Question

在三角形ABC中，∠A=60°，b=1,S△=根号3/2,（1）求边长a,c的值，（2）求a+b+c除以sinA+sinB+sinC的值

Answer 1

S△=1/2*bc*sinA=1/2*1*c*根号3/2=根号3/2
c=2
a^2=b^2+c^2-2bccosA=1+4-4*1/2=3
a=根号3
因为a/sinA=b/sinB=c/sinC=k
a=ksinA b=ksinB c=ksinC
所以(a+b+c)/(sinA+sinB+sinC)
=(ksinA+ksinB+ksinC)/(sinA+sinB+sinC)
=k
=a/sinA
=根号3/(根号3/2)
=2

Answer 2

S△=根号3/2=sinA*b*c/2
计算得c=2
a^2=c^2+b^2-2bccosA
=4+1-4*1/2
=3
a=根号3
因为a^2+b^2=c^2
所以三角形ABC是直角三角形
∠B=30°，∠C=90°
(a+b+c)/(sinA+sinB+sinC)
=(根号3+1+2)/(根号3/2+1/2+1)
=2

Answer 3

1.因为△高=2S/b=根号3，所以，tan∠A=根号3=a/b,所以△为直角三角形，a=根号3，c=2
2.a+b+c=3+根号3,sinA+sinB+sinC=根号3/2+1/2+1=(根号3+3)/2,所以前者除以后者=2