In the diagram, $P A D B, T B C$ and $N C D$ are straight line segments. If $\angle T B D=110^{\circ}$, $\angle B C N=126^{\circ}$, and $D C=D A$, determine the measure of $\angle P A C$.

![](https://cdn.mathpix.com/cropped/2024_04_30_3011c729c030a190170fg-1.jpg?height=398&width=677&top_left_y=1189&top_left_x=781)
Since $\angle T B D=110^{\circ}$, then $\angle D B C=180^{\circ}-\angle T B D=180^{\circ}-110^{\circ}=70^{\circ}$.

Since $\angle B C N=126^{\circ}$, then $\angle D C B=180^{\circ}-\angle B C N=180^{\circ}-126^{\circ}=54^{\circ}$.

Since the sum of the angles in $\triangle D B C$ is $180^{\circ}$, then $\angle B D C=180^{\circ}-70^{\circ}-54^{\circ}=56^{\circ}$.

Now $\angle B D C$ is an exterior angle of $\triangle A D C$.

Thus, $\angle B D C=\angle D A C+\angle D C A$.

But $\triangle A D C$ is isosceles with $\angle D A C=\angle D C A$.

Thus, $\angle B D C=2 \angle D A C$.

Since $\angle B D C=56^{\circ}$, then $\angle D A C=\frac{1}{2}\left(56^{\circ}\right)=28^{\circ}$.

Therefore, $\angle P A C=180^{\circ}-\angle D A C=180^{\circ}-28^{\circ}=152^{\circ}$.

ANSWER: $152^{\circ}$