Complicated mechanics resolving forces and angles question

[xllx]

Homework Statement

An object has a weight of 20N. It is supported by three strings to keep it in equilibrium, all attached to a smooth ring on the object. String A has a tension of 15N to the right of the block. String B is (a)degrees away from string A and String C is 90degrees away from String B. String B and C have the same tension. Show that the angle is 82 degrees

Homework Equations

cos(theta)/sin(theta)=tan(theta)
Forces in equilibrium cancel each other out.

The Attempt at a Solution

So far I've got this, but get stuck past this point:

15+Bcos(a)=Acos(90-a) and 20=Bsin(a) + Asin(90-a)
A and B then can be replaced by T and rearranged:

Tcos(a)=-Tsin(90-a) + 20
Tsin(a)= Tcos(90-a)-15

tan(a)= -Tsin(90-a)+20/Tcos(90-a)-15

But then I don't know how to resolve it. Any help at all would be greatly appreciated! Many Thanks!

 

[Redbelly98]

The Attempt at a Solution

So far I've got this, but get stuck past this point:

15+Bcos(a)=Acos(90-a) and 20=Bsin(a) + Asin(90-a)

Perhaps you have the benefit of a figure that we cannot see, but when I read the question I would say that rope C's angle is 90+a.

A and B then can be replaced by T and rearranged:

Tcos(a)=-Tsin(90-a) + 20

The LHS should be T sin(a) here.

Tsin(a)= Tcos(90-a)-15

The LHS here should be T cos(a)

Try making the corrections I indicated, then we'll have another look at what you have. Also, think about whether it's 90-a or 90+a for rope C. Again, you may be right about 90-a, if you have a figure that shows the rope arrangement.

 

[xllx]

Perhaps you have the benefit of a figure that we cannot see, but when I read the question I would say that rope C's angle is 90+a.


The LHS should be T sin(a) here.


The LHS here should be T cos(a)

Try making the corrections I indicated, then we'll have another look at what you have. Also, think about whether it's 90-a or 90+a for rope C. Again, you may be right about 90-a, if you have a figure that shows the rope arrangement.

Sorry I think I've got slightly confused with the letters. I think that it is 90-a because the angles are along a straight line and so will add up to 180 degrees. And so the angle between string B and C is 90 degrees so angle (a) plus the angle between A and the line has also got to add up to 90 degrees and the angle will be 90-a. Am I totally wrong? But that is where I was coming from to get the angle.

If I change the equations I then get:
Tsin(a)=-Tsin(90+a)+20
Tcos(a)=Tcos(90-a)-15

I've got to show that a=81.87 and t=17.68 and it only works if the second one is minus.

From there do I divide them to get tan(a) or do simultaneous equations?

 

[Redbelly98]

I'm continuing the discussion in your other thread on this same problem here: (where there is a figure)

https://www.physicsforums.com/showthread.php?t=304164

Please, in the future don't double-post questions. Things would become very cluttered in the forums if everybody did that.