Question

在三角形ABC中，角A，B，C的对边分别为a，b，c，已知角A=派/3，sinB=3sinC (1)求tanC的值 (2)若a= 根...

在三角形ABC中，角A，B，C的对边分别为a，b，c，已知角A=派/3，sinB=3sinC (1)求tanC的值 (2)若a= 根号7，求三角形ABC的面积

Answer 1

sinB=sin(π-π/3-C)=sin(2π/3-C)=3sinC
sin2π/3cosC-sinCcos2π/3=3sinC
（根号3）/2cosC=5/2sinC
tanC=（根号3）/5

2）a=√号7
sinB=3sinC
b=3c

利用余弦定理得cosA=1/2=(b^2+c^2-a^2)/2bc=(9c^2+c^2-7)/6c^2
9c^ 2+c^2-7=3c^2
c=1
b=3c=3 所以S=1/2bcsinA=1/2*3*1*（根号3）/2=（3根号3）/4

Answer 2

三角形ABC中.
1.A+B+C=派 sinB=sin(A+C)=sinAcosC+cosAsinc
=√3/2cosC+1/2sinc
=3sinC
即√3/2cosC=5/2sinC 解得tanC=√3/5
2.正弦定理：
a/sina=b/sinb=c/sinc
由sinB=3sinC则b=3c
根据余弦定理：cosA=（b^2+c^2-a^2）/2bc
即1/2=（10c^2-7）/6c^2
解得c=1 b=3
s=1/2bcsina=3√3/4

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Answer 3

(1)sinB=sin(A+C)=sinAcosC+cosAsinC=√3/2*cosC+1/2*sinC=3sinC,
√3/2*cosC=7/2*sinC,
tanC=√3/2*2/7=√3/7,
则sinC=√3/(2√13)=√39/26，sinB=3*√39/26
(2)a/sinA=c/sinC,√7/(√3/2)=c/(√39/26),
c=√7/(√3/2)* (√39/26)=√7/√13
则三角形ABC的面积=1/2*a*c*sinB
=1/2*√7*√7/√13*3*√39/26
=21√3/52

Answer 4

sqrt = 根号
sinB = sin(A+C) = 3sinC
sinAcosC + cosA sinC = 3sinC
sqrt(3) cosC + sinC = 6sinC
5sinC = sqrt(3) cosC
tanC = sqrt(3)/5
sinC = sqrt(3/28)

a = sqrt(7)
b = sinB/sinA a = 3sqrt(3/28)/sqrt(3) * 2 * sqrt(7) = 3
面积=1/2ab sinC = sqrt(7)*3/2 * sqrt(3/28) = 3sqrt(3)/4