In A B C , with usual notations, if a^4+ b ^4+ c ^4-2 a^2 c ^2-2 c ^2 ~b ^2=0 , then C = .

Step-by-step solution

Given a^4 + b^4 + c^4 - 2a^2c^2 - 2c^2b^2 = 0 , recognize that (a^2 + b^2 - c^2)^2 = a^4 + b^4 + c^4 - 2a^2c^2 - 2b^2c^2 + 2a^2b^2 . The equation is equivalent to (a^2 + b^2 - c^2)^2 - 2a^2b^2 = 0 . Isolate the squared term: (a^2 + b^2 - c^2)^2 = 2a^2b^2 . Taking square roots gives a^2 + b^2 - c^2 = 2 ab . From the cosine rule, c^2 = a^2 + b^2 - 2ab C , so a^2 + b^2 - c^2 = 2ab C . Equating both expressions for a^2 + b^2 - c^2 yields 2ab C = 2 ab . Since a > 0 and b > 0 , divide both sides by 2ab : C = 2 2 = 1 2 .