The answer is <ABC=90,<CAB=15 and <ACB=75.Now,let E be a point such that <CBE=<ABD and BE=DE,so from <CBD-<ABD=60 we have DBE is equilateral and by similar triangles ADB and BCE we have that ABCE is cyclic,so we have that <ADB=<BCE=180-<EAB,from which we have <EAB=30 and we have DB=DE and <EDB=60,so D is the center of ABCE,so AD=BD=CD,so we easy obtain angles,so we are finished.Although,an easy problem.

Call $\angle ABD = \alpha $. Hence $\angle CDB = \alpha+60^\circ; \angle ACB = 90^\circ-\alpha; \angle CAB = 30^\circ - \alpha$ By law of sines on triangle $ADB$ and $CDB$, we have: $$\frac{BC}{sin(30^\circ)} = \frac{BD}{sin(90^\circ - \alpha)}$$$$\frac{AB}{sin(30^\circ)}= \frac{BD}{sin(30^\circ-\alpha)}$$Multipling both and using the length condition: $$sin(30^\circ-\alpha)sin(90^\circ-\alpha) = \frac{1}{4}$$By simple computation we get that $\alpha = 15^\circ$ works and it is obvious that this is the only since the function $f(x) = sin(30^\circ-x)sin(90^\circ-x)$ is strictly decreasing in the interval $0 \leq x \leq 30^\circ$. So the internal angles are $(15^\circ, 75^\circ, 90^\circ)$

My solution at https://stanfulger.blogspot.com/2021/07/cono-sur-2011-httpsartofproblemsolvingc.html Best regards, sunken rock