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 $\hat{C}$
 A: 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 $\hat{C}$:
$$\hat{C}=\cos^{-1}\left(\frac{a^2+a^2(2+\sqrt{3})-2a^2}{2a^2\sqrt{2+\sqrt{3}}}\right)=\cos^{-1}\left(\frac{1+\sqrt{3}}{2\sqrt{2+\sqrt{3}}}\right)= 45^{\circ}$$
A: 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?
A: Elementary proof...
Drop a vertical line $CE$ onto $AB$. Notice that $E$ is in the middle of $AB$. (This is due to the $150°$ angle - the $\triangle ACD$'s height on base $AD$ is $a/2$). Thus, $\triangle ABC$ is isosceles, and knowing $\angle CAB$ is $75°$ (the right angle minus $\angle CAD$, which is $15°$), one concludes that $\angle ACB$ is $30°$, which adds to $\angle ACD$ of $15°$ to give you $\angle C$ of $45°$.
A: It is only the cosine rule to use:

