Question

已知三角形,cosB=根号3/6,根号3sinA-2sinC=0,(1)求tanA(2)若b=根

5,求三角形的面积

Answer 1

∵A、B、C为三角形内角
∴其正弦值均＞0
∵cosB=√3/6
∴sinB=√{1-cos²B} = √33/6
∴sinC=sin(π-(A+B))=sin(A+B) = sinAcosB+cosAsinB = √3/6sinA+√33/6cosA
∵√3sinA-2sinC=0
∴√3sinA-2√3/6sinA-2√33/6cosA=0
∴3sinA-√11cosA=0
∴3sinA=√11cosA
∴tanA = √11/3

∵tanA＞0
∴A为锐角
∴sinA=tanA/√(1+tan²A) = (√11/3)/√(1+11/9) = √55/10

cosA = √(1-sin²A) = 3√5/10
b=√5

a/sinA=b/sinB
a = bsinA/sinB = √5*(√55/10)/(√33/6) = √3
sinC=√3/6sinA+√33/6cosA = √3/6*√55/10 + √33/6 * 3√5/10 = √165/15
面积 = 1/2absinC = 1/2*√3*√5*√165/15 = √11/2