As title suggests, the goal is to find $\alpha$ from the given $\triangle ABC$ with some given angles and sides. I'm posting this here to look for better solutions, as my own, which I'll post as well, is rather complicated.
Find angle $\alpha$ in $\triangle ABC$
euclidean-geometrygeometrytrianglestrigonometry
Related Solutions
So this'll be my approach. I'm going to add a brief explanation too.
Here's how I did it:
1.) Mark the $\triangle ABC$ with all the appropriate angles and mark the midpoint of segment $AC$ as $D$.
2.) Connect point $B$ with midpoint $D$, because $D$ is circumcenter of $\triangle ABC$, $BD=AD=DC$. Locate point $F$ outside $\triangle ABC$ and join it with points $A$ and $B$ such that $AF=BD=AD=DC$ and $\angle BAF=15$. Thus implies that $\triangle ABF$ is congruent to $\triangle CED$ via the SAS property. This implies that $\angle AFB =\alpha$.
3.) Notice that $\angle FAD=60$. Connect $F$ and $D$, via $FD$, because $\triangle FAD$ becomes equilateral, this shows that $AF=BD=AD=FD=DC$. Notice also that $\triangle ADB$ is isosceles, implying $\angle ADB=30$. This also means that $\angle BDF=30$ and $\triangle BDF$ is isosceles as well congruent to $\triangle ADB$. This further implies that $\angle \alpha= \angle DFB-\angle DFA$. $\alpha=75-60$, therefore $\alpha=15$.
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
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Best Answer
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 $\alpha = 10^{\circ}$.