Bài tập 2.31 trang 101 SBT Hình học 10

Ta có \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{8 + 6 + 2 - 2\sqrt {12}  - 12}}{{4\sqrt 2 \left( {\sqrt 6  - \sqrt 2 } \right)}} =  - \frac{1}{2} \Rightarrow \widehat A = {120^0}\)

\(\cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ca}} = \frac{{6 + 2 - 2\sqrt {12}  + 12 - 8}}{{2\left( {\sqrt 6  - \sqrt 2 } \right).2\sqrt 3 }} = \frac{{\sqrt 2 }}{2} \Rightarrow \widehat B = {45^0}\)

\(\begin{array}{l}
{h_a} = \frac{{2S}}{a} = \frac{{ac.\sin B}}{a} = c.\sin B = \left( {\sqrt 6  - \sqrt 2 } \right).\frac{{\sqrt 2 }}{2} = \sqrt 3  - 1\\
\frac{b}{{\sin B}} = 2R \Rightarrow R = \frac{b}{{2\sin B}} = \frac{{2\sqrt 2 }}{{2.\frac{{\sqrt 2 }}{2}}} = 2\\
S = pr \Rightarrow r = \frac{S}{p} = \frac{{\frac{1}{2}ac.\sin B}}{{\frac{1}{2}\left( {a + b + c} \right)}} = \frac{{2\sqrt 3 \left( {\sqrt 6  - \sqrt 2 } \right).\frac{{\sqrt 2 }}{2}}}{{2\sqrt 3  + 2\sqrt 2  + \sqrt 6  - \sqrt 2 }} = \frac{{\sqrt 3 \left( {\sqrt 6  - \sqrt 2 } \right)}}{{\sqrt 6  + \sqrt 3  + 1}}
\end{array}\)

-- Mod Toán 10 HỌC247