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Was trying to help someone with a highschool geometry problem. Now I am stuck myself

Mathematics
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This ought to be interesting
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And what exactly were you asked to do with that?

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Other answers:

given that \(m\angle Q=42^{\circ}\), and the line NQ bisects \(\angle MNP\), and the line PQ bisects \(\angle MPR\) Find \(m\angle M\)
not sure if it's the notation that is throwing me off?
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the internal angle sum of a triangle is 180° the internal angle sum of a quadrilateral is 360°
are the lines QM and RN parallel by any chance?
I wasn't sure, but I tried under that assumption.
oh the answer is 84 by the way. This isn't what I got
yup just tried again. is anyone actually trying this?
im trying
i dont think those lines could be parallel
that is what I figured as well. so I attempted by just assuming that N,P and R all lie on one line. But then I can't figure out if there is a solution.
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|dw:1352876464319:dw| More assumptions
I have to give up! If someone figures it out please post it. I really don't know.
The question seems to be missing something. Is there a length somewhere? Or maybe it's cylic? It's brings me to a wild goose chase....
What's the answer? 84? if it's 84 then its parallel...
how did you work that out?
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@Shadowy, your logic appears to be flawed: 2(42) + 2x = 180? How?
Well,since \(\Delta \)MPN is an isosceles triangle, angle MPN=angle MNP, so at the point P, y=42, 2y+2x=180.
Bro, there's no evidence that suggests MPN is isosceles
If and only if the MQ and NP is parallel, then as shown above, QM=MP=MN. And the answer happens to be 84. So this assumption is right.

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