In this system, a particle on a horizontal surface is acted on by two forces, $F_1$ and $F_2$. Find the $\hat i$ and $\hat j$ components of the resultant force, where $\hat j$ represents due north and $\hat i$ due east. $F_1 = 3\hat i + 2\hat j$, $F_2$ has a magnitude $5$ N and acts at a bearing of 60 degrees. I thought that I had to resolve $F_2$ into horizontal and vertical components $5\cos(60^\circ)$ and $5\sin(60^\circ)$ and then add these to the $\hat i$ and $\hat j$ parts of $F_1$, giving me $5.5\hat i$ and $6.33\hat j$ for the resultant force, but the mark scheme says that the answer is $7.33\hat i + 4.5\hat j$. I don't know how to get to these, and I don't think I've made any stupid mistakes, but if I have, please let me know.
Any advice would be appreciated.