@ Simon Bridge spare me the comment! if u don't want to help don't reply!
I
cannot help without knowing how you are thinking about the problem.
From the information you supplied, you got a fair comment: but it's a bit terse. I edited so it reads more clearly. Better?
Your numbers don't mean anything - I need to see your reasoning: how are you thinking about the problem.
From what you have written I can make a guess about this - but it is unlikely to be very satisfying for you.
I wonder if you have misunderstood the nature of this place? We can help a student get unstuck but we do not do the work for the student. This creates a problem when someone asks "is this the right answer?" If I just say "yes" or "no" then some students will be cunning and keep guessing until I say "yes" and so get the right answer by making me do the work.
But it appears you can apply the rules of trigonometry OK :)
If ##\theta## is the angle some force ##\vec{F}## to the horizontal, then the horizontal component of that force is, indeed, ##F\cos(\theta)##
You kept going:
for 1b i got : 50xsin70=46,98 T
You appear to have computed the vertical component of T.
Is that correct?
1b asks for
torque
... what
is torque?
... what are the units for torque?
For (c) you have already asked in another thread.
But you already know - just consider what the term "translational equilibrium" means.
If in doubt, you can look in your class notes or use google ;)
For (d) it is the same process, but for rotation.