Question

三角形ABC中，A=60度，b=1,其面积根号3，则a+b+c/sinA+sinB+sinC等于

Answer 1

由面积公式,
得:S=0.5*b*c*sinA
代入已知:根号3=0.5*1*c*sin60度.
根号3=0.5*1*c*(根号3)/2
得:c=4.
由余弦定理:
a^2=b^2+c^2-2*b*c*cosA
得:a^2=1+16-2*1*4*(1/2)=13.
即:a=根号13.
由正弦定理:
a/sinA=b/sinB=c/sinC,
由比的性质,
有a/sinA=b/sinB=c/sinC=
=(a+b+c)/(sinA+sinB+sinC).
而已知:a/sinA=根号13/[(根号3)/2]
=2*(根号39)/3

即:(a+b+c)/(sinA+sinB+sinC)=2*(根号39)/3.

Answer 2

面积=1/2 b * c sinA = 1/2 c sqrt(3)/2 = sqrt(3)
所以c = 4
由余弦定理
a^2 = b^2 + c^2 - 2bc cosA = 1+16-4 = 13
a = sqrt(13)
b = 1
c = 4
a/sinA = c/sinC
sinC = csinA/a = 4*sqrt(3)/(2sqrt(13))
sinB = bsinA/a = sqrt(3)/(2sqrt(13))
sinA+sinB+sinC = 1/2 + 2sqrt(3)/sqrt(13)+sqrt(3)/(2sqrt(13)) = (sqrt(13)+4sqrt(3)+sqrt(3))/(2sqrt(13))

a+b+c = 5+sqrt(13)
a+b+c/sinA+sinB+sinC = (5+sqrt(13))*2sqrt(13)/(sqrt(13)+5sqrt(3)) = (10sqrt(13) + 26)/(sqrt(13)+5sqrt(3))

Answer 3

S=0.5bc×sinA 正弦求面积
∴根号3=0.5×1×c×sin60
2×根号3=c×(根号3)/2
c=4
a²=b²+c²-2bc×cos60=16+1-4=13 余弦定理
a=根号13
∵a/sinA=b/sinB=c/sinC 正弦定理
a/sinA=(a+b+c)/(sinA+sinB+sinC) 等比性质
=(根号13)/[(根号3)/2]=2/3倍的根号39

Answer 4

由S=1/2*bc*sin60 解得c=2 再根据余玄定理解得a=根号3
sinB=1/2 sinC=1
最终结果为自己算

Answer 5

解：
过点C作垂线CD垂直于AB于D
∵∠A=60°，b=1
∴AD=1/2，CD=√3/2
又∵S△ABC=1/2AB×CD=√3
∴c=AB=4
∴BD=AB-AD=7/2
又∵a=√BD²+CD²=√13
∴sinA=√3/2=a/2R
∴2R=2a/√3=2√13/√3
∴a+b+c/sinA+sinB+sinC= (a+b+c)/[(a+b+c)/2R]=2R=2√13/√3