If the measure of an ext. angle drawn at vertex M is....

Monkeyeatbutt

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Aug 12, 2006
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I'm having all-around trouble with this question:

If the measure of an exterior angle drawn at vertex M of triangle LMN is x, then m < L + m < N is what?

At first, I put "x", but I'm not sure that this is right. I have to show work, and I'm not sure how to work out the problem.
 
The measure of any exterior angle of a triangle is equal to the sum of the measures of the two opposite interior angles.
 
How does "m" relate to the triangle? ("M" is given as a vertex, so it's a point and "<M" would indicate the angle at that vertex, and presumably "m" would then be the side opposite. This is standard notation. But you seem to be adding a side to an angle, which doesn't make sense.)

Also, how does "X" relate? (In standard mathematical notation, "X" is not the same as "x", is why I ask.)

Thank you.

Eliz.
 
Re: Vertex

Hello, Monkeyeatbutt!

If the measure of an exterior angle drawn at vertex M\displaystyle M of ΔLMN\displaystyle \Delta\,LMN is x\displaystyle x,
then m( ⁣L)+m( ⁣N)  =  ?\displaystyle m(\!\angle L) \,+\,m(\!\angle N) \;= \;?
Code:
            N
            *
           /  \
          /     \
         /        \
        /           \
       /              \  x
      * - - - - - - - - * - - -
      L                 M

You're expected to know that an exterior angle of a triangle
    \displaystyle \;\;equals the sum of the two non-adjacent angles.

Therefore: L+N=x\displaystyle \,\angle L\,+\,\angle N\:=\:x

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If you didn't know that, you can still derive that fact . . .

In Δ<MN:  L+N+M=180o  \displaystyle \Delta <MN:\;\angle L + \angle N + \angle M\:=\:180^o\; [1]

We see that angles M\displaystyle M and x\displaystyle x are supplementary.
    \displaystyle \;\;That is: M+x=180o        M=180ox  \displaystyle \,\angle M + \angle x\:=\:180^o\;\;\Rightarrow\;\;\angle M \:= \:180^o - x\; [2]

Substitute [2] into [1]: L+N+(180ox)  =  180o\displaystyle \:\angle L\,+\,\angle N\,+\,(180^o\,-\,x)\;=\;180^o

Therefore: L+N  =  x\displaystyle \,\angle L\,+\,\angle N\;=\;x

 
M is the vertix of triangle LMN.

And I didnt know X and x were diffrent I just used caps on one and not the other. They should both be x
 
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