Find degree of arc with four points on a circle

As shown in the figure below, $\angle CAP = \angle CBP = 13^\circ$

$\overset{\huge\frown}{MA}=38^\circ$, find the degree measure for the arc $\overset{\LARGE\frown}{BN}$

Due to the fact that $\angle{CAP}$ and $\angle{CBP}$ are the same, we can make another circle passing over points $A,C,P,B$.

Now, my initial thought is that we can name one of the angles we have to be $\theta$, and try to make up an equation that can find $\theta$,

I tried a few, for example in the new circle we made:

$$\angle{CAB}+\angle{CPB}=180^\circ$$

where I named $\angle{APB}=\theta$

But After subsituting I get:

$$... + 2\theta + ... - 2\theta = 180 $$

Indicating I cannot find $\theta$ with the equation.

Note: I am bad at LaTeX, so if there's anything weird in my LaTex code feel free to improve it.

Solution 1:

We have $\angle APC = \angle ACM - \angle CAP = 38^\circ - 13^\circ =25^\circ$. Let us draw the circle passing through $A,C,P,B$ with center $D$.

One way is to angle chase is $\angle ADC = 2\angle APC = 50^\circ$.

In isosceles $\triangle ADC$, we get $\angle ACD = 65^\circ$. So $\angle ACB=2\angle ACD = 130^\circ$.

Finally $\overset{\huge \frown}{BN}=180^\circ - 38^\circ - 130^\circ = 12^\circ$.

Another way to angle chase is as follows (without requiring the center of second circle, $D$). Since $ACPB$ is cyclic quadrilateral, $\angle CAB = \angle ABC = \angle APC = 25^\circ$. Hence $\overset{\huge \frown}{BN}= \angle PCB = \angle PAB = \angle CAB - \angle CAP = 25^\circ - 13^\circ =12^\circ$