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