Question

三角形ABC中,2A=B,a:c=3:5,求角C的余弦值

Answer 1

A+B+C=180,B=2A
C=180-3A
a:c=3:5
sinA/sinC=3/5
sinA/sin(180-3A)=3/5
sinA/sin3A=3/5 ,因sin3A=3sinA-4(sinA)^3
sinA/{sinA(3-4(sinA)^2)]}=3/5
1/(3-4(sinA)^2)]=3/5
(sinA)^2=1/3
(sinC)^2=5/3*(sinA)^2=5/3*(1/3)=5/9
(cosC)^2=1-(sinC)^2=1-(5/9)=4/9
cosC=2/3 或 cosC=-2/3

Answer 2

解答：1、由余弦定理得：7²=3²＋5²－2×3×5cosC∴cosC=－½∴∠C=120°2、由正弦定理得：b／sinB=c／sinC∴5／sinB=7／sin120°解得：sinB=5√3／14∴由sin²B＋cos²B=1解得：cosB=11／14由sin﹙B＋π／3﹚=sinBcos﹙π／3﹚＋sin﹙π／3﹚cosB=﹙5√3／14﹚×½＋﹙√3／2﹚×11／14=4√3／7