Question

在三角形ABC中，已知a四次方+b四次方+c四次方=2c²(a²+b²),则角C为

在三角形ABC中，已知a四次方+b四次方+c四次方=2c²(a²+b²),则角C为 A,30° B.60° C.45°或135° D,120°

Answer 1

c^4-2(a^2+b^2)c^2+(a^4+b^4)=0
(c^2-(a^2+b^2))^2=(a^2+b^2)^2-(a^4+b^4)=2a^2b^2
(-2abcosC)^2=2a^2b^2
(cosC)^2=½
所以C＝45°或135°
唉，平方打不出来

Answer 2

a^4+b^4+c^4=2c^2(a^2+b^2)
c^4-2(a^2+b^2)c^2+(a^4+b^4)=0
(c^2-(a^2+b^2))^2=(a^2+b^2)^2-(a^4+b^4)=2a^2b^2
(-2abcosC)^2=2a^2b^2
(cosC)^2=½
所以C＝45°或135°