在三角形ABC中，内角ABC的对边分别是abc，若a方-b方=根3bc，sinC=2根3sinB，求A=？ 在三角形ABC中，内角ABC的对边分别是abc，若a的平方-...
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u4e09\u5185\u89d2ABC\u7684\u5bf9\u5e94\u8fb9\u5206\u522b\u662fabc,\u82e5a^2-b^2=\u6839\u53f73\uff08bc\uff09\uff0csinC=2\uff08\u6839\u53f73\uff09sinB,\u5219\u89d2A\u7684\u503c\u4e3a\uff1fb/sinB=c/sinC,b/c=sinB/sinC,
sinC=2\u221a3sinB,b/c=1/(2\u221a3),
c=2\u221a3b,
a^2-b^2=\u221a3bc=\u221a3b*(2\u221a3b)
a^2=7b^2
cosA=(b^2+c^2-a^2)/2bc
=(b^2+12b^2-7b^2)/[2b*(2\u221a3b)]
=\u221a3/2
A=\u03c0/6
60\u5ea6
A=π/6解析如下：sinC =2√3sinB
由正弦定理可知：c=2√3b
代入：a^2-b^2=√3bc
即：a^2=7b^2
由余弦定理：cosA=(b^2+c^2-a^2)/2bc =(b^2+12b^2-7b^2)/2b*2√3b = √3/2
所以A=π/6
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