can anyone give me a hand with this physics problem?

I could spend all day explaining you but that would take too long and you probably wouldn't understand. I personally like the parallelogram method and I found this webpage where they explain it quite well, you just have to use the la of Cosines.

Your resultant vector would be something like 80^2 + 40^2 - 2(80)(40)cos(105º). Then take the square root of all that and that should be your resultant Vector.

Once you form the parallelogram you need to calculate the inner angle, you do that by taking the greater angle and subtracting the minor one (I hope you've made your graphic already...). so in plain english you take the angle of B minus the angle of A or 135-60= 75º. That's one of the inner angles in the parallelogram. There's an opposite side that we know is congruent, so we know now that two out of the 4 angles are equal to 75º. Now we have two other angles we need to figure out... but we know that all four angles should be equal to 360º. So to figure out the other 2 angles left we can do the next:

360= A1+A2+75º+75º

A1=A2= X so we have 360= 2X + 150º------>X= (360-150)/2 ---->=105.

So now we know that within the the parallelogram we have two angles of 75º and two angles of 105º.

According to the law of cosines you resultant vectos AB should be directly opposed to that angle we just calculated, 105º, and your vectors A and B should be on either side.

So your final vector should look like:

AB^2 = A^2 + B^2 -2A*B*Cos(angle--->105º). There's your resultant vector AB.

I'll leave a link so you can see it for yourself. I think it'll be easier for you that way.

Scroll down to the botton http://www.mathwarehouse.com/vectors/resultant-vec...

First resolve them into x and y components. Cos for x and sin for y comp. The subtract both these components and you would get the desired vector. Finally find the magnitude by pythogoras theorem.