Resolving horisontally to the right
\displaystyle \begin{align} & {{T}_{2}}-{{T}_{1}}\cos {{40}^{\circ }}=0 \\ & \\ & {{T}_{2}}={{T}_{1}}\cos {{40}^{\circ }}=762\times 0\textrm{.}766=584\ \text{N} \\ \end{align}
