If A, B, C are acute angles and \sin(B+C-A)=\cos(C+A-B)=\tan(A+B-C)=1 then what is (A+B+C) equal to?

Explanation

From the given equations: B+C-A = 90^\circ, C+A-B = 0^\circ, and A+B-C = 45^\circ. Adding all three equations yields (B+C-A) + (C+A-B) + (A+B-C) = 90^\circ + 0^\circ + 45^\circ = 135^\circ. Simplifying the left side gives A+B+C = 135^\circ.

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