Question

在锐角△ABC中,a、b、c分别为角A、B、C所对的边,且根号3*a=2c*sinA

在锐角△ABC中,a、b、c分别为角A、B、C所对的边,且根号3*a=2c*sinA
(1)确定角C的大小
（2）若c=根号7，且△ABC的面积为（3根号3）/2，求a+b的值

Answer 1

对于任意三角形ABC，都有 a/sinA=b/sinB=c/sinC=2R ，你的题打错了。

Answer 2

用正弦定理 给出的等式两边a&c换成sinA,sinC .解得SINC等于三分之二

Answer 3

(1): ∵ 根号3*a=2c*sinA=2a*sinC
∴ sinC=（根号3）/2
∵△ABC为锐角三角形
∴C=60°
（2）： ∵S△ABC=（1/2）ab*sinC=(3根号3）/2
∴a*b=6
∵cosC=(a^2+b^2-c^2)/(2ab)=1/2
∴a^2+b^2=13
∴a+b=根号（(a+b)^2)=根号（a^2+b^2+2ab)=根号（13+2+6）=5

Answer 4

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