Question

在三角形ABC中,角A=60度,且AB/AC=4/3 ,求sinC？

Answer 1

余弦定理的应用：
AB/AC=4/3
令AB=4a,AC=3a
则cosA=(AB^2+AC^2-BC^2)/(2*AB*AC)=(25a^2-BC^2)/(24a^2)=cos60=1/2
解得BC=√13a
cosC=(AC^2+BC^2-AB^2)/(2*AC*BC)=(6a^2)/(6√13*a^2)=1/√13
sinC=√(1-cos^2C)=√(1-1/13)=√(12/13)=2√39/13,8,AB/AC=4/3 ,AB=4t,AC=3t ,由余弦定理BC^2=AB^2+AC^2-2AB*ACcos60度=13t^2
BC=根号13t 正弦定理:AB/sinC=BC/sinA 4t/sinC=根号13t/sin60度
sinC=2根号39/13,2,设AB=4，AC=3，余弦定理套进去，求出BC，再套正弦定理。,1,
A=60°
∴B+C=120°
根据正弦定理，得
AB/sinC=AC/sinB=AC/sin(120°-C)
∴AB/AC=sinC/sin(C+60°)
4/3=sinC/[(1/2)sinC+(√3/2)cosC]
3sinC=2sinC+2√3cosC
sinC=2√3cosC
tanC=2√3
∴sinC=2√3/√(12+1)=2√3/√13=2√39/13,1,