check this you smart people

[snakeeyes1000]

hey I've been messing around lately and I was playing with a few triangles when I came up with the formula(or did it already exist?)

For any trianlge ABC,



AB^2 = BC^2 = AC^2
m<C m<A m<B




Does that work? I am almost certain it will. it's not that great, I know, I'm just curious as to whether it would work. it looks right - but I'm not sure. thanks for your help.

[Yonatan]

Which lines are m<A, m<B, m<C?
If they're the angle bisectors, that seems incorrect at a first glance, but I might be wrong.

[bugzpodder]

first i gotta figure out what you are talking about

AB^2
------
m<C


the top i take it is length of side a (opposite of angle A) times length of side b squared
and the bottom is the measure of angle C in radians??

[bugzpodder]

hey snakeeyes what happened to your chinese snake symbol that i remember seeing?

[snakeeyes1000]

sorry for the confusion. triangle ABC has three verteces(Sp?): A, B, and C.

AB^2 is the length of the segment between points A and B raised to the second power. BC^2 is the same for points B&C, and AC^2 for A and C.

m<A stands for the measure of angle A. m<B is for B, m<C is for C. they are not angle bisectors. what do you guys think?

btw bugz, you must be thinking of someone else, I've never had one.

[Yonatan]

Ah, in that case it is definitely wrong. :o

The value of the angle (especially in degrees) itself doesn't say much about the ratio between lines and areas in a triangle.
Trigonometric functions applied to the angles do.

[snakeeyes1000]

try it though. i've not found a single triangle it won't work for. i've tried scalene, regular, isoceles, right, acute, and obtuse - but to no avail. it works with them all. I just don't know why. try it man - would it work for a 30/60/90 degree right triangle? what about equilateral. and 45-45-45? all of them. Just try before you pronounce judgement(oh and btw it doesn't matter whther they are in degrees or radians, since it would remain proportional - you could just have easily done the measure of the angles in radians or degrees)

[Yonatan]

Quick counter example:

I used a right triangle to simplify the trigonometry.

AB = 3, BC = 4, AC = 5
<B = 90
<A = arctan(4/3) = ~53.130102354155978703144387440907

AC^2 / <B = 25 / 90 = 0.2777777....
BC^2 / <A = 16 / <A = ~0.30114754707880658181785543011763

Pretty close, but not equal.

Btw, the "closeness" of the ratios comes from the use of degrees. Degrees are typically "large" numbers. Had you used radians, you would probably have not as close ratios.

[Yonatan]

Angles can't represent ratios. They have measuring units - "degrees", "radians", and other less important ones.
Results of trigonometric functions don't have measuring units. The sine of 90 degrees is just 1, not 1 of some unit. This is because they're used as ratios.

[bugzpodder]

snakeeyes first tell me, are you talking about the measurement of angles in degrees or radians? and why don't you give an example of 30-60-90 triangle like you mentioned?

[Kalkewl8ter]

snakeeyes1000,

How did you come up with a 45-45-45 triangle?

[bugzpodder]

LOL!!! :D :D :D

[A$$Bandit]

You need to use the bog standard triangle labelling rules:

3 vertices, A B and C. The side opposite A is labelled a, the side opposite B is labelled b, and the side opposite C is labelled c. That way, the line AB is 'c', and the angle CAB is 'A', etc. Now what you originally said comes to:

a2/A = b2/B = c2/C

Compare that with:

a/sinA = b/sinB = c/sinC

square all of those and u get

a2/sin2A = b2/sin2B = c2/sin2C.

Now they can't both be true, and I'm inclined to stick with what mathematicians have been using in triangle problems for hundreds of years.

That proves one of them's wrong, and combined with the proof that the old one is right, that proves your forumla is incorrect.

Sorry!

[stewie]

i feel really stupid because i actualy have no idea at all what you guys are on about

[Yonatan]

Simple trigonometry? :(
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