We claim the answer is $90^{\circ}$. First, we make the following claim: Claim: $TY\perp BC$. Proof: [asy][asy] size(10cm); pair A = dir(150); pair C = dir(330); pair B = dir(210); pair Y = dir(270); pair X = dir(180); pair T = (0,0); draw(circumcircle(B,T,C)); draw(unitcircle); draw(A--B--C--A--cycle); draw(X--T--Y); pair P = 2*circumcenter(B,T,C)-T; draw(T--P); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$T$",T,dir(45)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(315)); dot("$P$",P,dir(P)); [/asy][/asy] We note that since $\angle BTY=\angle YTC$, this means that $TY$ passes through the midpoint of $\widehat{BC}$ in $(BTC)$. Since $TY$ passes through the midpoints of the arcs $\widehat{BC}$ in both circles, this means that $TY$ is the perpendicular bisector of $BC$. $\square$ Similarly, $TX\perp AB$, and since $\angle XTY=90^{\circ}$ from the two angle bisector conditions, it follows that $AB\perp TX\perp TY\perp BC$, hence $AB\perp BC$.

Since $TX$ angle bisector of $\angle{ATB}$ and $X$ midpoint of $AB$ then $AT=BT$. Hence $\angle{TAB}=\angle{TBA}$. Since $TY$ angle bisector of $\angle{CTB}$ and $Y$ midpoint of $CB$ then $CT=BT$.Hence $\angle{TCB}=\angle{TBC}$. We have $180^\circ=\angle{TAB}+\angle{TBA}+\angle{TCB}+\angle{TBC}$ $\angle{B}=90^\circ$