Three wires $AP,BP$ and $CP$  are made of same materials and have same cross sectional area. They are fixed to points A, B and C on a ceiling and their free ends are knotted together at  P. Lengths of the three wires are $AP=CP=l_2$ and $BP=l_1.$ A weight W is suspended at P and the wires get taut. Assume the deformations to be small.

After the weight is added, the point P moves to $P_1$ extension in BP is (say)

Drop a perpendicular from P on the line $A{P}_1.$ Extention of wire $AP$  is $Q{P}_1=\Delta l_2\left(say\right)$
[This is because  $AQ\simeq AP$]
Since extension are small,
$$\begin{gathered} \angle APB\simeq \angle A{P}_1=B=\theta \left(say\right) \\ {P}_{1}Q=P_{1}P\cos \theta \Rightarrow \Delta l_2=\Delta l_1\cos \theta \mathrm{......}\left(i\right) \end{gathered}$$ 
$\therefore \Delta l_1>\Delta l_2.$  It means extension in $BP$ is more. 

$$\begin{gathered} Stress\left(s\right)=Y\frac{\Delta l}{l} \\ \frac{S_1}{S_2}=\frac{\Delta l_1}{\Delta l_2}\times \frac{l_2}{l_1} \\ =\frac{1}{\cos \theta }\times \frac{1}{\cos \theta } \\ =\frac{1}{{\cos }^2\theta }>1 \\ \therefore S_{BP}>S_{AP} \end{gathered}$$

For BP, stress = Y strain 

$$\begin{gathered} \therefore \frac{T_1}{A}=Y.\frac{\Delta l_1}{l_1}\Rightarrow T_1=YA\frac{\Delta l_1}{l_1} \\ For AP,\frac{T_2}{A}=Y\frac{\Delta l_2}{l_2}\Rightarrow T_2=YA\frac{\Delta l_2}{l_2} \\ 2T_2\cos \theta +T_1=W \\ 2\left(T_1{\cos }^2\theta \right)\cos \theta +T_1=W \\ {T}_1=\frac{W}{1+2{\cos }^3\theta }=\frac{W}{1+2{\left(\frac{l_1}{l_2}\right)}^3} \end{gathered}$$