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.一、任意三角形外接圆半径设三角形各边边长分别为a,b,c外接圆半径为R,(如右图所示)222R则 cos()abccoscossinsincb2ab(余弦定理)bb2bR2而 cos2, sin4aRR2RaR2a 22a4cos, sinRR2Ra2b2c22 b 22a2即有:b aR4R42ab2R 2RRR即有: a2b 2c2ab(4R 2b 2 )(4R2a2 )ab2R2所以: ab2R 2 ( a 2b2c 2 )(4R 2b 2 )(4R2a2 )ab即有: ( ab)24R 2 ( a 2b2c 2 )4R 4 ( a 2b2c2) 216R 44(a 2b 2 )R 2a 2b 2ab所以: c 2R24(a 2b2c2 )2 ,即: a 2b2 c2R2 4a 2 b 2( a2b2c 2 ) 2 ab所以: Rabc(a b c)( a b c)(a c b)(b c a)而三角形面积:4S( abc)(abc)( acb)(b ca) (海伦公式)所以,有:Rabc4S 另一求法,可用正弦定理,即:所以:ab 2c 2a 2sin A2R,而 cos A2bcaaaabcR2b 2222 222222 sin A(cos A)2 12 1 (ca ) 24b c(bca )2bc.二、任意三角形内切圆的半径设三角形各边边长分别为a,b,cyy内切圆半径为r,(如右图所示)cRb因为内切圆的圆心为各角的角平分线的交点,所以,会有zrxxzaabcxyb ,解得 x2zaxyzc显然: rx tan,而 tansin 21(cos2) 21cos21cos2而由余弦定理有:cos2a 2b2c 22aba 2b 2c 224(ab) 2(a2b2c 2 )2所以: tan1(2ab)1a 222(abc)(a bc)bc2ab即有 : rab c4(ab) 2(a 2b2c 2 ) 24(ab) 2(a2b2c2 ) 22(abc)(abc)2(abc)即: r(abc)(ab c)(acb)(bca)4S2S2(abc)2(abc)a b c
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