Problema 2
\(=\sqrt{ax^2+ay^2}=\sqrt{\left(-25\right)^2+\left(40\right)^{2\ }}\)=\(\sqrt{625+1600}\)
\(\sqrt{2225}\)\(=47.16\)
\(Tan\theta=\frac{Ay}{Ax}=\theta\tan^{-1}\left(\frac{40}{-25}\right)=-57.99\)°