# Thread: confusing vector problem

1. ## confusing vector problem

< is angle symbol

R is the resultant of the three forces A, B , C, that is R= A + B+C

If A =A<210 , B =200< Theta , C = 200 <65 , and R = 250<125.

use the component method to determine A and Theta(there may be 2 solutions)

So

Ax= A cos 30
Ay= A sin 30
Bx=200 cos Theta(possibly could be switched)
By=200 sin Theta
Cx=200 cos 65 = 84.5
Cy=200 sin 65 = 181
Rx=250 sin 35 = 114.7
Ry=250 cos 35 = 204.7

A cos 30 + 200 cos Theta = 30.2

is this the correct way to do this problem? Where would I go from here since I have 2 variables? any help to give me a shove would be appreciated.

2. Originally Posted by ur5pointos2slo
< is angle symbol

R is the resultant of the three forces A, B , C, that is R= A + B+C

If A =A<210 , B =200< Theta , C = 200 <65 , and R = 250<125.

use the component method to determine A and Theta(there may be 2 solutions)

So

Ax= A cos 30
Ay= A sin 30
Bx=200 cos Theta(possibly could be switched)
By=200 sin Theta
Cx=200 cos 65 = 84.5
Cy=200 sin 65 = 181
Rx=250 sin 35 = 114.7
Ry=250 cos 35 = 204.7

A cos 30 + 200 cos Theta = 30.2

is this the correct way to do this problem? Where would I go from here since I have 2 variables? any help to give me a shove would be appreciated.
A cos(210) + 200 cos(theta) + 200 cos(65) = 250 cos(125) :: x-vectors
A sin(210) + 200 sin(theta) + 200 sin(65) = 250 sin(125) :: y-vectors

You have 2 unknows & 2 equations
It is (my opinion) not a good idea to reduce the angle to first quadrant.

.

3. Originally Posted by aidan
A cos(210) + 200 cos(theta) + 200 cos(65) = 250 cos(125) :: x-vectors
A sin(210) + 200 sin(theta) + 200 sin(65) = 250 sin(125) :: y-vectors

You have 2 unknows & 2 equations
It is (my opinion) not a good idea to reduce the angle to first quadrant.

.
Thank you but I still do not know how to solve the system. I have never dealt with systems of equations involving trig functions. I went ahead and calculated the cos and sins and got this:

-.866A + 200 cos (Theta) +84.52 = -143.39
-1/2A + 200 sin(Theta)+181.26 = 204.8

could someone please help me with the first few steps at least. I am not sure how to solve this.

4. Originally Posted by ur5pointos2slo
Thank you but I still do not know how to solve the system. I have never dealt with systems of equations involving trig functions. I went ahead and calculated the cos and sins and got this:

-.866A + 200 cos (Theta) +84.52 = -143.39
-1/2A + 200 sin(Theta)+181.26 = 204.8

could someone please help me with the first few steps at least. I am not sure how to solve this.
Let B= cos(Theta). Your equations become
-.866A + 200 B +84.52 = -143.39
-1/2A + 200 B+181.26 = 204.8

or -.866A+ 200 B= -227.91
-.5A+ 200B= 23.54
with no trig functions! Solve for A and B and then use B= cos(THETA) so THETA= arccos(B) to find THETA.