Side AC is aligned with the $x$ axis; therefore, ${\theta }_{AC}=0^{\circ }$

Write a strain transformation equation for side AC:

$+1,250\mu \epsilon ={\epsilon }_x {\cos }^2\left(0\right)+{\epsilon }_y {\sin }^2\left(0\right)+{\gamma }_{xy} \sin \left(0\right)\cos \left(0\right)$

Since $\sin \left(0\right)=0$ and $\cos \left(0\right)=1$, this equation simplifies to:

Therefore, the normal strain in the $x$ direction can be found directly from the original and deformed lengths of side AC.

The initial length of side AC is $360 mm$ and the deformed length is $360.45 mm$. The normal strain along side AC is therefore:

$$\begin{gathered} {\epsilon }_{AC}=\frac{360.45-360.00}{360.00} \\ =+0.00125 mm/mm \\ =+1,250 \mu \epsilon \end{gathered}$$