Given: A,B,C are acute angles sin(B+C−A)=cos(C+A−B)=tan(A+B−C)=1 Concept used:
0°
30°
45°
60°
90°
sin
0
1
2
1
√2
√3
2
1
cos
1
√3
2
1
√2
1
2
0
tan
0
1
√3
1
√3
cosec
∞
2
√2
2
√3
1
sec
1
2
√3
√2
2
∞
cot
∞
√3
1
1
√3
0
sin(B+C−A)=1 So, sin(B+C−A)=sin90∘ ⇒(B+C−A)=90∘⋯(i)cos(C+A−B)=1 So, cos(C+A−B)=cos0∘ ⇒(C+A−B)=0∘⋯ (ii) tan(A+B−C)=1 So, tan(A+B−C)=tan45∘ ⇒(A+B−C)=45∘ By adding (i), (ii), and (iii) we get B+C−A+C+A−B+A+B−C=90∘+0∘+45∘ ⇒A+B+C=135∘ ∴A+B+C is equal to 135∘