# Quick Resultant Forces Question

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1. Oct 24, 2016

### Pablo1122

1. The problem statement, all variables and given/known data

So we have 65N @ 30*, 30N @ 180* and 20N at 250*

2. Relevant equations

Ncos(angle)i + Nsin(angle)j

3. The attempt at a solution

Using this video

I did the following.

F1(65N) = 65cos(30)i+65sin30i

Then using the radian circle I saw that 30* has x,y coordinates of sqrt3/2 and 1/2 respectively. So as shown in the video I did

65(root3/2)i+65(1/2)j

= 56.3i + 32.5j

I did this for the left one too (30N).

But for the one at the bottom (20N) I get the degree of it to be 250. (270-20) = 250.

250* is not on the radian circle so how can I write 250* in terms of x,y coordinates?

If we look at this radian circle. We see that 240* = (-1/2, -sqrt3/2) and then it goes to 270* = (0,-1). How can I find the x,y coordinates for 250* and use it in the equation?

2. Oct 24, 2016

### Staff: Mentor

I don't believe that trig functions of that particular angle can be simply expressed. (Like the others can.) You'll need a calculator.

3. Oct 24, 2016

### Pablo1122

Ah yes, I just found this out. Thanks!

4. Oct 24, 2016

### haruspex

If it's a matter of expressing it in surds, you can use the usual expansion of sin(3x), with x=20o, and apply the formula for solving cubics.