I have attached a file , would u pls give me adetailed solution for the problem

Answer
Let m is the mass of the rigid frame.
Since the horizontal bar is not equal in length on both sides so , let x is the location of center of gravity
from left end of the horizontal bar.
Now, If we resolve the Tension in to two perpendicular component, we get
Tcos200 along horizontal in right hand side direction
Tsin200 Vertically downward.
For the rigid frame in equili
ium
Sum of translational forces must be equal to zero.
And sum of Moment on both sides must be zero.
Then the frame...