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Question

If a, b and c are the sides of a triangle such that a4 + b4 + c4 = 2c2 (a2 + b2) then the angles opposite
to the side C is:
(A) 45° or 135° (B) 30° or 100° (C) 50° or 100° (D) 60° or 120°

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Solution

a4+b4+c4=2c2a2+b2a4+b4+c4=2c2a2+2c2b2a4+b4+c4-2c2a2-2c2b2=0a4+b4+c4-2c2a2-2c2b2+2a2b2=2a2b2a22+b22+-c22+2-c2a2+2-c2b2+2a2b2=2a2b2a2+b2-c22=2a2b2a2+b2-c2=±2a2b2a2+b2-c22=±2abBy cosine ruleCos C=a2+b2-c22ab=±2ab2ab=±12Cos C=12 or cos C=-12Cos C=Cos 45° or cos C= -cos 45°Cos C=Cos 45° or cos C= cos 180°-45°C=45° or C= 180°-45°=135°

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