A community for students.
Here's the question you clicked on:
 0 viewing
richyw
 4 years ago
Was trying to help someone with a highschool geometry problem. Now I am stuck myself
richyw
 4 years ago
Was trying to help someone with a highschool geometry problem. Now I am stuck myself

This Question is Closed

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0This ought to be interesting

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0And what exactly were you asked to do with that?

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0given that \(m\angle Q=42^{\circ}\), and the line NQ bisects \(\angle MNP\), and the line PQ bisects \(\angle MPR\) Find \(m\angle M\)

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0not sure if it's the notation that is throwing me off?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352873934959:dw

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0the internal angle sum of a triangle is 180° the internal angle sum of a quadrilateral is 360°

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0are the lines QM and RN parallel by any chance?

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0I wasn't sure, but I tried under that assumption.

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0oh the answer is 84 by the way. This isn't what I got

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0yup just tried again. is anyone actually trying this?

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0i dont think those lines could be parallel

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0that 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.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352875825955:dw

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352876464319:dw More assumptions

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0I have to give up! If someone figures it out please post it. I really don't know.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The 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....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What's the answer? 84? if it's 84 then its parallel...

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0how did you work that out?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1352887526794:dw

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0@Shadowy, your logic appears to be flawed: 2(42) + 2x = 180? How?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well,since \(\Delta \)MPN is an isosceles triangle, angle MPN=angle MNP, so at the point P, y=42, 2y+2x=180.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Bro, there's no evidence that suggests MPN is isosceles

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.