In an isosceles triangle `ABC, AB = AC, XY ||BC`. If `/_A = 30^@`, then `/_BXY = ?`

Options

A. `75^@`
B. `30^@`
C. `150^@`
D. `105^@`

Answer

To solve the problem, we need to find the measure of angle \( \angle BXY \) in the isosceles triangle \( ABC \) where \( AB = AC \) and \( XY \) is parallel to \( BC \). Given that \( \angle A = 30^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Angles in Triangle \( ABC \)**:
- Since \( ABC \) is an isosceles triangle with \( AB = AC \), the angles opposite these sides are equal. Therefore, we have:
\[ \angle B = \angle C = \theta \]

2. **Use the Triangle Sum Property**:
- The sum of the angles in any triangle is \( 180^\circ \). Thus, we can write:
\[ \angle A + \angle B + \angle C = 180^\circ \]
- Substituting the known values:
\[ 30^\circ + \theta + \theta = 180^\circ \]
- This simplifies to:
\[ 30^\circ + 2\theta = 180^\circ \]

3. **Solve for \( \theta \)**:
- Rearranging the equation gives:
\[ 2\theta = 180^\circ - 30^\circ \]
\[ 2\theta = 150^\circ \]
\[ \theta = \frac{150^\circ}{2} = 75^\circ \]

4. **Determine the Angles \( \angle AXY \) and \( \angle BXY \)**:
- Since \( XY \) is parallel to \( BC \), the angles formed by the transversal \( AX \) are equal:
\[ \angle AXY = \angle ABC = \theta = 75^\circ \]
- Similarly, since \( XY \) is parallel to \( BC \):
\[ \angle CXY = \angle ACB = \theta = 75^\circ \]

5. **Use the Straight Angle Property**:
- The angles along line \( AX \) must sum to \( 180^\circ \):
\[ \angle AXY + \angle BXY = 180^\circ \]
- Substituting the known angle:
\[ 75^\circ + \angle BXY = 180^\circ \]
- Solving for \( \angle BXY \):
\[ \angle BXY = 180^\circ - 75^\circ = 105^\circ \]

### Final Answer:
\[ \angle BXY = 105^\circ \]