Finding angle in a given triangle. 
In the picture above:

  
*
  
*$\overset{\Delta}{ACD}$ is a triangle.
  
*$B$ is a point on $[CD]$.
  
*$m(\widehat{ABC})=140^\circ$
  
*$|AB|=|BC|$.
  
*$|AC|=|BD|$.
  
*What is $\color{red}{m(\widehat{ADB})}$?
  

There is probably a short answer, but i can't find it.
[Answer is $\color{red}{30^\circ}$.]
 A: Obviously, $\angle BAC = \angle BCA = 20^{\circ}$, and $\angle DBA = 40^{\circ}$.
Choose point $E$ such that $\triangle EAB$ forms an equilateral triangle.
Since $\angle DBA = 40^{\circ}$, and $\angle ABE = 60^{\circ}$, then $\angle DBE = 20^{\circ}$.
Now notice: $\overline {AC} = \overline {DB}$; $\angle ACB = \angle DBE = 20^{\circ}$; and $\overline {CB} = \overline {BE}$.  We have side-angle-side equivalence, so, if we draw in segment $DE$, we can say that $\triangle ACB \cong \triangle DBE$.
Hence, $\angle DEB = 140^{\circ}$.  Since $\angle AEB = 60^{\circ}$, then $\angle DEA = 80^{\circ}$.  Since side $\overline {DE} = \overline {EA}$ in $\triangle DEA$, then $\angle EDA = \angle EAD = 50^{\circ}$.
Now, $\angle EDA$ ($50^{\circ}$) - $\angle EDB$ ($20^{\circ}$) = $\angle CDA$ ($30^{\circ}$).
A: Hint :
$$sin(140-A)=sin(140)cos(A)-cos(140)sin(A)=ksin(A)$$
with $k=\frac{1}{2sin(70)}$
can be transformed in
$$\frac{sin(140)}{k+cos(140)}=tan(A)$$
A: By Law of Sines ( using angles in degrees)
$$\dfrac {\sin D }{\sin A}= \dfrac {singlestripledline}{doublestripledline}=\dfrac {singlestripledline }{2\; halfline}=\dfrac {1}{2 \sin 70}$$
$$ 140 = D +A $$
$$ \dfrac{\sin D}{\sin (140-D) } = \dfrac{1/2}{\sin 110} =\dfrac{\sin 30}{\sin 110}$$
Compare arguments,
$ D= 30, 180 + 30, $ the latter angle is discarded.
