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If Sin(A+B)=1 and Cos(A-B)=root 3/2, A>B find the acute angles A & B |
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Answer» We have Sin (A+B)=1 (A+B)=sin-1 1 We know sin-1 1 = 90 A+B=90 --------(1) Cos (A-B) =√3/2 A-B=cos-1 √3/2 We know cos-1 √3/2=30 A-B=30 -----------(2) Solve (1) & (2) A+B=90 A-B=30 add 2A=120 A=60 & B=30 Sin (A+B) = 1 and Cos (A-B) = √3/2 Sin (A+B) = 1 Sin (A+B) = Sin 90o A + B = 90o A = 90o - B ............. (i) Cos (A-B) = √3/2 Cos (A-B) = Cos 30o A - B = 30o A = 30o + B .............(ii) In equation (i) and (ii), 90o - B = 30o + B 90o - 30o = B + B 60o = 2B B = 60o/2 = 30o Putting the value of B is eq (i) A = 90o - B = 90o - 30o = 60o Hence, A = 60o and B = 30o |
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