A, B, C are three point on a circle. The tangent at A meets BC produced at T,`angleBTA=40^(@),angleCAT=44^(@)` . The angle subtended by BC at the centre of the circle is -

Options

A. `84^(@)`
B. `92^(@)`
C. `96^(@)`
D. `104^(@)`

Answer

To solve the problem, we need to find the angle subtended by chord BC at the center of the circle, given the angles at point T where the tangent at point A meets line BC produced.

### Step-by-Step Solution:

1. **Identify Given Angles:**
- Angle \( BTA = 40^\circ \)
- Angle \( CAT = 44^\circ \)

2. **Find Angle \( ACT \):**
- Since \( T \) is a point on the tangent and \( A \) is a point on the circle, we can use the property of angles in a triangle.
- The angle \( ACT \) can be found using the fact that the sum of angles in triangle \( ACT \) is \( 180^\circ \).
- Therefore,
\[ ACT = 180^\circ - (BTA + CAT) = 180^\circ - (40^\circ + 44^\circ) = 180^\circ - 84^\circ = 96^\circ \]

3. **Find Angle \( ACB \):**
- Angle \( ACB \) is the angle subtended by chord \( BC \) at point \( A \).
- Angle \( ACB \) is equal to angle \( CAT \) because they are in the same segment of the circle.
- Thus,
\[ ACB = CAT = 44^\circ \]

4. **Find Angle \( BAC \):**
- To find angle \( BAC \), we can use the triangle \( ABC \).
- The sum of angles in triangle \( ABC \) gives:
\[ BAC + ACB + ABC = 180^\circ \]
- We already know \( ACB = 44^\circ \) and \( ABC = 40^\circ \) (since \( ABC \) is equal to \( BTA \)).
- Therefore,
\[ BAC + 44^\circ + 40^\circ = 180^\circ \]
\[ BAC + 84^\circ = 180^\circ \]
\[ BAC = 180^\circ - 84^\circ = 96^\circ \]

5. **Find Angle at the Center:**
- The angle subtended by chord \( BC \) at the center of the circle is twice the angle subtended at any point on the circumference.
- Therefore, the angle subtended by \( BC \) at the center \( O \) is:
\[ \text{Angle at center} = 2 \times BAC = 2 \times 96^\circ = 192^\circ \]

### Final Answer:
The angle subtended by chord \( BC \) at the center of the circle is \( 192^\circ \).

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