Question

第三题求解，高中数学

Answer 1

根据余弦定理，cosA=(b^2+c^2-a^2)/(2bc)=1/2
∠A=π/3

∠B+∠C=2π/3

cosBcosC=-1/8
cosBcos(2π/3-B)=-1/8
cosB(cos(2π/3)cosB+sin(2π/3)sinB)=-1/8
cosB(-1/2*cosB+sqrt(3)/2*sinB)=-1/8
-1/2*(cosB)^2+sqrt(3)/2*sinBcosB=-1/8
-1/4*cos(2B)-1/4+sqrt(3)/4*sin(2B)=-1/8
记2B=x
sqrt(3)*sinx-cosx=1/2
又因为(sinx)^2+(cosx)^2=1
解得x=0.7763
B=0.38815，C=1.706245
根据正弦定理a/sinA=b/sinB=c/sinC=k
sinBsinC=0.375,bc=1
sinBsinC*k^2=1
k=1.633
a=ksinA=1.414
b=ksinB=0.618
c=ksinC=1.618
a+b+c=3.65(精确值可能为sqrt(5)+sqrt(2))

Answer 2

√2 + √15

追问

怎么算