Two forces of 200N are acting at a common point. The first force acts due East and the second force is acting 40 degrees North of East. Find the magnitude of the resultant force by cosine law and the direction using sine law.
First of all, I was thinking it would look like this: thinking of the origin as the common point. But I thought of it another way like this: and I was really confused on where I would put my resultant force because I really don't know where to put the angle of the resultant force(or the direction of the resultant force). I mean 40° cannot be that angle right?
Your first diagram corresponds to the forces acting at one point, as the problem stated. The second diagram could be used to construct the vector sum of those two forces, but it does not represent where the forces are acting. The problem asks you to use the laws of cosines and sines. Can you state those for us, and then see where you might go from there?
Ok here we go with my computation(Please correct me if I'm wrong): Is this correct? I'm really having doubts on where to put my resultant force and its angle you know.
You've got the right idea, although it sounds like you're just plugging in numbers without being sure about why it works. I'm also not sure how that term under the square root in your first part came out to be positive - it doesn't look like it would, given what you've written. Look at your second diagram. Can you add the resultant force to it? That should give you a triangle, and the calculations you've done apply directly to that. Keep in mind that the angle you've called theta is an external angle to the triangle, so you have to be careful about how the sine and cosine of that angle change, compared to what you'd get for the internal angle, which is what the sine and cosine laws refer to. The force diagram - i.e. a picture of real space with the force vectors drawn from the points where they are applied - is your first diagram. After you've used the second diagram to find the magnitude and direction of the resultant force, you can copy it to the first diagram to see how each of the forces acts. You could also have done the whole problem there (which would be my preference), using the parallelogram rule for adding vectors. You get the same triangle as in the second diagram. I think you've got the right ideas here (and your answers look right to me); you just need to be clearer on why it all works. Keep posting questions until you're sure you've got it.
its very rare around here that i get to help out, so here i go: as for the law of sines bit, your formula in there is 100% right, but for the real answer you wanna use the new number you get from plugging in whatever i just gave you above and you're good!