cosC=(a^2+b^2-c^2)/2ab
=[(√3+1)^2+2^2-(√2)^2]/[2*(√3+1)*2]
=[4+2√3+4-2]/[4(√3+1)]
=[6+2√3]/[4(√3+1)]
=2√3(√3+1)/[4(√3+1)]
=√3/2
C=30°
用余弦定理得
c^2=a^2+b^2-2abcosC
(√2)²=(√3+1)²+2²-2(√3+1)*2*cosC
2=3+2√3+1+4-4(√3+1)cosC
4(√3+1)cosC=6+2√3=2√3(√3+1)
cosC=(√3)/2
C=60°
不明白,可以追问如有帮助,记得采纳,
谢谢 祝学习进步!
cosc=a方 b方-c方/2ab求出cosc为更好3/2为30度,,,,,