在三角形ABC中,abc分别是角ABC的对边,C=2A .cosA=3╱4

答：

C=2A，cosA=3/4
cosC=cos2A=2cos²A-1=2*(9/16)-1=1/8
求得：sinA=√7/4，sinC=3√7/8
cosB=-cos(A+C)
=-cosAcosC+sinAsinC
=-(3/4)*(1/8)+(√7/4)*(3√7/8)
=-3/32+21/32
=18/32
=9/16
所以：
cosB=9/16，cosC=1/8

(1)
A+B+C =π
B=π-3A
(2)
cosB= cos(π-3A)
= -cos3A
= -[4(cosA)^3-3(cosA)]
= -[ 4(27/64) -3(3/4) ]
= - ( 27/16 -9/4 )
= 9/16
cosC= cos2A
= 2(cosA)^2 -1
= 2(9/16)-1
= 1/8
(2)

BA.BC = 27/2
accosB = 27/2
a/sinA =b/sinB =c/sinC
=> a= bsinA/sinB = b (√7/4)/( 5√7/16) = 4b/5
=> c = bsinC/sinB =b (3√7/8)/( 5√7/16) = 6b/5

b^2=a^2+c^2- 2accosB
= 16b^2/25 + 36b^2/25 -27/2
27b^2/25 = 27/2
b^2 = 25/2
b = 5√2/2
AC=b = 5√2/2

cosC = cos2A = 2cos²A -1 = 1/8

cosB = cos(180-A-C) = -cos(A+C) = -(cosAcosC - sinAsinC)
= -(3/4 * 1/8 - √5/4 *√63/8)
= -3/32 + √315/32
= -3/32 + 3√35/32

cosB=9/16,cosC=1/8