ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and `angleADC = 150^(@)`. Then. `angleBAC` is equal to:

Options

A. `40^(@)`
B. `60^@`
C. `38^@`
D. `50^@`

Answer

To solve the problem, we need to find the angle \( \angle BAC \) in the cyclic quadrilateral \( ABCD \) where \( AB \) is the diameter of the circumcircle and \( \angle ADC = 150^\circ \).

### Step-by-Step Solution:

1. **Understanding the Properties of Cyclic Quadrilaterals**:
- In any cyclic quadrilateral, the opposite angles sum up to \( 180^\circ \). Therefore, we have:
\[ \angle ADC + \angle ABC = 180^\circ \]

2. **Substituting the Given Angle**:
- We know \( \angle ADC = 150^\circ \). Substituting this into the equation gives:
\[ 150^\circ + \angle ABC = 180^\circ \]

3. **Calculating \( \angle ABC \)**:
- Rearranging the equation to find \( \angle ABC \):
\[ \angle ABC = 180^\circ - 150^\circ = 30^\circ \]

4. **Using the Diameter Property**:
- Since \( AB \) is the diameter of the circle, any angle subtended by the diameter at the circumference is \( 90^\circ \). Therefore:
\[ \angle ACB = 90^\circ \]

5. **Applying the Triangle Sum Property**:
- In triangle \( ABC \), the sum of angles is \( 180^\circ \):
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]
- Substituting the known angles:
\[ \angle BAC + 30^\circ + 90^\circ = 180^\circ \]

6. **Solving for \( \angle BAC \)**:
- Simplifying the equation:
\[ \angle BAC + 120^\circ = 180^\circ \]
- Therefore:
\[ \angle BAC = 180^\circ - 120^\circ = 60^\circ \]

### Conclusion:
Thus, the angle \( \angle BAC \) is equal to \( 60^\circ \).