Find the angle between the forces given the magnitude of their resultant. (Hint: Write force 1 as a vector in the direction of the positive x-axis and force 2 as a vector at an angle theta with the positive x-axis. Force 1=45 pounds, Force 2=60 pounds, Resultant Force= 90 pounds.

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Find the angle between the forces given the magnitude of their resultant. (Hint: Write force 1 as a vector in the direction of the positive x-axis and force 2 as a vector at an angle theta with the positive x-axis. Force 1=45 pounds, Force 2=60 pounds, Resultant Force= 90 pounds.

Mathematics
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Did you attempt to make a force diagram?
|dw:1433812186168:dw|
Well, that is what I have so far.

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Other answers:

This involves the section of "Vectors in a plane."
correct so far
|dw:1433819768895:dw|
Ops, forgot to put a 60 in there. Now, I'm not sure what to do next.
The directions state that the magnitude of the resultant is 90 pounds.
If we call the resultant vector R then magnitude of R= sqrt( Rx^2 + Ry^2)
What is the magnitude of R= sqrt(Rx^2+Ry^2) formula/name called?
that comes from pythagorean theorem
the length of a vector is the length of the hypotenuse
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Just to check, R= sqrt (Rx^2+Ry^2) is the same as C= sqrt (Ax^2+By^2) ?
yes
|dw:1433813159626:dw|
|dw:1433820470702:dw|
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=\[2025+3600\cos^2t+5400cost+3600\sin^2=8100\]
is that the correct next step?
ops forgot to put the t for 3600sin^2t.
yes
we can simplify this by factoring out 3600 from sin^2 t and cos^2 t
$$\large {2025+3600\cos^2t+5400\cos t+3600\sin^2 t=8100 \\ 2025+3600( \cos^2t +\sin^2 t ) +5400\cos t=8100 \\2025+3600( 1) +5400\cos t=8100 } $$
\[5625+5400\cos t = 8100. ->5400\cos t =2475. ->\cos t =2475/5400. -> \cos t = .4583 -> arc \cos t = 62.7 degrees.\]
now solve for t, the angle
\[\cos t = 0.4583 -> \arccos t (0.4583)\]
= 62.7 degrees?
Ok, that is the answer. Thanks! One small question, so dealing with these type of problems, this will always involve the Pythagorean Theorem of R=sqrt(Rx^2+Ry^2)?
if you need to find magnitude of a vector, then yes :)
Ok, thanks again. :D

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