Question

在三角形中a/cosA=b/2cosB=c/3cosC 1.求A 2.若三角形面积为3 求a 第二问

Answer 1

∵a/cosA=b/(2cosB)=c/(3cosC)
又∵根据正弦定理：a/sinA=b/sinB=c/sinC=2R
∴ sinA/cosA=sinB/(2cosB)=sinC/(3cosC)
∴ tanA=tanB/2=tanC/3＞0，三角形为锐角三角形
∴tanB=2tanA，tanC=3tanA
∴tanA=tan[π-(B+C)]=-tan(B+C)=(tanB+tanC)/(tanBtanC-1)
∴ tanA=(2tanA+3tanA)/(2tanA*3tanA-1) = 5tanA/(6tan²A-1)
∴ 1=5/(6tan²A-1)
∴tan²A=1
∴tanA=1
∴A=45°

∴tanB=2，tanC=3
∴sinB=tanB/√(1+tan²B)=2√5/5
sinC=tanC/√(1+tan²C)=3√10/10
根据正弦定理：a/sinA=b/sinB=c/sinC
b=asinB/sinA，c=asinC/sinA
S=1/2bcsinA=3
1/2 * asinB/sinA * asinC/sinA * sinA = 3
a²sinBasinC = 6sinA
a² * 2√5/5 * 3√10/10 = 6*√2/2
a² * 6/√50 = 6*√2/2
a² = √100 = 10
a = √10