Question
If sin (3A − 2B) = (√3/2) and cos (A + B) = (1/2),
where 0° < A, B < 90°, then find the value of ‘A’.Solution
sin (3A − 2B) = (√3/2)
Or, sin (3A − 2B) = sin 60°
Or, 3A − 2B = 60° ----------- (I)
And, cos (A + B) = (1/2)
Or, cos (A + B) = cos 60°
Or, A + B = 60° -------------- (II)
On solving equation I + 2 × equation II, we get;
3A − 2B + 2 × (A + B) = 60 + 2 × 60
Or, 3A − 2B + 2A + 2B = 180°
Or, 5A = 180°
So, ‘A’ = 36°
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