Law of cosines:
The side you are looking for is opposite the 30 degree angle.
I get a magnitude of about 9
For the angle, Iwould use law of sines. Draw a sketch first.
Well, I am taking the resultant to be along the x axis and the 8 N vector at 30 degrees above the x axis.
Then our vector of length x will lie below the x axis an angle y
Now there are two things that have to be true
8 cos 30 + x cos y = 15
8 sin 30 - x sin y = 0
6.93 + x cos y = 15
4 - x sin y = 0
x cos y = 8.07
x sin y = 4
but we know sin y / cos y = tan y so divide
tan y 4/8.07
y = 26.4 degrees below x axis
26.4 + 30 = 56.4 degrees from original vector
x = 4/sin26.4 = 9 Newtons
You can also solve this problem purely algebraically without using geometry or drawing diagrams. If we denote the resultant force by R, the force of magnitude 8 N at an angle of 30° to the resultant by y and the unknown force by x, we have:
x + y = R ------->
x = R - y
Square both sides:
x^2 = R^2 + y^2 - 2 R dot y
The inner product R dot y can be expressed as:
R dot y = |R||y| cos(30°)
This then gives |x| = about 9 N
To find the angle of x to the resultant (let's call this theta), you can take the inner product of x with R and use that
x dot R = |x| |R| cos(theta)
x = R - y
taking the inner product of both sides with R gives:
x dot R = R^2 - y dot R =
If we divide both sides by |x||R| we get cos(theta) on the l.h.s.:
cos(theta) = [|R| - |y|cos(30°)]/|x|
this gives theta = 26.4°
My diagram doesn't work out, I have done it to scale, but I can't figure out how to draw it. I got the same answers as everyone above, I used the cosine and sine law to do mine (I got my answer after I posted :) ). But my diagram looks weird. Any thoughts?
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