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- Feb 14, 2012
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In the triangle ABC, we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, $\angle MCB=55^\circ$ and $\angle NBC=80^\circ$. Find $\angle NMC$.
$ \angle A=15^o$In the triangle ABC, we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, $\angle MCB=55^\circ$ and $\angle NBC=80^\circ$. Find $\angle NMC$.
Hi Albert, I want to thank you for participating in this problem but your answer doesn't match mine.$ \angle A=15^o$
using MC as a mirror we got triangle A' MC as a symetric image of triangle AMC
then it is very easy to see that :
$ \angle NMC=25^o=\angle A+(65^o - 55^o)$
Yes, you're right, I labeled the angle incorrectlyanemone:
Your diagram is not correct ,please check it
note :angle MCB=$55^o$,
but your diagram angle MCB=$10^o$
do it again ,I will post my solution later
Ah, but you don't know how much paper and messy equations he spent before giving a short answer.Your approach must be one of the smartest one because you solved it by observation but I approached it using lots of messy equations and therefore, my hat is off to you, Albert!
It's the angle at point B between the sides AB and BC.Maybe off topic but what does exemple $\angle ABC=100^\circ$ means?
Asking for clarification about a step taken or notation used is encouraged, as this is how we learn, and is not considered off-topic at all.Maybe off topic but what does exemple $\angle ABC=100^\circ$ means?