三角形ABC,角A,B,C所对的边分别为a,b,c,acosB+bcosA=2ccosC,1)求内角C;(2)若a=5,b=8.求边c的长
问题描述:
三角形ABC,角A,B,C所对的边分别为a,b,c,acosB+bcosA=2ccosC,1)求内角C;(2)若a=5,b=8.求边c的长
答
acosB = c1bcosA = c2acosB + bcosA = c1 + c2 = c = 2ccosCcosC = 1/2C = 60°c² = a² + b² - 2abcosC = 5² + 8² - 5*8 = 25 + 64 - 40 = 49c = 7