Sin (A – B) = 1/2; Cos (A + B) = 1/2 then ∠A isA) 60° B) 15°C

1.	Sin (A – B) = 1/2; Cos (A + B) = 1/2 then ∠A isA) 60° B) 15°C) 30°D) 45°
Answer» Correct option is: D) 45° We have sin (A-B) = \(\frac 12 = sin \, 30^\circ\) = A - B = \(30^\circ\) ...(1) and cos (A+B) \(\frac 12\) = cos \(60^\circ\) = A + B = \(60^\circ\).....(2) By adding equations (1) & (2), we get (A-B) + (A+B) = \(30^\circ\) + \(60^\circ\) = \(90^\circ\) = 2A = \(90^\circ\) = A = \(\frac {90^\circ}2 = 45^\circ\) Hence, \(\angle\)A = \(45^\circ\) Correct option is: D) 45°

Sin (A – B) = 1/2; Cos (A + B) = 1/2 then ∠A isA) 60° B) 15°C) 30°D) 45°

Answer»

Correct option is: D) 45°

We have sin (A-B) = \(\frac 12 = sin \, 30^\circ\)

= A - B = \(30^\circ\) ...(1)

and cos (A+B) \(\frac 12\) = cos \(60^\circ\)

= A + B = \(60^\circ\).....(2)

By adding equations (1) & (2), we get

(A-B) + (A+B) = \(30^\circ\) + \(60^\circ\) = \(90^\circ\)

= 2A = \(90^\circ\)

= A = \(\frac {90^\circ}2 = 45^\circ\)

Hence, \(\angle\)A = \(45^\circ\)

Correct option is: D) 45°