If any triangle `A B C`, that:`(asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-a^2)=(csin(A-B))/(a^2-b^2)`

If any triangle `A B C`, that:`(asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-

1.	If any triangle `A B C`, that:`(asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-a^2)=(csin(A-B))/(a^2-b^2)`
Answer» From sine law, we have, `a/sinA = b/sinB = c/sinC = k` `=> a = ksinA, b = ksin B, c = ksinC` Now, `(asin(B-C))/(b^2-c^2) = (ksinAsin(B-C))/(k^2(sin^2B-sin^2C))` `= sin(pi-(B+C)sin(B-C))/(k(sin^2B-sin^2C))` `= (sin(B+C)sin(B-C))/(k(sin^2B-sin^2C))` We know, `sin(B+C)sin(B-C) = sin^2B-sin^2C` So, it becomes, `=1/k(sin^2B-sin^2C)/(sin^2B-sin^2C) = 1/k` `:. (asin(B-C))/(b^2-c^2) = 1/k` Now, `(bsin(C-A))/(c^2-a^2) = (ksinBsin(C-A))/(k^2(sin^2C-sin^2A))` `= (sin(pi-(C+A))sin(C-A))/(k(sin^2C-sin^2A))` `= (sin(C+A)sin(C-A))/(k(sin^2C-sin^2A))` `=1/k(sin^2C-sin^2A)/(sin^2C-sin^2A) = 1/k` `:.(bsin(C-A))/(c^2-a^2) = 1/k` Now, `(csin(A-B))/(a^2-b^2) = (ksinCsin(a-b))/(k^2(sin^2A-sin^2B))` `= (sin(pi-(A+B))sin(A-B))/(k(sin^2A-sin^2B))` `= (sin(A+B)sin(A-B))/(k(sin^2A-sin^2B))` `=1/k(sin^2A-sin^2B)/(sin^2A-sin^2B) = 1/k` `:.(csin(A-B))/(a^2-b^2) = 1/k` `:. (asin(B-C))/(b^2-c^2) = (bsin(C-A))/(c^2-a^2) =(csin(A-B))/(a^2-b^2) = 1/k`