A chain of weight W is slung between two walls such that the two ends are level. Given that the chain is slack, and that the angle formed between the normal to the wall and the chain at each end is $\displaystyle \theta$, find the tension at the bottom of the chain.

This is part b of the same question.

Part a asks for the the tension at each end of the chain. That is obvious because the sum of the vertical components of the tension (T) at each end = weight of the chain. This implies that W=2Tsin$\displaystyle \theta$, hence T=W/(2sin$\displaystyle \theta$)

Can anyone help with part b?