Question

数学题：在△ABC中,角A，B，C的对边分别为abc,且asinC/1-cosA=√3c.多谢！

在△ABC中,角A，B，C的对边分别为abc,且asinC/1-cosA=√3c.(1)若a=2，求△ABC外接圆的半径.(2)若b+c=10,S△ABC=4√3,求a的值.

Answer 1

asinC/(1-cosA)=√3c
asinC = √3c(1-cosA)
asinC = √3c-√3ccosA
a=2RsinA，c=2RsinC
2RsinAsinC = 2√3RsinC-2√3RsinCcosA
sinA = √3-√3cosA

sinA+√3cosA = √3
1/2sinA+(√3/2)cosA = √3/2

sinAcos(π/3)+cosAsin(π/3) = √3/2
sin(A+π/3)=√3/2
0＜A＜π
∴A+π/3=2π/3
∴A=π/3
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（1）
∵a=2
又：a=2RsinA
∴R=a/(2sinA)=2/(2*√3/2) = 2√3/3
即：△ABC外接圆的半径 2√3/3
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（2）
S△ABC=4√3,
(1/2)bcsinA=4√3
bc=8√3/sinA= 8√3/(√3/2) = 16

b+c=10
a²=b²+c²-2bccosA = b²+c²-2bc*1/2 = b²+c²-bc = (b+c)²-3bc = 10²-3×16 = 52

a=√52 = 2√13