Given
In an acute angled triangle ABC .
sin 2(A+B-C) = 1
tan(B + C -A) =√3
To proof
As given in the question
In an acute angled triangle ABC .
first solving the equation
sin 2(A+B-C) = 1
As we know
1 = sin90°
put this in the above equation
we get
2A + 2B - 2C = 90
A+B -C =45 ( first equation )
now solving the equation
we get
tan(B + C -A) =√3
B + C -A = 60 ( second equation )
As given acute angled triangle ABC
thus
∠A + ∠ B +∠ C = 180° ( Angle sum property of a triangle )
than the third equation becomes
A + B + C = 180 ( third equation)
Now solve the equation
A+B -C =45
and B + C -A = 60
Now subtract B + C -A = 60 from A+B -C =45
we get
(A+B -C) - (B + C -A) = 45-60
2A -2C = -15
Put this value in the equation 2A + 2B - 2C = 90
-15 + 2B = 90
2B = 90 + 15
B = 52.5
now subtracted -A +B +C = 60 from A + B + C =180
A + B + C +A - B - C =180 - 60
2A = 120
A = 60
Put the value of A , B in the equation A + B + C =180
60 + 52.5 + C = 180
C = 180 - 112.5
C = 67.5
Thus ΔABC is an acute angle triangle
therefore
∠A = 60°
∠B = 52.5°
∠C = 67.5°
Hence proved