If sin(A + B) = 1/√2 and cos (A – B) = 1/√2 then ∠B =A) 60�

1.	If sin(A + B) = 1/√2 and cos (A – B) = 1/√2 then ∠B =A) 60° B) 45° C) 30° D) 0°
Answer» Correct option is: D) 0° We have sin (A + B) = \(\frac 1{\sqrt2} = sin \, 45^\circ\) = A + B = \(45^\circ\) .....(1) and cos (A-B) = \(\frac 1{\sqrt2} = cos\, 45^\circ\) = A - B = \(45^\circ\) .....(2) Subtract equation (1) from (2), we get (A + B) - (A-B) = \(45^\circ\) - \(45^\circ\) = 2B = \(0^\circ\) = B = \(0^\circ\) Hence, < B = \(0^\circ\) Correct option is: D) 0°

If sin(A + B) = 1/√2 and cos (A – B) = 1/√2 then ∠B =A) 60° B) 45° C) 30° D) 0°

Answer»

Correct option is: D) 0°

We have sin (A + B) = \(\frac 1{\sqrt2} = sin \, 45^\circ\)

= A + B = \(45^\circ\) .....(1) and

cos (A-B) = \(\frac 1{\sqrt2} = cos\, 45^\circ\)

= A - B = \(45^\circ\) .....(2)

Subtract equation (1) from (2), we get

(A + B) - (A-B) = \(45^\circ\) - \(45^\circ\)

= 2B = \(0^\circ\)

= B = \(0^\circ\)

Hence, < B = \(0^\circ\)

Correct option is: D) 0°