Respuesta :
check the picture below. Notice that we can always solve for the cosine and simply use the sides alone to get an angle.
[tex]\bf cos^{-1}\left(\cfrac{10^2+(\sqrt{604})^2-18^2}{2(10)(\sqrt{604})} \right)=\measuredangle C \\\\\\ cos^{-1}\left(\cfrac{100+604-324}{20\sqrt{604}} \right)=\measuredangle C \\\\\\ cos^{-1}(0.77309903583)\approx \measuredangle C\implies 39.366998916677^o\approx \measuredangle C[/tex]
[tex]\bf cos^{-1}\left(\cfrac{10^2+(\sqrt{604})^2-18^2}{2(10)(\sqrt{604})} \right)=\measuredangle C \\\\\\ cos^{-1}\left(\cfrac{100+604-324}{20\sqrt{604}} \right)=\measuredangle C \\\\\\ cos^{-1}(0.77309903583)\approx \measuredangle C\implies 39.366998916677^o\approx \measuredangle C[/tex]
