In the square ABCD, ∠BAC = ……………..A) 70° B) 45° C) 80�

1.	In the square ABCD, ∠BAC = ……………..A) 70° B) 45° C) 80° D) 60°
Answer» 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)) \(\Rightarrow\) \(2\angle BAC\) \(=180^\circ-90^\circ\) \(=90^\circ\) \(\Rightarrow\) \(\angle BAC\) \(=\frac{90^\circ}2\) = \(45^\circ\) Correct option is B) 45°

In the square ABCD, ∠BAC = ……………..A) 70° B) 45° C) 80° D) 60°

Answer»

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))

\(\Rightarrow\) \(2\angle BAC\) \(=180^\circ-90^\circ\) \(=90^\circ\)

\(\Rightarrow\) \(\angle BAC\) \(=\frac{90^\circ}2\) = \(45^\circ\)

Correct option is B) 45°

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