precalculus
posted by Iamxe .
A force of 369 lbs makes an angle of 18 degrees 20 minutes with a force 427 lbs. Find the angle made by the equilibrant with the 369 lbs force.
I know I can find this using the law of cosines, but I think my teacher wants me to use some vector thing, but I do not know how to relate the two vectors to the equilibrant. Is it like uv?

Hmmm. use the 427 Lbs at 0 degrees, then the cos18'20" * 369 is the force in the direction of zero degrees, and 369*sin18'20"is the force 90 degrees to that. Now add these two to see the resulatant, and then add180 degrees to get the equilibrant.

Well, either way it is a "vector thing.
The "equilibrant" is equal and opposite to the "resultant" or sum of the two vectors. In other words it is the vector needed to cancel them.
I will find the resultant, then we can do minus signs.
I am going to say we have 369 pounds along the x axis and 427 pounds at 18 1/3 degrees above the x axis.
In vector notation that is
Rx = 369 + 427 cos 18.33
Ry = 427 sin 18.33
or
Rx = 369 + 405 = 774
Ry = 134
Our equilibrant has components then
Ex = 774
Ey = 134
That is in the third quadrant at
tan A = 134/774 where A is below the x axis
A = 9.82 degrees below x axis
or
189.8 degrees counterclockwise from x axis where our original 369 lb force was
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