The normal strain along diagonal BD is computed from:

$$\begin{gathered} {\epsilon }_{BD}={\epsilon }_x {\cos }^2 \theta + {\epsilon }_{y }{\sin }^{2 }\theta + {\gamma }_{xy} \sin \theta \cos \theta \\ =(-700\mu ) {\cos }^{2 }(-26.{565}^{\circ }) + (-500\mu ) {\sin }^{2 }(-26.{565}^{\circ }) \\ + \left(900\mu \right) \sin (-26.{565}^{\circ }) \cos (-26.{565}^{\circ }) \\ = -1,020.03 \mu m/m \end{gathered}$$

Finished