Question
If sin (2A − 3B) = (1/2) and cos (A + B) = (√3/2),
where 0° < A, B < 90°, then find the value of ‘A’.Solution
sin (2A − 3B) = (1/2)
Or, sin (2A − 3B) = sin 30°
Or, 2A − 3B = 30° ----------- (I)
And, cos (A + B) = (√3/2)
Or, cos (A + B) = cos 30°
Or, A + B = 30° -------------- (II)
On solving equation I + 3 × equation II, we get;
2A − 3B + 3 × (A + B) = 30 + 3 × 30
Or, 2A − 3B + 3A + 3B = 120°
Or, 5A = 120°
So, ‘A’ = 24°
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