△ABC中,内角A,B,C的对边为a,b,c,C=2A,cosA=3/4,向量BA•向量BC=27/2,求c
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△ABC中,内角A,B,C的对边为a,b,c,C=2A,cosA=3/4,向量BA•向量BC=27/2,求cosB及b
(1)
cosA =3/4
sinA =√7/4
sin2A = 2sinAcosA = 2(3/4)(√7/4) = 3√7/8
cos2A = 2(cosA)^2-1 = 18/16-1=1/8
B= π-A-C
=π-3A
cosB=-cos3A
= -(cos2AcosA- sin2AsinA)
=-((1/8)(3/4)- (3√7/8)(√7/4))
=-(3/32-21/32)
=18/32
= 9/16
(2)
BA.BC=27/2
|BA||BC|cosB = 27/2
ac(cosB) = 27/2
ac = (27/2)(16/9) = 24
c= 24/a
by sine rule
a/sinA = c/sinC
a/(√7/4)= (24/a)/(3√7/8)
a^2= 16
a = 4
c = 24/4 = 6
b^2 = a^2+c^2-2ac(cosB)
= 16+36-27
= 25
b= 5
cosA =3/4
sinA =√7/4
sin2A = 2sinAcosA = 2(3/4)(√7/4) = 3√7/8
cos2A = 2(cosA)^2-1 = 18/16-1=1/8
B= π-A-C
=π-3A
cosB=-cos3A
= -(cos2AcosA- sin2AsinA)
=-((1/8)(3/4)- (3√7/8)(√7/4))
=-(3/32-21/32)
=18/32
= 9/16
(2)
BA.BC=27/2
|BA||BC|cosB = 27/2
ac(cosB) = 27/2
ac = (27/2)(16/9) = 24
c= 24/a
by sine rule
a/sinA = c/sinC
a/(√7/4)= (24/a)/(3√7/8)
a^2= 16
a = 4
c = 24/4 = 6
b^2 = a^2+c^2-2ac(cosB)
= 16+36-27
= 25
b= 5
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