$\triangle ABC$ 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 $100 ^\circ$.
Best Answer
$\angle ABC=\angle ABD+\angle DBC=80^\circ$.
\begin{align*} AB&=BC\\ \implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ. \end{align*}
Erect an equilateral triangle $ACE$ on base $AC$. Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$
$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus $$\angle DEC=\angle DBC=20^\circ.$$
\begin{align*} \angle ECB&=\angle ECA-\angle BCA=10^\circ\\ \implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC. \end{align*}
Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus
\begin{align*} \angle CAD&=\angle DAE=30^\circ\\ \angle BAE&=\angle CAE-\angle CAB=10^\circ\\ \implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\ \implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ. \end{align*}