Question
If sin (3A − 4B) = (1/2) and cos (A + B) = (√2/2),
where 0° < A, B < 90°, then find the value of ‘A’.Solution
sin (3A − 4B) = (1/2)
Or, sin (3A − 4B) = sin 30°
Or, 3A − 4B = 30° ----------- (I)
And, cos (A + B) = (√2/2)
Or, cos (A + B) = cos 45°
Or, A + B = 45° -------------- (II)
On solving equation I + 4 × equation II, we get;
3A − 4B + 4 × (A + B) = 30 + 4 × 45
Or, 3A − 4B + 4A + 4B = 210°
Or, 7A = 210°
So, ‘A’ = 30°
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