Tension in a string, at an angle 
This was a question on a mechanics exam. Part i wants me to assume that the tension in  both parts of the string is the same. Even though I got the correct answer( 18.9N) by assuming so, I don't understand how this assumption makes sense. Consider the bit of string under the ring. The net force on it must be zero ( light string ) so that implies the net vertical component is zero. This is not possible if I assume the tensions to be the same, or at least that's how I see it ( see image)
Basically in my mind: If the tensions are equal, the net force on the bit in contact with the ring can not be zero since the angles are different. But since the string is light, this does not add up. ( Or the system is at rest. Net force on that bit of rope must be zero, regardless of light/ not light)  
But the question assumes the opposite. I would like to know what part of my argument here is flawed. 
 A: First answering to your doubt of tension being same in both parts of string.Young's modulus of a material is constant irrespective of shape length etc. (at least from the point of view of solving high school questions ) , you can study about it a bit on the internet if you don't know it already. $$Y={{F\over A}\over {\Delta L\over L}} $$where $Y$ is the Young's modulus of elasticity of your string.From here $$F={YA\over L}(\Delta L)$$As you can see from this equation,  the force or the tension for a realistic string depends on its extension and the constants beside it.Moreover the Force equation depends inversely on the actual length of string too (see L in the denominator ), so unless you cut the string where the ring is and then attach both parts above and below the ring , the force equations are not going to change and the string as a continuous entity will have same tension everywhere , because $\Delta L$ here accounts for whole string.
The purpose of invoking elasticity in the picture is to explain the tension aspect in near ideal conditions , there is nothing like a mass less string but the above explanation is a close analysis.
Coming to the actual question (tension is now same  throughout the string). HORIZONTAL EQUILIBRIUM $$T(cos50+cos20)=X$$ VERTICAL EQUILIBRIUM $$T(sin50)=T(sin20)+(0.8)g$$
Upon solving these two , $T=18.66 N$ and $X=29.29 N$ approx. 


A: First, since the ring doesn't involve any corners, sharp angles, etc. we can safely assume the tension is uniform throughout the massless string that is threaded through the ring.
Assuming that the vertical components of tension is equal to the vertical force the ring experiences due to each string, then $X$ must take a specific value consistent with Newton's laws in the horizontal direction. There is no issue in having a constant tension throughout the string.
Indeed, if you do the work for general angles (you can do the work yourself) you will find
$$X=\left(\frac{\cos\theta_A+\cos\theta_B}{\sin\theta_A-\sin\theta_B}\right)mg$$
If we were to change the applied force $X$, then this would just result in different angles. Although notice that, for angles less between $0^\circ$ and $90^\circ$, we cannot have $\sin\theta_A<\sin\theta_B$ or else $X<0$. In other words, as we pull harder and harder ($X$ gets larger and larger) the angles will both approach $45^\circ$.
