For the $n$ direction, $\theta =+{50}^{\circ }$. Therefore, the normal strain ${\epsilon }_n$ is computed in the following manner:

$$\begin{gathered} {\epsilon }_n={\epsilon }_x {\cos }^2\theta +{\epsilon }_y {\sin }^2\theta +{\gamma }_{xy} \sin \theta \cos \theta \\ =\left(900\mu \right) {\cos }^2\left({50}^{\circ }\right)+(-600\mu ) {\sin }^2\left({50}^{\circ }\right) \\ +(-850\mu ) \sin \left({50}^{\circ }\right) \cos \left({50}^{\circ }\right) \\ =-398.78 \mu m/m \end{gathered}$$