A great start! You have the wrong answer, but don't worry.
First thing, let's improve your diagram,
i) write down the magnitude of each force (14N or 9N) by each arrow. Remember that your vectors in this problem represent force, not velocity, so you should write the units as N (for Newtons).
ii) draw a set of axes at the point the two forces originate (basically just a vertical line and a horizontal line). Label the axes x and y. Use a ruler! If you get puzzled about the horizontal vector because it overlaps, make the horizontal vector bold by making the arrow a little thicker. You should always draw the axes first.
iii) We know that the 14N force is off to one side, so we can draw a little triangle and put the 65 degree angle in the diagram. This helps a lot with the calculations later.
You should have something that looks like the attached file. I prefer to hand-draw these things but I don't have a scanner to hand
Also, a couple of things to clarify before we move on
i) The magnitude is, in simple terms, how big the vector is. The magnitudes of the forces are 14N and 9N.
ii) The direction is where the arrow is pointing. In our question we are using angles or descriptions (horizontally to the right) to mark the direction.
Every vector must have a magnitude
and a direction. If you don't put the direction down, you can lose marks in an exam!
That's step 1 done from my first post. Now let's look at the second step:
2. Determine the horizontal and vertical components of each vector (use trigonometry for this)
I recommend following through each calculation bit by bit on paper as you read this. It will really help, so make sure you have pencil, paper and calculator at the ready and that you have your newly improved diagram in front of you
You see the triangle we drew with the angle 65 degrees? We need to calculate the lengths of the sides of that triangle. That will give us the horizontal and vertical components of the vector. We use trigonometry for this.
Well.... \(\displaystyle \sin \theta = \frac{opp.}{hyp.}\) from SOHCAHTOA (if you're not sure what this is, look it up on Google!). In our triangle the opposite is the horizontal side, which we're after, and the hypotenuse is 14N. So, we can rearrange it a bit:
\(\displaystyle opp. = hyp. \times \sin\theta\)
and we can put the numbers in:
\(\displaystyle opp. = 14 \times \sin 65^{\circ} = 14 \times 0.9063 = 12.688\)
So the horizontal component of the 14N force is 12.688N
to the left. It's important to notice that it goes to the left
You now need to get the vertical component of the 14N force (hint: use cosine...)
The horizontal component of the 9N force is just 9N, because it is horizontal already
The vertical component of the 9N force is 0N, because the arrow is horizontal.
Once you do this step, you're nearly done as this is the annoying bit. Also, the 9N force is a trick question. Let me know what answer you get and, if you're feeling adventurous, have a go at step 3 in my first post.