SOLUTION: in a acute angle triangle ABC if SIN (B+C-A) = (SQUARE ROOT 3) /2 and TAN (C+A-B)=1 then C is equal to

From the first condition

B + C - A = 60 degrees.       (1)


From the second condition

C + A - B = 45 degrees.       (2)


Add equations (1) and (2). You will get

2C = 60 + 45 = 105 degrees  ====>  C =  = 52.5 degrees.


Substitute this value of C into equation (1). You will get

B - A = 60 - 52.5 = 7.5       (3)    (from (1))   and


An additional equation is

A + B + C = 180  degrees      (4).


Substitute  C = 52.5 into this equation. You will get

A + B + 52.5 = 180  degrees,   or

A + B = 180 - 52.5 = 127.5    (5)


Now add equations (3) and (5). You will get

2B = 7.5 + 127.5 = 135 degrees  ====>  B =  = 67.5 degrees.


Now you know the angle  B = 67.5 degrees  and the angle C = 52.5 degrees.


From this point find the angle A on your own.