Substitute ${\epsilon }_x=+1,250\mu \epsilon$ into the strain transformation equations for sides AB and BC.

$$\begin{gathered} +1,000\mu \epsilon =(+1,250\mu \epsilon ) {\cos }^2(53.{13}^{\circ })+{\epsilon }_y {\sin }^2(53.{13}^{\circ }) \\ +{\gamma }_{xy}\sin (53.{13}^{\circ })\cos (53.{13}^{\circ }) \end{gathered}$$

and

$$\begin{gathered} +1,000\mu \epsilon =(+1,250\mu \epsilon ) {\cos }^2(126.{87}^{\circ })+{\epsilon }_y {\sin }^2(126.{87}^{\circ }) \\ +{\gamma }_{xy}\sin (126.{87}^{\circ })\cos (126.{87}^{\circ }) \end{gathered}$$