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    Question

    If sin (3A − 2B) = (√3/2) and cos (A + B) = (1/2),

    where 0° < A, B < 90°, then find the value of ‘A’.
    A 36° Correct Answer Incorrect Answer
    B 24° Correct Answer Incorrect Answer
    C 42° Correct Answer Incorrect Answer
    D 30° Correct Answer Incorrect Answer

    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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