[tex]Voi~nota~AB=c;~BC=a~;~AC=b~si~m_a=~lungimea~medianea \\ \\ dusa~din~A. \\ \\ Din~teorema~cosinusului~avem: \\ \\ cosA= \frac{b^2+c^2-a^2}{2bc} \Leftrightarrow cos 60= \frac{4^2+2^2-a^2}{2 \cdot 4 \cdot 2} \Leftrightarrow \frac{1}{2} = \frac{20-a^2}{2 \cdot 8} \Leftrightarrow8=20-a^2 \Rightarrow \\ \\ \Rightarrow a^2=12 \Rightarrow a= 2 \sqrt{3}~(cm). \\ \\ Din~teorema~medianei~avem: [/tex]
[tex]m_a^2= \frac{2(b^2+c^2)-a^2}{4}= \frac{2(4^2+2^2)-(2 \sqrt{3})^2}{4}= \frac{40-12}{4} = 7 \Rightarrow m_a= \sqrt{7}~(cm). \\ \\ Solutie:~\sqrt{7}~cm.[/tex]
