In an acute angled triangle ABC,if sin (A+B-C)= 1/2 and cos (B+C-A)= 1

1.	In an acute angled triangle ABC,if sin (A+B-C)= 1/2 and cos (B+C-A)= 1/root 2
Answer» Sin (A+B-C) = 1/2 => Sin (A+B-C) = Sin 300Cos (B+C-A) = 1/√2 => Cos (B+C-A) = Cos 450=> A + B - C = 300 ........ (Eq. 1)AND, B + C - A = 450 ........ (Eq. 2)ALSO, A + B + C = 1800 ........ (Eq. 3)Adding Eq. 1 & 2, we get: 2 B = 750 => B = 37.50Subtracting eq. 2\xa0from eq. 3, we get: 2 A = 1350\xa0=> A = 67.50Using value of A & B, in Eq. 3, we get C = 750\xa0\xa0\xa0\xa0

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