`/_\ABC ` is made on diameter in semicircle. Such that `/_BAC = 30^(@)`, the value of `/_BCA` is

Options

A. `30`
B. `45^(@)`
C. `60^(@)`
D. `80^(@)`

Answer

To solve the problem, we need to find the value of angle \( \angle BCA \) in triangle \( \triangle ABC \) where \( \triangle ABC \) is inscribed in a semicircle with \( AB \) as the diameter and \( \angle BAC = 30^\circ \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Since \( \triangle ABC \) is inscribed in a semicircle with \( AB \) as the diameter, we can use the property of angles in a semicircle. This property states that the angle opposite the diameter (in this case, \( \angle ACB \)) is a right angle (90 degrees).

**Hint**: Recall that any triangle inscribed in a semicircle has one angle equal to 90 degrees when the opposite vertex lies on the semicircle.

2. **Identifying the Angles**:
- We know \( \angle BAC = 30^\circ \) and \( \angle ACB = 90^\circ \).

**Hint**: Write down the known angles of the triangle to keep track of what you have.

3. **Using the Triangle Angle Sum Property**:
- The sum of the angles in any triangle is \( 180^\circ \). Therefore, we can write the equation:
\[ \angle BAC + \angle ACB + \angle BCA = 180^\circ \]
Substituting the known values:
\[ 30^\circ + 90^\circ + \angle BCA = 180^\circ \]

**Hint**: Remember the triangle angle sum property: \( \text{Sum of angles} = 180^\circ \).

4. **Solving for \( \angle BCA \)**:
- Now we can simplify the equation:
\[ 120^\circ + \angle BCA = 180^\circ \]
Subtract \( 120^\circ \) from both sides:
\[ \angle BCA = 180^\circ - 120^\circ = 60^\circ \]

**Hint**: Isolate the variable to find the unknown angle.

5. **Conclusion**:
- Therefore, the value of \( \angle BCA \) is \( 60^\circ \).

### Final Answer:
\[ \angle BCA = 60^\circ \]