Correct option is (B) 45°
\(\because\) AB = BC (sides of square ABCD)
\(\therefore\) \(\triangle ABC\) is an isosceles triangle.
\(\therefore\) \(\angle BAC\) = \(\angle ACB\) ______(1) (Opposite angles of equal sides are equal)
Now in \(\triangle ABC\), \(\angle ABC\) = \(90^\circ\) (angle in square)
Also, \(\angle ABC+\angle BAC+\angle ACB\) \(=180^\circ\) (Sum of angles in a triangle)
\(\Rightarrow\) \(90^\circ+\) \(2\angle BAC\) \(=180^\circ\) (From equation (1))
\(\Rightarrow\) \(2\angle BAC\) \(=180^\circ-90^\circ\) \(=90^\circ\)
\(\Rightarrow\) \(\angle BAC\) \(=\frac{90^\circ}2\) = \(45^\circ\)