For the $t$ direction, the angle from the $x$ reference axis is $\theta =+{50}^{\circ }+{90}^{\circ }={140}^{\circ }$. Use this value to compute the normal strain ${\epsilon }_t$:

$$\begin{gathered} {\epsilon }_t={\epsilon }_x {\cos }^2\theta +{\epsilon }_y {\sin }^2\theta +{\gamma }_{xy} \sin \theta \cos \theta \\ =\left(900\mu \right) {\cos }^2\left({140}^{\circ }\right)+(-600\mu ) {\sin }^2\left({140}^{\circ }\right) \\ +(-850\mu ) \sin \left({140}^{\circ }\right) \cos \left({140}^{\circ }\right) \\ =+698.78 \mu m/m \end{gathered}$$