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 $\hat{C}$

Hint. Draw the diagonal $BD$. Find $BC$ using:


*

*The Theorem of Pythagoras in $\triangle ADB$, which gives $BD=a\sqrt{2}$.

*The fact that $\angle ADB = \left(\frac{180-90}{2}\right)^{\text{o}} = 45^{\text{o}}$. Hence $\angle BDC=105^{\text{o}}$.

*Using the law of cosines, $BC^2=DB^2 + DC^2 - 2\cdot DB\cdot  DC\cos(105)$. Thus:$$BC=\sqrt{2a^2+a^2-2a^2\sqrt{2}\cdot\cos(105)}=\sqrt{3a^2-2a^2\sqrt{2}\cdot\frac{\sqrt{2}-\sqrt{6}}{4}}$$
$$BC=\sqrt{a^2\left(3-\frac{2-\sqrt{12}}{2}\right)}=\sqrt{a^2\left(2+\sqrt{3}\right)}=a\sqrt{2+\sqrt{3}}$$

*You can simply apply the law of cosines again to find the exact value of $\angle C$. Can you continue from here?
