4. How do tangents from a common external point relate to each other in geometry?
5. How would the angles change if the circle's center was not known?

Tip: In cyclic quadrilaterals, always remember that opposite angles sum to $180^\circ$. This property is often key to solving angle problems in circles.