Answer:
[tex]1000\; {\rm N}[/tex] at approximately [tex]36.9^{\circ}[/tex] east of north.
Explanation:
The forces in this question are in a two-dimensional plane. Let the first component denote forces in the north-south direction (with north being the positive direction.) Let the second component denote forces in the east-west direction (with east being the positive direction.)
The [tex]600\; {\rm N}[/tex] force points to the north: [tex]\langle 600,\, 0\rangle \; {\rm N}[/tex].
The [tex]800\; {\rm N}[/tex] force points to the east: [tex]\langle 0,\, 800\rangle \; {\rm N}[/tex].
The net force is the vector sum of the components:
[tex]\langle 600,\, 0\rangle \; {\rm N} + \langle 0,\, 800\rangle \; {\rm N} = \langle 600,\, 800\rangle \; {\rm N}[/tex].
The magnitude of the net force would be:
[tex]\sqrt{600^{2} + 800^{2}}\; {\rm N} = 1000\; {\rm N}[/tex].
Both components of the net force are positive, meaning that this force points in a direction that is east of north. To find the angle between this force and north, take the inverse tangent of the ratio between the second and the first component of the force vector:
[tex]\displaystyle \arctan \left(\frac{600}{800}\right) \approx 36.9^{\circ}[/tex].
In other, the net force points at approximately [tex]36.9^{\circ}[/tex] east of north.
