在三角形ABC中,tanA=2,tanB=3 设AB=根号2,求AC
发布网友
发布时间：2024-10-24 00:20
共1个回答
热心网友
时间：2024-11-09 07:28
c=AB=√2
tanA=2, tanB=3
cosA = 1/√5
tanA + tanB =5
sinA/cosA + sinB/cosB =5
sin(A+B)/(cosAcosB)=5
sin(A+B) =5cosAcosB
sinC =5cosAcosB
AC=b
b/sinB= c/sinC
b/sinB=√2/(5cosAcosB)
b= (√2/5)tanB/cosA
= (√2/5)(3)√5
= (3/5)√10
AC=b=(3/5)√10
