TITLE: Geometry Problem on $\triangle ABC$ and Angle Chasing
QUESTION [5 upvotes]: $\triangle ABC$ is an isosceles triangle with $AB=BC$ and $\angle ABD=60^{\circ}$, $\angle DBC=20^{\circ}$ and $\angle DCB=10^{\circ}$. Find $\angle BDA$.

My approach: Let $\angle BDA=x$. Let $AB=BC=p$. Applying sine law in $\triangle ADB$, $\dfrac{p}{\sin x}=\dfrac{BD}{\sin (60+x)}$. Applying sine law in $\triangle BDC$, $\dfrac{p}{\sin150^{\circ}}=\dfrac{BD}{\sin 10^{\circ}}$. Using the two equations, we get $\dfrac{1}{2\sin 10^\circ}=\dfrac{\sin x}{\sin (60^\circ +x)} \implies 2\sin 10^\circ=\dfrac{\sqrt{3}}{2}\cot x + \dfrac{1}{2} \\ \implies x = \text{arccot} \left(\dfrac{4\sin 10^\circ-1}{\sqrt{3}}\right)$. 

Now I am stuck. I know that the answer is $100^\circ$ but no matter how hard I try I cannot seem to simplify it any further. Please help. If anybody has a better solution (involving simple Euclidean Geometry), I would be grateful if you provide it too.
Edit: I am extremely sorry. The original problem was when $AB=BC$. Sorry for the inconvenience caused. I have rectified my mistake. Also, I have changed the answer to $100 ^\circ$.

REPLY [4 votes]: Let $E$ be the circumcenter of $BCD$. Then $\angle BED=2\angle BCD=20^\circ$ and $\angle DEC =2\angle DBC =40^\circ$. Hence $\angle BEC=60^\circ$. This and $BE=EC$ shows that $BEC$ is equilateral. So $BC=BE$ and $\angle CBE=60^\circ$. By assumption $AB=BC$, so $AB=BE$ and $$\angle BEA = 90^\circ -\frac 12 \angle ABE =90^\circ -\frac 12 \cdot 140^\circ =20^\circ =\angle BED.$$ Therefore $A,D,E$ are collinear and we find $$\angle BDA =180^\circ -\angle EDB = \angle BED+\angle DBE= 20^\circ+80^\circ =100^\circ.$$