Physics: How to add and subtract vectors

Here are two sample problems that I don't know how to answer. It'd be nice to explain it method by method. Most understandable explanation gets awarded.

Problems:
1. Fishing boat:
6m/s due North = wind velocity
10m/s due 240 degrees is the motor velocity
and the current velocity happens to be 3m/s due east.

Add the above equation up :D


The one below is for subtraction:
A spacecraft of mass 800kg is maneuvered by 2 rockets. The first rockets fires from the south west with a thrust of 1000N. The second fires from the south east with a thrust of 1000N. What is the:
Net force on the spaceship
and the net acceleration of the spaceship?

The simplest method is to split the vectors into Cartesian co-ordinates (North/South and East/West).

Problem 1:
Vector 1 is 6m/s North and 0m/s East/West
Vector 2 is at 240 degrees, aka West 30 degrees South. It breaks down into 10m/s * sin 30 degrees in the South direction plus 10m/s * cos 30 degrees in the West direction. These are 5.0 South and 8.66 West respectively.
Vector 3 is 0m/s North/South and 3m/s East.

Adding up the North/South components of the vectors, we get 6m/s North + 5m/s South + 0, which comes out to a total of 1m/s North.

Adding up the East/West components, we get 0 + 8.66m/s West + 3m/s East, which comes out to 5.66m/s West.

So the components of the final vector will be 1m/s North and 5.66m/s West. These are the vertical and horizontal lines in a triangle, with the third line being the final vector.

The size of the final vector will be the square root of the sum of the squares of the two components. sqrt (5.66^2 + 1^2) is 5.74m/s.

We already know (from the components) that the direction of the final vector will be somewhere between North and West, and will be more West than North. This means that the compass bearing will be between 270 degrees (West) and 315 degrees (North-West).