知道两条边长和夹角求对角线,怎么求?

2025-02-22 12:10:22
推荐回答(2个)
回答1:

用余弦定理: c^2=a^2+b^2-2*a*b*CosC
即AB^2
= BC^2+AC^2-2*BC*AC*cos(∠ACB)
=50^2+60^2-2*50*60*cos145
=2500+3600-6000*cos145
=6100+4914.91
=11014.91

所以AB= |根号11014.91| = 104.95

回答2:

余弦定理