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June 15th, 2014, 01:07 AM   #1
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June 15th, 2014, 09:49 AM   #2
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Chasing angles around the picture quickly yields that $\displaystyle \angle PCM = \theta$. Thus $\displaystyle PC=PM=BC$.

Without loss of generality let $\displaystyle AM=1$. Then $\displaystyle MC=1$, and $\displaystyle BC=PC=PM=2\sin \theta$

Writing the law of cosines on $\displaystyle \triangle PMC$ gives you that $\displaystyle \sin 2\theta=1/2$ and $\displaystyle \theta=15$.
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