Question

知道两条边长和夹角求对角线,怎么求?

Answer 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