Question

Consider a 626N weight held by two cables. The left hand cable had tension T and makes an angle of theta with the wall. The right hand cable had tension 720N and makes an angle of 27 degrees with the ceiling.

What is the tesion T in the left hand cable slanted at an angle of theta with respect to the wall? Answer in N.

What is the angle theta which the left hand cable makes with respect to the wall? Answer in degrees.

Here is my work, but i am not sure if i did it correctly:

Tension 1 y axis + Tension 2 y axis=626N
T1y=626-720N sin 27=299.1

T1x+T2x=0
T1x+720 cos27=0
T1x= -720cos27= -641.5

T1 = square root of (T1x)^2 + (T1y)^2= 707.8

T1=left side tension =707.8N

theta=arc tan (T1x/T1y) =
-641.5/299.1= 2.145 degrees <- ( that should not be a negative right?)

Answer 1

You simply calculated T1x/T1y, you didn't take the arctan of this number.
It's difficult to explain, especially without a diagram, but ignore the negative sign:
theta = arctan(641.5/299.1) = arctan(2.14) = 65 degrees
This angle is relative to the wall. Relative to the horizontal, theta = 180 - (90-65) = 155 degrees.
So,
T1 cos theta1 + T2 cos theta2 = 0
707.8 cos 155 + 720 cos 27 does indeed equal zero.