1. ## Help!!

ok this kind of problem would usually seem easy to me but for some reason i cant get it: If mAB= 4x+1 mBC=3(x-3) mAC=6x+1 then determin each: mAB, mBC, and mAC. If anyone could give me help on this it would be greatly appreciated as it is due tomorow lol. Thank you!

2. Hello, luvjonandmunky!

We need a lot more information . . .

If $\displaystyle m(AB) \,= \,4x+1,\;m(BC) \,= \,3(x-3),\;m(AC)\,=\,6x+1$
then determine: $\displaystyle m(AB),\;m(BC),\;m(AC).$
Can we assume that we have a triangle
. . and those are the lengths of the three sides?

Is $\displaystyle x$ supposed to be an integer?

Is there some relationship between the three sides?

3. Originally Posted by luvjonandmunky
ok this kind of problem would usually seem easy to me but for some reason i cant get it: If mAB= 4x+1 mBC=3(x-3) mAC=6x+1 then determin each: mAB, mBC, and mAC. If anyone could give me help on this it would be greatly appreciated as it is due tomorow lol. Thank you!
I'm guessing, based on the looks of the notation that this is a circle. If so, Arc AB + Arc BC = Arc AC so
4x+1+3(x-3)=6x+1
7x-8=6x+1
x-8=1
x=9

So mAB=37, mBC=18 and mAC=55.

Dave