a: \(AM=\sqrt{\dfrac{6^2+9^2}{2}-\dfrac{11^2}{4}}=\dfrac{\sqrt{113}}{2}\left(cm\right)\)
\(cos\widehat{A}=\dfrac{6^2+9^2-11^2}{2\cdot6\cdot9}=\dfrac{-1}{27}\)
=>góc A=92 độ
\(sinA=\sqrt{1-\left(-\dfrac{1}{27}\right)^2}=\dfrac{2\sqrt{182}}{27}\)
\(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot\dfrac{2\sqrt{182}}{27}\cdot6\cdot9=2\sqrt{182}\left(cm^2\right)\)
\(p=\dfrac{6+9+11}{2}=13\left(cm\right)\)
\(S=p\cdot r\)
=>r=2căn 182/13(cm)
\(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}=\dfrac{BC}{sinA}=2R\)
=>\(\dfrac{6}{sinC}=\dfrac{9}{sinB}=\dfrac{11}{\dfrac{2\sqrt{182}}{27}}=2R\)
=>\(4R\cdot\dfrac{\sqrt{182}}{27}=11\)
=>\(R=11:\dfrac{4\sqrt{182}}{27}=\dfrac{297}{4\sqrt{182}}\left(cm\right)\)
\(\dfrac{6}{sinC}=\dfrac{9}{sinB}=2\cdot\dfrac{297}{4\sqrt{182}}=\dfrac{297}{2\sqrt{182}}\)
=>góc C=33 độ; góc B=180-33-92=55 độ
\(S=2\sqrt{182}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot AH\cdot11=5,5\cdot AH\)
=>\(AH=\dfrac{2\sqrt{182}}{5.5}=\dfrac{4\sqrt{182}}{11}\left(cm\right)\)
b: \(S_{ABC}=\dfrac{1}{2}\cdot AB\cdot AC\cdot sinA=\dfrac{1}{2}\cdot6\cdot9\cdot sin60\)
\(=3\cdot9\cdot\dfrac{\sqrt{3}}{2}=\dfrac{27\sqrt{3}}{2}\left(cm^2\right)\)
\(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{6^2+9^2-BC^2}{2\cdot6\cdot9}=\dfrac{1}{2}\)
=>117-BC^2=54
=>BC^2=63
=>\(BC=3*\sqrt7(cm)\)
\(p=\dfrac{6+9+3\sqrt{7}}{2}=\dfrac{3\sqrt{7}+15}{2}\)
\(S=p\cdot r\)
=>\(r\cdot\dfrac{3\sqrt{7}+15}{2}=\dfrac{27\sqrt{3}}{2}\)
=>\(r=\dfrac{27\sqrt{3}}{3\sqrt{7}+15}\left(cm\right)\)
\(\dfrac{BC}{sinA}=2\cdot R=\dfrac{AC}{sinB}=\dfrac{AB}{sinC}\)
=>\(\dfrac{9}{sinB}=\dfrac{6}{sinC}=\dfrac{3\sqrt{7}}{\dfrac{\sqrt{3}}{2}}=2\cdot R\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{sinB}=\dfrac{6}{sinC}=2\sqrt{21}\\2\cdot R=2\sqrt{21}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}R=\sqrt{21}\left(cm\right)\\\widehat{B}=79^0;\widehat{C}=180^0-79^0-69^0=41^0\end{matrix}\right.\)
\(S=\dfrac{27\sqrt{3}}{2}=\dfrac{1}{2}\cdot AH\cdot BC=\dfrac{1}{2}\cdot AH\cdot3\sqrt{7}\)
=>\(AH=\dfrac{9\sqrt{21}}{7}\left(cm\right)\)
