Can someone solve a word problem determining force placed on a barge

One rope pulls a barge directly east with a force of 100 newtons. Another rope pulls the barge to the northeast with a force of 200 newtons. Find the resultant force acting on the barge, to the nearest unit, and the angle between the resultant and the first rope, to the nearest tenth.

Break down each pull into "orthogonal" vectors.
(orthogonal = fancy word for mutually perpendicular)

In such problems, involving cardinal direction, the easiest is to use East as the "x-direction" and North as the "y-direction).

Then figure out which portion of each pull gets assigned to each direction, using trigonometry.

In this problem, things are easy:

100 N due East.
Obviously, all of it is to the East, and none of it to the North.

200 N to Northeast.
That is exactly halfway between East and North.
Since the two directions are perpendicular (90 degrees or pi/2 radians), then the direction is 45 degrees (counting from East, going northward as in mathematics).

The component going in the x-direction is
200 * cos(45) = 141.42 N
The component going in the y-direction is
200 * sin(45) = 141.42 N

Once you have finished separating all vectors into their components (here it was easy because we only had two), you add together all the x-components, and separately all the y components

x-components (towards the East):
100 + 141.42 = 241.42 N

y-component (towards North):
0 + 141.41 = 141.42

Vectorial addition -- use the Pythagorean Theorem
(yes, the same one you used for right-angle triangles)
(didn't they tell you it gets used a lot?)

Resultant pull = sqrt( East^2 + North^2)
R = sqrt( 241.42^2 + 141.42^2)
R = sqrt(78,567) = 280.3 Newtons

Direction: use tangent
Tan(direction) = opposite/adjacent
since we decided to measure angles from East:
opposite = North
adjacent = East
Tan(D) = 141.42 / 241.42 = 0.58578...