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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Identify the Properties of the Cyclic Quadrilateral:**
In a cyclic quadrilateral, the sum of the opposite angles is \( 180^\circ \). Therefore, we can write:
\[ \angle ADC + \angle ABC = 180^\circ \]

2. **Substitute the Given Angle:**
We know that \( \angle ADC = 140^\circ \). Substituting this value into the equation gives:
\[ 140^\circ + \angle ABC = 180^\circ \]

3. **Solve for \( \angle ABC \):**
Rearranging the equation to find \( \angle ABC \):
\[ \angle ABC = 180^\circ - 140^\circ = 40^\circ \]

4. **Use the Property of Angles Subtended by the Diameter:**
Since \( AB \) is the diameter of the circle, the angle subtended at the circumference by the diameter is a right angle. Therefore:
\[ \angle ACB = 90^\circ \]

5. **Find \( \angle BAC \):**
Now, we can find \( \angle BAC \) using the fact that the sum of angles in triangle \( ABC \) is \( 180^\circ \):
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]
Substituting the known angles:
\[ \angle BAC + 40^\circ + 90^\circ = 180^\circ \]

6. **Solve for \( \angle BAC \):**
Rearranging gives:
\[ \angle BAC = 180^\circ - 40^\circ - 90^\circ = 50^\circ \]

### Final Answer:
Thus, \( \angle BAC = 50^\circ \).

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