Hint to find angle $\hat{C}$ excuse me ! I put right picture ...sorry $\hat {D}=150$ my typing was wrong $105$
I need some hint to find the angle $\hat{C}$

All we know is that $$AB=DA=DC\\\hat{D}=150$$
I get stuck to find $CB$ or angle $\hat{C}$

Drawing the diagonal $BD$, we find that $BD=a\sqrt{2}$ and $CDB=105^{\circ}$. We can use the Cosine Rule to find $BC$:
$$BC^2=a^2+(a\sqrt{2})^2-2a^2\sqrt{2}\cos105$$
Thus, $BC=a\sqrt{2+\sqrt{3}}$.
We can now use the Cosine Rule again to find $\hat{C}$:
$$\hat{C}=\cos^{-1}\left(\frac{a^2+a^2(2+\sqrt{3})-2a^2}{2a^2\sqrt{2+\sqrt{3}}}\right)=\cos^{-1}\left(\frac{1+\sqrt{3}}{2\sqrt{2+\sqrt{3}}}\right)= 45^{\circ}$$
