1. The problem statement, all variables and given/known data 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] 2. Relevant equations 3. 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??