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在三角形abc中,已知向量m=(2sinb,2-cos2b),n(2sin^2(π/4+b/2),-1),m垂直于n,(

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在三角形abc中,已知向量m=(2sinb,2-cos2b),n(2sin^2(π/4+b/2),-1),m垂直于n,(1)求角b的大小,(2
一、在三角形abc中,已知向量m=(2sinb,2-cos2b),n(2sin^2(π/4+b/2),-1),m垂直于n,
(1)求角b的大小,(2)若a=根号3,b=1,求角c的值
二、在三角形abc中,ABC所对的边分别为abc,向量m=(cosA,1)n=(1,1-根号3sinA)且m垂直于n,求若b+c=根号3a,求角b,角c的大小
在三角形abc中,已知向量m=(2sinb,2-cos2b),n(2sin^2(π/4+b/2),-1),m垂直于n,(
(1)求角b的大小:m垂直于n,
m.n=0=2sinb*2sin^2(π/4+b/2)+(2-cos2b)*(-1)
=2sinb[1-cos(π/2+b)]+cos2b-2
=2sinb(1+sinb)+(1-2sin^2b)-2
=2sinb+2sin^b-2sin^2b-1
=2sinb-1=0
sinb=1/2,角b=π/6
(2)求角c的值:a=√3,b=1
cosb=√3/2,b^2=a^2+c^2-2ac*cosb=3+c^2-2√3c*√3/2=c^2-3c+3=1
c^2-3c+2=0,c=1 or c=2
c=1:b=1,角c=角b=π/6
c=2:sinc/c=sinb/b,sinc=csinb/b=2*1/2=1,角c=π/2
向量m=(cosA,1)n=(1,1-根号3sinA)且m垂直于n
m.n=0=cosA*1+1*(1-√3sinA)=cosA-√3sinA+1=2(cosA*1/2-sinA*√3/2)+1=1-2sin(A-π/6)
sin(A-π/6)=1/2,A-π/6=π/6,or A-π/6=5π/6 A=π/3 or A=π(舍去)
b/sinB=a/sinA,b=asinB/sinA,c=asinC/sinA,b+c=a(sinB+sinC)/sinA=√3*a
sinB+sinC=√3sinA=3/2,sinB=3/2-sinC
根据余弦定理可得:a^2=b2+c^2-2bc*cosA=a^2(sin^2B+sin^2C-2sinBsinC*1/2)/sin^2A
sin^2B+sin^2C-sinBsinC=sin^2A=3/4=9/4-3sinC+sin^2C+sin^2C-3sinC/2+sin^2C
3sin^2C-9sinC/2+6/4=0,sin^2C-3sinC/2+1/2=0,(sinC-3/4)^2=9/16-1/2=1/16=(1/4)^2
sinC=3/4+1/4=1 or sinC=3/4-1/4=1/2 ,C=π/2 or C=π/6
sinB= 1/2 or sinB=1,B=π/6 or B=π/2