Re: Law of sine vector problem...

[Robert Mathieson <rmathieson@culver.edu>]

I am guessing you are a student so I will answer assuming that.


> 	Here are my steps to solving it.  I solved for the first angle
> (theta 1) as the angle between the two vectors as 135 degrees.  I drew
> the parallelogram and solved for the acute angles as 45 degrees (theta
> 2).  I drew the resultant vector and solved its length using the law of
> cosine:  Resultant = square root (30.556^2 + 16.667^2 -
> (2*30.556*16.667*cos 45)) = 22.164m/s
> 	To solve for the angle I chose the angle between the westward
> vector and the resultant vector (theta 3).  I used the law of sine:
> Resultant velocity / sin (theta 2) = northeast velocity / sin (theta 3)
> 22.164/ sin 45 = 30.556/ sin (theta 3)


You are ok to here.  Solve for sin theta3

                         sin theta 3 = 0.97484

Here is your difficulty.  You correctly pushed the inverse sin button
(sin^-1) and the calculator correctly gave you 77.12 degrees.  You just need
to know one more thing.  There are many angles for  which the sin is
0.97484.  Your calculator only gives you one of them.  You need to learn a
bit more about trigonometry to find the others.  Check a trig book out.  For
your problem you just need to know that one of the other angles is 180
degrees minus the angle your calculator gave you:

                        180 - 77.12 = 102.88 degrees

which agrees with your other calculation.

Look in a trig book.  It will be pretty easy to see what's going on when you
have some diagrams to look at.  Good luck.

Robert Mathieson
Culver-Stockton College
Canton, MO 63435
(217)233-6000
rmathieson@culver.edu