已知△ABC的三个内角A,B,C所对的边分别为a,b,c,√3b=2asinB,且向量AB•向量AC>0. (1)求

已知△ABC的三个内角A,B,C所对的边分别为a,b,c,√3b=2asinB,且向量AB•向量AC>0.(1)求∠A的度数(2)若cos(A-C)+cosB... 已知△ABC的三个内角A,B,C所对的边分别为a,b,c,√3b=2asinB,且向量AB•向量AC>0.
(1)求∠A的度数
(2)若cos(A-C)+cosB=√3/2,a=6,求△ABC的面积.
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解:
(1)
∵a/sinA=b/sinB=2R
∴a=2RsinA,b=2RsinB
∵(√3)b=2asinB
∴(√3)2RsinB=2×2RsinA×sinB
∴(√3)sinB=2sinAsinB
∵在△ABC中,0<∠B<180°
∴sinB≠0
∴(√3)=2sinA
∴sinA=(√3)/2.
∵向量AB•向量AC>0
∴|向量AB|×|向量AC|×cosA>0
∵|向量AB|>0,|向量AC|>0
∴cosA>0
∴∠A=60°.
(2)
∵在△ABC中,A+B+C=π
∴B=π-(A+C)
∵cos(A-C)+cosB=(√3)/2
∴cos(A-C)+cos[π-(A+C)]=(√3)/2
∴cos(A-C)-cos(A+C)=(√3)/2
∴(cosAcosC+sinAsinC)-(cosAcosC-sinAsinC)=(√3)/2
∴2sinAsinC=(√3)/2
∵∠A=60°
∴sinA=(√3)/2
∴2[(√3)/2]sinC=(√3)/2
∴2sinC=1
∴sinC=1/2.
∵∠A=60°
∴0°<∠C<120°
∴∠C=30°
∴∠B=180°-∠A-∠C=90°
∴△ABC是以∠B为直角的直角三角形,sinB=1
∵(√3)b=2asinB,a=6
∴(√3)b=12
∴b=4√3.
∵△ABC是以∠B为直角的直角三角形
∴a^2+c^2=b^2
∴36+c^2=48
∴c^2=12
∴c=2√3.
∴S△ABC=(1/2)ac=6√3.
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(1)
√3b=2asinB
sinB = √3b/(2a)
By sine-rule
a/sinA = b/sinB
a/sinA = b/(√3b/(2a))
[√3b/(2a)].a = bsinA
sinA =√3/2
A = π/3 ( AB.AC >0 )
(2)
cos(A-C)+cosB= √3/2
cos(A-(π-A-B)+ cosB = √3/2
cos(B-π/3)+ cosB = √3/2
(1/2)cosB+ (√3/2)sinB + cosB = √3/2
(√3/2)sinB-(1/2)cosB = √3/2
sin(B-π/6)= sin(π/3)
B = π/2
C = π/6

△ABC的面积
(1/2)a. (atanA)
= (1/2) a. (atanA)
=(1/2).6 (6. tan(π/3))
= 18√3
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