⒈假设a≤b≤c,根据余弦定理:cosA=(b²+c²-a²)/2bc,sin²A=1-cos²A=1-(b²+c²-a²)²/4b²c²=[4b²c²-2b²c²+2a²b²+2a²c²-(a²)²-(b²)²-(c²)²]/4b²c²=[2b²c²+2a²b²+2a²c²-(a²)²-(b²)²-(c²)²]/4b²c²
正弦定理有:2R=a/sinA,∴R=a/2sinA=abc/√[2b²c²+2a²b²+2a²c²-(a²)²-(b²)²-(c²)²]=abc/√[(a+b+c)(a+b-c)(c+a-b)(c-a+b)]