The question is as the title states. In the following figure, with some given angles and two sides, the goal is to find the measure of $\angle ABC$. This problem is inspired by one which appeared in a local Math contest in Turkey. I'll post my approach to this below as an answer! Please show your own methods of solving this! 
Geometry – Find $\angle ABC$ in $\triangle ABC$ with Given Angles and Segment Ratios
contest-matheuclidean-geometrygeometrytrianglestrigonometry
Related Solutions
I think my solution is quite simple. Since $\triangle BDC$ is isosceles, it follows that $\angle BCD = \angle CBD = 2 \alpha$, and as such, $\angle BCA = 3\alpha$. We then compute that $$ \angle ABD = 180 - \angle CBD - \angle BCA - \angle BAC = 180 - 2\alpha - 3\alpha - 7\alpha = 180 - 12 \alpha. $$ Because also $\triangle ABD$ is isosceles, we decude from this that $$ \angle BAD = \frac{180 - \angle ABD}{2} = 6\alpha. $$ In particular, we find that $\angle CAD = 7\alpha - 6\alpha = \alpha$. This implies that also $\triangle CDA$ is isosceles; whence $|AD| = |DC| = |DB| = |BA|$. It follows that $\triangle BDA$ is even equilateral and as such, its angles are all $60^{\circ}$. Hence $6\alpha = 60^{\circ}$, and thus $\alpha = 10^{\circ}$.
This is my own approach. I'll add a brief explanation too:
This is the procedure I followed:
1.) Label the triangle $ABC$ and mark all the appropriate angles and side lengths. Notice that since $\triangle ABC$ is obtuse, its circumcenter must lie outside of the triangle itself. Locate circumcenter of $\triangle ABC$ outside and call it point $E$. Connect all the vertices of $\triangle ABC$ to $E$ via $AE$, $BE$ and $CE$. Notice that this means $AB=BE=AE=CE=CD$ because $\triangle ABE$ is equilateral.
(for further explanation of above, point $E$ is a circumcenter of $\triangle ABC$, in that case, $\angle AEB$ must be twice the measure of $\angle ACB$ (inscribed angle theorem), which means it'll be $60$. Because $\angle AEB=60$ and know that $AE=BE$, it follows that $\triangle ABE$ is equilateral)
2.) Notice also that $\angle EAC=\angle ECA=6$. Connect point $D$ with $E$ via segment $DE$. Notice that because segment $CD=CE$ and $\angle ECD=36$, this implies that $\angle EDC=\angle DEC=72$. However, notice that $\angle EBD=96-60=36$, therefore, $\angle BED$ is also $36$. But this implies that segment $BD=DE$
3.) Lastly, notice that this means $\triangle ABD$ is congruent to $\triangle AED$ via the SAS property. This means $\angle BAD=\angle EAD=x$. This means that $2x=60$, therefore, $x=30$.
Related Question
- In $\triangle ABC$, $BD$ is a median, $\angle DAB=15$ and $\angle ABD=30$. Find $\angle ACB$.
- $\triangle ABC$ is a triangle with internal point $O$. Find $\angle x$.
- In the quadrilateral $ABCD$, find the measure of length $AB$.
- In Quadrilateral $ABCD$, $AB=BC=AC$, $DC=3$, $AC=5$ and $\angle ADC=60$. Find the measure of $BD$
- Two overlapping triangles $\triangle ABC$, $\triangle DBE$. Given some angles and sides, find the area of the Blue triangle $\triangle ABC$
- In $\triangle ABC$ with an internal point $D$, find the measure of $\angle DCA$
- Given a triangle $\triangle ABC$, with an internal point $K$, find $\angle AKC$.
Best Answer
Here's my approach. I'll post a brief explanation below!
1.) Draw a perpendicular from $D$ onto line segment $AD$ such that it meets at point $E$. Join point $E$ with $C$ via $EC$. Notice that $\triangle BED$ is a $30-60-90$ triangle, therefore, $ED=DC=a$, and $\angle EBD=\angle DEC=\angle DCE=30$. We also know that $BE=EC=a\sqrt3$.
2.) Note than $\angle DAC=\angle ACE=15$, therefore $BE=EC=AE=a\sqrt3$. However, this proves that $\triangle AEB$ is an isosceles right angle triangle, therefore $\angle ABE=45$, and thus, $\angle ABC=45+30=75$