In triangle $ABC$, $AA_1$ and $BB_1$ are bisectors of angles $\angle BAC$ and $\angle ABC$, respectively. Find $\angle BAC, \angle ABC,$ and $\angle ACB$ if $\angle AA_1B_1=24^\circ$ and $\angle BB_1A_1=18^\circ$.
I found that angle $ACB=96^\circ$, but I don't know how to continue.
Let $AA_1\cap BB_1=\{I\},$ $A_1B_1\cap CI=\{D\}$ and $\measuredangle DA_1C=x$.
Thus, $$18^{\circ}<x=18^{\circ}+\frac{\beta}{2}<18^{\circ}+\frac{84^{\circ}}{2}=60^{\circ}$$ and by the law of sinuses we obtain: $$\frac{ID}{DC}=\frac{\frac{ID}{DA_1}}{\frac{DC}{DA_1}}=\frac{\frac{\sin24^{\circ}}{\sin(108^{\circ}-x)}}{\frac{\sin{x}}{\sin48^{\circ}}}=\frac{\sin24^{\circ}\sin48^{\circ}}{\sin(108^{\circ}-x)\sin{x}}.$$ Also, we have: $$\frac{ID}{DC}=\frac{\frac{ID}{DB_1}}{\frac{DC}{DB_1}}=\frac{\frac{\sin18^{\circ}}{\sin(30^{\circ}+x)}}{\frac{\sin(84^{\circ}-x)}{\sin48^{\circ}}}=\frac{\sin18^{\circ}\sin48^{\circ}}{\sin(30^{\circ}+x)\sin(84^{\circ}-x)}.$$ Id est, $$\frac{\sin24^{\circ}}{\sin(108^{\circ}-x)\sin{x}}=\frac{\sin18^{\circ}}{\sin(30^{\circ}+x)\sin(84^{\circ}-x)}$$ or $$\sin24^{\circ}(\cos(54^{\circ}-2x)-\cos114^{\circ})=\sin18^{\circ}(\cos(108^{\circ}-2x)-\cos108^{\circ})$$ or $$\sin24^{\circ}\cos(54^{\circ}-2x)-\sin18^{\circ}\cos(108^{\circ}-2x)=\sin24^{\circ}\cos114^{\circ}-\sin18^{\circ}\cos108^{\circ}$$ or $$(\sin24^{\circ}\cos54^{\circ}-\sin18^{\circ}\cos108^{\circ})\cos2x+(\sin24^{\circ}\sin54^{\circ}-\sin18^{\circ}\sin108^{\circ})\sin2x=$$ $$=-\sin^224^{\circ}+\sin^218^{\circ}.$$ Now, $$\sin24^{\circ}\cos54^{\circ}-\sin18^{\circ}\cos108^{\circ}=\frac{1}{2}(\sin78^{\circ}-\sin30^{\circ}-\sin126^{\circ}+\sin90^{\circ})=$$ $$=\frac{1}{2}(\sin78^{\circ}+\sin30^{\circ}-\sin54^{\circ})=\frac{1}{2}(2\sin54^{\circ}\cos24^{\circ}-\sin54^{\circ})=$$ $$=\sin54^{\circ}(\cos24^{\circ}-\cos60^{\circ})=2\cos36^{\circ}\sin18^{\circ}\sin42^{\circ}=\frac{1}{2}\sin42^{\circ}.$$ Also, by the similar way prove that $$\sin24^{\circ}\sin54^{\circ}-\sin18^{\circ}\sin108^{\circ}=\frac{1}{2}\tan6^{\circ}\sin42^{\circ}$$ and $$-\sin^224^{\circ}+\sin^218^{\circ}=-\sin6^{\circ}\sin42^{\circ}.$$ Thus, we need to solve $$\cos2x+\tan6^{\circ}\sin2x=-2\sin6^{\circ}$$ or $$\cos(2x-6^{\circ})=-\sin12^{\circ}$$ or $$\cos(2x-6^{\circ})=\cos102^{\circ}$$ and since $18^{\circ}<x<60^{\circ},$ we obtain $$x=54^{\circ}$$ and from here $$\alpha=12^{\circ}$$ and $$\beta=72^{\circ}.$$.
Let's denote the intersection of bisectors as M. We use following statement:
In triangle ABC the obtuse angle $<AMB=\alpha$ between the bisectors of BAC and CBA is :
$$\alpha=90^o+\frac{<ACB}{2}$$.
$$<BMA_1=18^o+24^o=42^o$$
$$\alpha=180^o-42^o=138^o$$
$$<BCA/2=138-90=48$$
⇒ $$<BCA= 96^o$$
That is, for certain values of $<BB_1A_1$ and $<AA_1B_1$ angles $\alpha$ and $<ACB$ are constant. What may vary are positions of point $A_1$ and $B_1$ on sides $BC$ and $AC$. In a particular position where $A_1B_1$ is parallel with $AB$, we have:
$<BB_1A_1=<BAA_1=24^o$ ⇒ $<BAC=2\times 24=48^o$
$<ABB_1<BB_1A_1=18^o$ ⇒ $<ABC=2\times 18=36^o$
If the position of $A_1$ and $B_1$ varies ($A_1B_1$ rotates) $B_1M$ and $A_1M$ remain bisectors of new angles $<ABC$ and $<BAC$ due to above statement, hence there can be numerous values for angles $<ABC$ and $<BCA$.
This is ISL'92, so you may google it, basically on this site: https://artofproblemsolving.com/. I have a different solution, my own, but I am not allowed to post it, due to copyright. OK, this is the solution from the site and, if I am not wrong, the official one:
Take $M$ reflection of $D$ about $CE$, $N$ the one of $E$ about $BD$; both belong to $BC$. Take $\{\ R\ \}\in ME\cap BD$. Clearly $\widehat{DEN}=66^\circ, \widehat{DEM}=36^\circ\implies \widehat{MEN}=30^\circ$, i.e. $\widehat{BRE}=\widehat{NRB}=\widehat{MRN}=60^\circ, RN$ is external bisector of $\angle DRM$; also, since $\widehat{MDE}=72^\circ, \widehat{RDE}=\widehat{NDR}=24^\circ\implies\widehat{MDN}=24^\circ$ and $N$ is $D-$excenter of $\triangle DMR$, so $CN$ is external bisector of $\widehat{RMD}=72^\circ$, i.e. $\widehat{DMC}=\widehat{NMR}=54^\circ$, that is, $\widehat{DCM}=72^\circ$, a.s.o.
