How do you work out the tension and angle of this block in equilibrium?

[xllx]

Homework Statement

A block that has a weight of 20N, is in equilibrium. It is attached to 3 strings, A, B and C. String A has a tension of 15N to the right of the block. Between String A and B is the angle (a) and between B and C is 90degrees. The tension in String B and C is the same. Calculate the tension and the angle (a).

Homework Equations

sum of the forces=0

The Attempt at a Solution

For the horizontal forces I got:
15+Tcos(a)=Tcos(90+a)

And then for the vertical:
20=Tsin(a)+Tsin(90+a)

But I don't know how to figure it out from there, or whether they are even right. Any help at all would be greatly apprecaited!

 

[rootX]

cos(90+a) = sin(a)
sin (90+a) = -cos(a)

But, why the angle is 90+a?

 

[xllx]

cos(90+a) = sin(a)
sin (90+a) = -cos(a)

But, why the angle is 90+a?

The angle of String C is 90degrees plus (a) away from the 15N to the right. I've attached a diagram of the situation, sorry I aren't very good at expalining. Am I totally wrong with the angles??

Thanks by the way!

 

[rootX]

It is easier to visualize if you use 90-a (triangle left to the force vector C).

I would recommend to divide both equations and eliminate T. Then you will get Acos (a) = B sin(a) relationship..which is easier to solve

 

[xllx]

So my two equations are:

15+Bcos(a)=Ccos(90-a)
20=Bsin(a)+Csin(90-a)

Which because the tension is the same in both means the A and C can be replaced by T:

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

From this do I rearrange to get Tcos(a) and Tsin(a) as the subjects? And if they are divided they equal tan(a), or is this wrong?

 

[Redbelly98]

I would next simplify the cos(90-a) and sin(90-a). You can use the trig formulas for sin(x-y) and cos(x-y) to do this.