在三角形ABC,已知内角A,B,C,的对边分别是a,b,c且满足2^(1/2)asin[B+(1/4*π)]=c

1.求角A的大小
2.当三角形ABC为锐角三角形时，求sinBsinC的取值范围 详细过程

1
2^(1/2)asin[B+(1/4*π)]=c
√2sinA*(sinBcosπ/4+cosBsinπ/4)=sinC
sinAsinB+sinAcosB=sin(A+B)
sinAsinB+sinAcosB=sinAcosB+cosAsinB
∴sinAsinB=cosAsinB
∵sinB>0
∴sinA=cosA
tanA=1
∴A=π/4
2
∵三角形ABC为锐角三角形
A=π/4
∴B+C=π-π/4=3π/4
∴B=3π/4-C<π/2
∴π/4 ∴sinBsinC
=sin(3π/4-C)sinC
=(sin3π/4cosC-cos3π/4sinC)sinC
=√2/2(sinCcosC+sin²C)
=√2/2(1/2sin2C-1/2*cos2C+1/2)
=1/2(√2/2sin2C-√2/2cos2C)+√2/2
=1/2sin(2C-π/4)+√2/2
∵π/4 ∴π/2<2C<π
π/4<2C-π/4<3π/4
√2/2 ∴1/2+√2/2<√2/2sin(2C-π/4)+√2/2≤√2
即sinBsinC的取值范围是(1/2+√2/2,√2]

1,s=(1/2)*a*b*sinc=√3/4ab=√3
ab=4
c^2=a^2+b^2-2ab*cosc
4=a^2+b^2-8*1/2
a^2+b^2=4+4=8
(a+b)^2=8+8=16
(a-b)^2=8-8=0
a=b=22,
s=acsinb*1/2=bcsina*1/2
asinb=bsina
a=2b
s=absinc*1/2=b*b/2因为c^2=a^2+b62-2ab*cosc
cosc=(a*a+b*b-c*c)/2ab=(5b^2-4)/4b^2=1/2
所以5/4-1/b^2=1/2
b^2=4/3
所以s=b^2/2=4/3/2=2/3 网上找的 希望对你有帮助