ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and `angleADC= 140^@,` then `angleBAC` is equal to

Options

A. `80^@`
B. `50^@`
C. `40^@`
D. `30^@`

Answer

To solve the problem, we need to find the measure of angle \( \angle BAC \) in the cyclic quadrilateral \( ABCD \) where \( AB \) is the diameter of the circumscribing circle and \( \angle ADC = 140^\circ \).

### Step-by-Step Solution:

1. **Understanding the Properties of a Cyclic Quadrilateral**:
- In a cyclic quadrilateral, the sum of the opposite angles is \( 180^\circ \).
- Since \( AB \) is the diameter, \( \angle ACB \) will be \( 90^\circ \) (angle subtended by a diameter).

2. **Given Information**:
- \( \angle ADC = 140^\circ \).

3. **Finding \( \angle ABC \)**:
- Using the property of cyclic quadrilaterals:
\[ \angle ABC + \angle ADC = 180^\circ \]
- Substitute \( \angle ADC \):
\[ \angle ABC + 140^\circ = 180^\circ \]
- Solving for \( \angle ABC \):
\[ \angle ABC = 180^\circ - 140^\circ = 40^\circ \]

4. **Finding \( \angle ACB \)**:
- Since \( AB \) is the diameter, we have:
\[ \angle ACB = 90^\circ \]

5. **Finding \( \angle BAC \)**:
- In triangle \( ABC \), the sum of the angles is \( 180^\circ \):
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]
- Substitute the known angles:
\[ \angle BAC + 40^\circ + 90^\circ = 180^\circ \]
- Simplifying:
\[ \angle BAC + 130^\circ = 180^\circ \]
- Solving for \( \angle BAC \):
\[ \angle BAC = 180^\circ - 130^\circ = 50^\circ \]

### Final Answer:
\[ \angle BAC = 50^\circ \]