Solving tension equations with trig.

[sdoi]

Homework Statement

There are four pieces of rope tied between four trees. Given the forces of tension in three of the strands, find the force of tension in the remaining rope.
Rope 1: 2.1x10^-3N [20°E of N]
Rope 2: 1.6x10^-3N [60° E of S]
Rope 3: 1.9x10^-3N [40° W of S]

Homework Equations

The Attempt at a Solution

I tried using trig to find the tension in rope 4, but I have no idea if I'm approaching this in the right way:
1st I drew out all of the ropes and ended up with something that looks like a distorted square. I drew a line down the middle so I would end up with two separate triangles. This line would represent the sum of the tension in rope 2 and 3.

(a/sinA) x sin C =c
(1.5x10^-3N/ sin 40) sin 80= c
c= 2.45x10^-3N

Then I went on to find the tension in the third rope:
Because I know two sides of the triangle (Rope 1, and the sum of rope 2&3) I used the cosine law:
a^2=b^2+c^2-2bccosθ
= (2.1x10^-3N)^2 + (2.45x10^-3N)^2 - 2(2.1x10^-3N)(2.45x10^-3N)cos20
a= √7.4x10^-7N
a= 8.62x10^-4N
Therefore the tension in the fourth rope would be 8.62x10^-4N.

I know this isn't correct, but I don't know why, wait unless the 8.62x10^-4N is the sum of all of the tensions in the ropes? and all i have to do is subtract Rope 1, 2, &3 from 8.62x10^-4N.

Any Ideas??

 

[Spinnor]

Its not clear but I'm thinking the four ropes are connected at a common point in the center. If so figure out the x and y components of the three forces and add the x and y components and call that sum vector S. The fourth force must be equal to -S

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