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半角与边长的关系公式

\[ \sin \frac{A}{2} = \sqrt {\frac{{\left( {p - b} \right)\left( {p - c} \right)}}{{bc}}} \] \[ \sin \frac{B}{2} = \sqrt {\frac{{\left( {p - c} \right)\left( {p - a} \right)}}{{ca}}} \] \[ \sin \frac{C}{2} = \sqrt {\frac{{\left( {p - a} \right)\left( {p - b} \right)}}{{ab}}} \] \[ \cos \frac{A}{2} = \sqrt {\frac{{p\left( {p - a} \right)}}{{bc}}} ,\cos \frac{B}{2} = \sqrt {\frac{{p\left( {p - b} \right)}}{{ca}}} ,\cos \frac{C}{2} = \sqrt {\frac{{p\left( {p - c} \right)}}{{ab}}} \] \[ \tan \frac{A}{2} = \frac{r}{{p - a}} = \sqrt {\frac{{\left( {p - b} \right)\left( {p - c} \right)}}{{p\left( {p - a} \right)}}} \] \[ \tan \frac{B}{2} = \frac{r}{{p - b}} = \sqrt {\frac{{\left( {p - c} \right)\left( {p - a} \right)}}{{p\left( {p - b} \right)}}} \] \[ \tan \frac{C}{2} = \frac{r}{{p - c}} = \sqrt {\frac{{\left( {p - a} \right)\left( {p - b} \right)}}{{p\left( {p - c} \right)}}} \] 式中\[ p = \frac{1}{2}\left( {a + b + c} \right) \] r为△ABC的内切圆半径,且 \[ r = \sqrt {\frac{{\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)}}{p}} = \frac{S}{p} = p\tan \frac{A}{2}\tan \frac{B}{2}\tan \frac{C}{2} \] 式中S为△ABC的面积. 


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