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http://www.mathisfunforum.com/viewtopic.php?id=15250

It is not possible to say you do not like what I did. I did not do it, Geogebra did. That is the point.

Your numbers are all a tad high.

]]>The resultant vector is

(120 + 200COS52, 200SIN52)= (243.132, 157.602)

The magnitude of this vector is

Sqrt(243.1322 +157.6022) = 289.744

The angle measurement of BAC is 52 degrees and the vector AD is the resultant vactor. The angle you want is BAD and you know the lengths of the sides of the triangle BAD so you can find the angle you need using the Law of Coaines.

]]>http://www.mathisfunforum.com/viewtopic … 17#p168917

We want to solve for P and Q. We want to solve this using Geogebra.

Define the origin as A

Start by putting in the initial vectors. On the command line enter:

u = (0,-5)

Now create a point at (-8 , 0 ) called B. Get the angle of B and A of 15 degrees. That point will be called B'. Draw a vector from A to B' it should be called v. It will have a magnitude of 8 N.

Add the two vectors u and v by inputting u + v in the command line. It should be called w or w'. Negate that vector to reflect it into the first quadrant by inputting:

w = -w'

This is the reflection vector.

Put a point at ( 0 , 8 ) called C. Create an angle of 20 degrees using C and A. That will reate the point C'. Draw a vector from A to C'. Relabel C' to Q.

Create a vector from A to ( 8 , 0 ) and call it P. Get the resultant of P and Q by adding in the command the name of two vectors they will be small letters. On mine they are a,z

a + z

The resultant is called b. Recolor this b vector red. Now move P ( the head of the horizontal vector ) towards A until it covers the reflection vector from the 3rd quadrant.

Now if you drag C (0,8) towards A the red vector c will shorten. Keep adjusting P and C until the red vector exactly superimposes over the reflection vector.

When you eyeballed it as best you can measure the distance between A and Q and A and P. I get Q = 7.55 and P = 5.18 That is quite close.

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