jmc

In $△ACT$, the three angles sum to $180^{\circ }$, and $\angle C=\angle T$ $\angle CAT+\angle ATC+\angle ACT=180$ $36+\angle ATC+\angle ATC=180$ $2\angle ATC=144$ $\angle ATC=72$ Since $\angle ATC$ is bisected by $\overline{TR}$, $\angle RTC=\frac{72}{2}=36$ Now focusing on the smaller $△RTC$, the sum of the angles in that triangle is $180^{\circ }$, so: $\angle RTC+\angle TCR+\angle CRT=180$ $36+\angle ACT+\angle CRT=180$ $36+\angle ATC+\angle CRT=180$ $36+72+\angle CRT=180$ $\angle CRT=72^{\circ }$, giving the answer $\boxed{72^{\circ }}$