∵a:b:c=2:√6:(√3+1)
∴可设a=2k,b=√6k,c=(√3+1)k,其中k≠0,
根据余弦定理:
cosA=(b²+c²-a²)/2bc
=[2k²+(4+2√3)k²] / [2√6(√3+1)k²]
=(6+2√3)/[2√6(√3+1)]
=(3+√3)/[√6(√3+1)]
=[(3+√3)(√3-1)] / 2√6
=√2/2
∵0
∵a:b:c=2:√6:(√3+1)
∴可设a=2k,b=√6k,c=(√3+1)k,其中k≠0,
根据余弦定理:
cosA=(b²+c²-a²)/2bc
=[2k²+(4+2√3)k²] / [2√6(√3+1)k²]
=(6+2√3)/[2√6(√3+1)]
=(3+√3)/[√6(√3+1)]
=[(3+√3)(√3-1)] / 2√6
=√2/2
∵0