Answer:
[tex]x=16[/tex]
Step-by-step explanation:
[tex]We\ are\ given:\\PNM\ is\ a\ triangle\\Point\ P\ belongs\ to\ or\ lies\ on\ the\ line\ MQ.\\Now,\\We\ observe\ that,\\\angle QPN\ and\ \angle MPN\ form\ a\ linear\ pair\ and\ hence,\ are\ supplementary.\\Hence,\\\angle QPN+ \angle MPN=180\\Hence,\\\angle MPN=180 -\angle QMN\\ \angle MPN=180 -6x \\Now,\ lets\ consider\ the\ \triangle MPN\\From\ the\ Angle\ Sum\ Property\ Of\ A\ Triangle,\\'The\ sum\ of\ all\ the\ interior\ angles\ of\ a\ triangle\ is\ 180'\\Hence,\\\angle MPN + \angle NMP + \angle PNM=180\\[/tex]
[tex]Substituting\ \angle PMN=32, \angle PNM=64, \angle MPN=(180-6x),\\32+64+(180-6x)=180\\Hence,\\32+64-6x=180-180\\96-6x=0\\96=6x\\x=\frac{96}{6} \\x=16[/tex]
