A=[tex] \frac{1}{2}*AB*BC*sin120= \frac{1}{2}*10*16* \frac{ \sqrt{3} }{2} = 40 \sqrt{3}cm^2 \\ [/tex]
A=[tex] \frac{1}{2}*AC*h [/tex]
[tex]40 \sqrt{3} = \frac{AC*h}{2} \\ h= \frac{80 \sqrt{3} }{AC} [/tex]
AC il aflam folosind teorema cosinusului
[tex] AC^{2}=100+256-2*10*16*(- \frac{1}{2}) \\ AC^{2}=356+160=516 \\ AC=2 \sqrt{129} [/tex]
[tex]h= \frac{80 \sqrt{3} }{2 \sqrt{129} } = \frac{40 \sqrt{3} }{ \sqrt{129} } = \frac{40 \sqrt{3} }{ \sqrt{3}* \sqrt{43} } = \frac{40}{ \sqrt{43} } [/tex]
