As title suggests, compute angle $\alpha$ from the given triangle.
I'm going to post my own approach here, please post your own answers as well, both geometric and trigonometric, I'm curious to see the different methods to solve this.
Find Angle $\alpha$ from the triangle
euclidean-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
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Best Answer
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 $\alpha=15$.