[Kenny C.]

Ok, I'm having a brain lapse right now. Here are my two problems I am having with an honors assignment. They are simple and I could do them under most conditions except with the fact that this is due tommorow or I get no credit for all the honors things I have done over the semester.

a)There are two adjacent angels, KJL and LJM, that make up angle KJM. Knowing that, how would I solve this? (FYI /_ is the angle symbol. It's as close as I can get)

If 3(m/_LJM) = 2(m/_KJL) and m/_KJM = 140*, find m/_KJL.

I know how do this but my brain isn't functioning right now. Help would be nice.

b)There are two adjacent angels, KJL and LJM, that make up angle KJM.

If m/_KJM = (2x2 + 6x – 5)*, m/_KJL = 85*, and m/_LJM = (x2 + 1)*, find x

FYI the 2's after the x's are supposed to be superscripts. (ex. 2xsquared)

Help would be beautiful.

[75cents]

okay so lets say that KJL is y and LJM is x (for simplicities sake)

1) look at the givens
a) 3x=2y
b) x + y = 140
2) make part b so it is all in one term. You can choose either x or y.
I'll choose x.

1)3x=2y
2) then divide both sides by 2, now we have that y = (3/2)x

3) now plug your value for y in terms of x into part b
3) x + (3/2)x = 140
To make things simpler multiply each term by 2
4) 2x + 3x = 280
5) 5x = 280
6) x= 56
To find angle y we subract 56 from 140 and get 84
BLAMO

Well I assissted with the first, but I'll let you struggle through the second.

[75cents]

Oh, and dude/dudette, it helps to draw out the scenario.
Draw out the second one and write the variables into their perspective places.
You know what both angles added together equal.
So, now think about it. Look at your drawing. You should be able to figure it out from there.

[Kenny C.]

Thanks so much for the help here and on AIM!

[twinner2]

dont you just LOVE geometry