Diện tích tam giác ABC là:
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinBAC=\dfrac{1}{2}\cdot3\cdot6\cdot sin60\)
\(=3\cdot3\cdot\dfrac{\sqrt{3}}{2}=\dfrac{9\sqrt{3}}{2}\)
Xét ΔABC có \(cosBAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(3^2+6^2-BC^2=2\cdot3\cdot6\cdot cos60=6\cdot6\cdot\dfrac{1}{2}=18\)
=>\(BC^2=9+36-18=9+18=27\)
=>\(BC=3\sqrt{3}\)
Xét ΔABC có \(\dfrac{BC}{sinA}=2R\)
=>\(2R=3\sqrt{3}:sin60=6\)
=>R=3
\(S_{ABC}=\dfrac{1}{2}\cdot h_A\cdot BC\)
=>\(h_A\cdot\dfrac{3\sqrt{3}}{2}=\dfrac{9\sqrt{3}}{2}\)
=>hA=3
Nửa chu vi tam giác ABC là:
\(p=\dfrac{AB+AC+BC}{2}=\dfrac{3+6+3\sqrt{3}}{2}=\dfrac{9+3\sqrt{3}}{2}\)
\(S=p\cdot r\)
=>\(r=\dfrac{9\sqrt{3}}{2}:\dfrac{9+3\sqrt{3}}{2}=\dfrac{9\sqrt{3}}{9+3\sqrt{3}}=\dfrac{-3+3\sqrt{3}}{2}\)
\(S_{ABC}=\dfrac{1}{2}AB.AC.sinA=\dfrac{1}{2}.3.6.\dfrac{\sqrt[]{3}}{2}=\dfrac{9\sqrt[]{3}}{2}\left(đvdt\right)\)
\(BC^2=AB^2+AC^2-2bcCosA=3^2+6^2-2.3.6.\dfrac{1}{2}=27\)
\(\Rightarrow BC=3\sqrt[]{3}\) (Định lý Cô sin)
\(S_{ABC}=\dfrac{1}{2}h_aBC\Rightarrow h_a=\dfrac{2S_{ABC}}{BC}=\dfrac{\dfrac{9\sqrt[]{3}}{2}}{3\sqrt[]{3}}=\dfrac{3}{2}\)
\(p=\dfrac{AB+AC+BC}{2}=\dfrac{3+6+3\sqrt[]{3}}{2}=\dfrac{9+3\sqrt[]{3}}{2}=\dfrac{3\sqrt[]{3}\left(\sqrt[]{3}+1\right)}{2}\)
\(S_{ABC}=pr\Rightarrow r=\dfrac{S_{ABC}}{p}=\dfrac{\dfrac{9\sqrt[]{3}}{2}}{\dfrac{3\sqrt[]{3}\left(\sqrt[]{3}+1\right)}{2}}=\dfrac{3}{\sqrt[]{3}+1}\)
\(S_{ABC}=\dfrac{AB.AC.BC}{4R}\Rightarrow R=\dfrac{AB.AC.BC}{4S_{ABC}}=\dfrac{3.6.3\sqrt[]{3}}{4.\dfrac{9\sqrt[]{3}}{2}}=3\)
