Question

Qs: If sin(A + B)=1 and cos(A-B) =√3/2, find A and B.​

Answer 1

Answer:

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Answer 2

$Given \: Question \: Is \: \: \\ \\ \sin(a + b) = 1 \: \: \: and \: \: \: \cos(a - b) = \frac{ \sqrt{3} }{2} \\ \\ Answer \: \\ \\ \sin(a + b) = sin(90) \\ \\ a + b = 90 \: \: ....Equation \: \: \: i \\ \\ \cos(a - b) = \cos(30) \\ \\ a - b = 30 \: \: \: ....Equation \: \: ii \\ \\ Add \: \: both \: the \: Equations \: \: we \: have \\ \\ 2a = 120 \\ \\ a = \frac{120}{2} \\ \\ a = 60 \\ \\ put \: value \: of \: a \: in \: Equation \: \: i \: we \: have \\ \\ 60 + b = 90 \\ \\ b = 90 - 60 \\ \\ b = 30\\ \\ therefore \: \: a = 60 \: \: and \: \: b = 30 \\ \\ Note \: \\ \\ \sin(90) = 1\\ and \\ \cos(30) = \frac{ \sqrt{3} }{2}$