Question

求这道题思路以及过程

Answer 1

图一

追答

Answer 2

1)sinA(sinA+sinB)=6(sinB)^2
由正弦定理得:
a^2+ab=6b^2
等式两边除以b^2得
(a/b)^2+(a/b)=6
(a/b)^2+(a/b)-6=0
[(a/b)-2][(a/b)+3]=0
a/b=2 或a/b=-3(舍去)
所以:a/b=2
2)cosC=3/4
(sinC)^2=1-(cosC)^2=1-(9/16)=7/16
(sinC)^2=7/16
因,sinB=sin(A+C)=sinAcosC+cosAsinC,
因a/b=sinA/sinB=2
所以,sinA=2sinB
sinB=2sinB*(3/4)+cosAsinC,
-sinB=2cosAsinC,
(sinB)^2=4(cosA)^2*(sinC)^2,
(sinB)^2=4[1-4(sinB)^2]*(sinC)^2,
(sinB)^2=4[1-4(sinB)^2]*[7/16]
4(sinB)^2=7[1-4(sinB)^2]
(sinB)^2=7/32
sinB=√14/8 或sinB= -√14/8(舍去)
sinB=√14/8