ABCD is a cycle quadrilateral such that is the diameter of the circle circumscribing it and `/_ ADC = 150^(@)`. Then, `/_ BAC` is equal to `:`

Options

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

Answer

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

### Step 1: Understand the properties of cyclic quadrilaterals
In a cyclic quadrilateral, the sum of the opposite angles is \( 180^\circ \). Therefore, we can write:
\[ \angle ADC + \angle ABC = 180^\circ \]

### Step 2: Substitute the known angle
We know that \( \angle ADC = 150^\circ \). Substituting this into the equation gives:
\[ 150^\circ + \angle ABC = 180^\circ \]

### Step 3: Solve for \( \angle ABC \)
Now, we can solve for \( \angle ABC \):
\[ \angle ABC = 180^\circ - 150^\circ = 30^\circ \]

### Step 4: Use the property of angles in a semicircle
Since \( AC \) is the diameter of the circle, by the property of angles subtended by a diameter, we know that:
\[ \angle ABC = 90^\circ \]

### Step 5: Analyze triangle \( ABC \)
In triangle \( ABC \), we can use the angle sum property:
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]
We already have \( \angle ABC = 30^\circ \) and \( \angle ACB = 90^\circ \). So we can substitute these values:
\[ \angle BAC + 30^\circ + 90^\circ = 180^\circ \]

### Step 6: Solve for \( \angle BAC \)
Now, simplify the equation:
\[ \angle BAC + 120^\circ = 180^\circ \]
Subtract \( 120^\circ \) from both sides:
\[ \angle BAC = 180^\circ - 120^\circ = 60^\circ \]

### Conclusion
Thus, the measure of \( \angle BAC \) is \( 60^\circ \).