Answer:
force of 279 N
53.75 degrees from east to north
Explanation:
let net force be F , Fn - north force , Fe - east force
[tex]F^{2} =F_e^{2} +F_n^{2} \\=225^{2} +165^{2} \\=77850\\\\F=\sqrt{77850} \\=279 N[/tex]
let the direction be α angle from east to north
[tex]tan( \alpha)=\frac{F_n}{F_e} =\frac{225}{165} \\=1.3636\\\\\alpha =tan^{-1} (1.3636)\\=53.75 degrees[/tex]