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looks to me like; a/2 + (180-b/2) + c = 180 would account for all the angles in one of those relevant triangles
how did you get a b c? I just have x
20/2 + 180-(x+60)/2 + x = 180 they are just generic variables to help me sort thru it; i found them on the keyboard.
funny. so is that how I solve it now? just 20/2=10+180 how do I divide with x and 60?
or can we just multiply 180 by 2?
there is a "thrm" that deals with this that i can never remember so I tend to have to build it anew by simpler ideas every time i do it :)
10 - x/2 + 30 + x = 0 40 = x/2 - x 40 = x(1/2 - 1) 40 = x(-1/2) -80 = x ; im gonna assume i made a sign error someplace so I have to rechk it
180-(x+60)/2 + 20/2 + x = 180 180-x/2+30 + 10 + x = 180 -x/2 + x = 180 - 10 - 30 - 180 -x/2 + x = -40 -x + 2x = -80 x(-1 + 2) = -80 x(1) = -80 .... well, thats what i get :)
how is it possible to be a negative number?? -_-
http://www.mathplanet.com/education/geometry/circles/advanced-information-about-circles lets see if this comes up better; it says the angle is equal to 1/2 (outter - inner)
x = 1/2 (x+60 - 20) x = 1/2 (x+40) 2x = x+40 x = 40
that sounds better, still dunno what was in error with my first thought tho
okay thank you. so for future problems like this, I should just set it up like this one?
having everything on opposite side of x?
its best to try to remember the thrm, which i can never do. but yes,
okay thank you.
I see my mistake; i forgot to carry the "-" thru the paranthesis. 180-(x+60)/2 + 20/2 + x = 180 ^ that thing there 180 -x/2 - 30 + 10 + x = 180 160 +x(-1/2 + 1) = 180 160 +x(1/2) = 180 x(1/2) = 20 x = 40