Math, asked by unknown785, 9 months ago

If Sin (A+B) = 1 and cos(A-B) = √3/2
0 < A+B < 90° A>B then find
A and B.
20

Answers

Answered by Utkarshkesharwani933

Answer:

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Answered by abdul9838

$<b> <body \: bgcolor = "yellow">$

$\bf \red{hey \: mate \: here \: is \: answer} \\ \\ \bf \blue{ \huge \: solution} \\ \\ \bf \pink{as \: given} \\ \\ \bf \pink{ \sin(A+B) = 1 } \\ \\ \bf \red{and} \\ \\ \bf \pink{ \cos(A - B) = \frac{ \sqrt{3} }{2} } \\ \\ \bf \green{ \huge \: now} \\ \\ \bf \pink{ \sin(A+B) = 1} \\ \\ \bf \pink{ = > \sin(A+B) = \sin(90) \degree } \\ \\ \bf \pink{A+B = 90 \degree \: ........... \: equation \: 1st} \\ \\ \bf \green{and} \\ \\ \bf \pink{ \cos(A - B) = \frac{ \sqrt{3} }{2} } \\ \\ \bf \pink{ \cos(A - B) = \cos(30) \degree } \\ \\ \bf \pink{A - B = 30 \degree \: \: ..........equation \: 2nd} \\ \\ \\ \bf \orange{adding \: both \: equations} \\ \\ \bf \pink{ A+ \cancel{B }= 90 \degree } \\ \bf \pink{A \cancel{ - B }= 30 \degree} \\ \\ \bf \pink{ = > 2A = 120 \degree} \\ \\ \bf \pink{ = > A = \frac{ \cancel{120}}{ \cancel{2} }} \\ \\ \bf \pink{ = >A = 60 \degree } \\ \\ \bf \red{now} \\ \\ \bf \green{putting \: the \: value \: of \: A \: in \: equation \: 1st} \\ \\ \bf \pink{A+B = 90 \degree} \\ \\ \bf \pink{B = 90 \degree - A} \\ \\ \bf \pink{ = >B = 90 \degree - 60 \degree } \\ \\ \bf \pink{B = 30 \degree \: \: answer}$

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If Sin (A+B) = 1 and cos(A-B) = √3/20 < A+B < 90° A>B then findA and B.20​

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If Sin (A+B) = 1 and cos(A-B) = √3/2
0 < A+B < 90° A>B then find
A and B.
20