Question

数学第16题怎么做

Answer 1

(sinA)^2+(sinB)^2-sinAsinB=(sinC)^2
由正型简尺弦定理得
a^2+b^2-ab=c^2
a^2+b^2-c^2=ab
cosC=(a^2+b^2-c^2)/2ab=ab/2ab=1/2
cosC=1/2
C=π/3
A+B=2π/3
B=2π/3-A
2R=a/sinA=b/sinB=c/sinC=2/sinπ/3=2/(√3/2)=4√3/3
a=2R*sinA ,b=2R*sinB ,
a+b=2R*sinA +2R*sinB
=2R(sinA +sinB )
=2R[sinA+sin(2π/3-A)]
=4√3/3[sinA+√3/2cosA+1/2sinA]
=4√3/3(3/2sinA+√3/2cosA)
=4(√3/2sinA+1/2cosA)
=4sin(A+π/6)
0<A<2π/3
π/6<A+π/6<5π/6
sin(A+π/6)在(π/咐液6,5π/6)上值域为:(1/2,1]
4sin(A+π/6)在(π/6,5π/6)上值域卜高为:(2,4]
所以,a+b的范围:(2,4]

Answer 2

答案是2.4

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过程发一下。 麻烦你了～