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September 1, 2014

Posted by **Derek** on Sunday, June 29, 2008 at 8:30pm.

- URGENT SUMMER SCHOOL CALCULUS -
**bobpursley**, Sunday, June 29, 2008 at 9:06pmLaw of cosines:

The side you are looking for is opposite the 30 degree angle.

c^2=15^2+8^2-2*15*8cos30

I get a magnitude of about 9

For the angle, Iwould use law of sines. Draw a sketch first.

- URGENT SUMMER SCHOOL CALCULUS -
**Damon**, Sunday, June 29, 2008 at 9:16pmWell, 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

or

6.93 + x cos y = 15

4 - x sin y = 0

or

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

which is

26.4 + 30 = 56.4 degrees from original vector

x = 4/sin26.4 = 9 Newtons

- URGENT SUMMER SCHOOL CALCULUS -
**Count Iblis**, Monday, June 30, 2008 at 10:23amYou 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)

We have:

x = R - y

taking the inner product of both sides with R gives:

x dot R = R^2 - y dot R =

R^2 -|y||R|cos(30°)

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°

- URGENT SUMMER SCHOOL CALCULUS -
**Derek**, Monday, June 30, 2008 at 1:25pmMy 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?

- URGENT SUMMER SCHOOL CALCULUS -
**Tapiwa Mudzori**, Thursday, January 24, 2013 at 3:52pm15 cos

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