We have $\triangle ABC$ and we must find $PM$

Question

Here we have $\triangle ABC$ and some information related: $$\angle A=60^\circ$$ $$\angle B=45^\circ$$ $$AC=8$$ $$CM=MB$$ The vertical line $PM$ is perpendicular to $BC$. So now we want to calculate $PB$. $$PB=?$$ I have tried some different ways to calculate it. But it was unsuccesfully not true. What I tried was to calculate $\angle C=75^\circ$ and dividing the $\triangle ABC$ into 4 pieces and try to find the side $PM$.like this:

Do you have any ideas? Please explain your answer briefly.

Answer 1

Guide:

I would construct line $PC$ and show that $\angle APC$ is $90^\circ$.

After that I can use trigonometry to obtain $PC$ and I can use trigonometry one more time to obtain the length of $PM$.

Edit:

$PC$ is the line connecting $P$ to $C$. Notice that we have $PM = CM$. Then, we can compute $\angle ACP = 75^\circ - 45^\circ=30^\circ$. Hence $\angle APC = 90^\circ$. Hence $PC= AC \cos 30^\circ$.

Answer 2

With the information of the second picture, you can calculate $CH$ and $PS$.

If you know $PS$, you know $HM$. Than $HM + CH = MB$

And $MB = PM$

Knowing $PM$ and $MB$ you can calculate $PB$