Math Q

[ssjskipp]

In the pythagorean theorem (a^2 + b^2 = c^2), given 'c', how would I find 'a' and 'b'?

[Toast]

You can't, you need to know at least two of them.

Impossible.

[Craige]

At 1/30/06 02:37 PM, ssjskipp wrote: In the pythagorean theorem (a^2 + b^2 = c^2), given 'c', how would I find 'a' and 'b'?

because A^2 + B^2 = C^2, then B^2 = C^2 - A^2

[Craige]

At 1/30/06 02:42 PM, Dezmerkt wrote:

At 1/30/06 02:37 PM, ssjskipp wrote: In the pythagorean theorem (a^2 + b^2 = c^2), given 'c', how would I find 'a' and 'b'?

because A^2 + B^2 = C^2, then B^2 = C^2 - A^2

Ignore that. I missed the " 'a' and..." You do need to know 2 sides.

[ssjskipp]

righ,t so given two points (both ends of a line segment), being the hypotenuse, is there then a way to get A and B?

[Toast]

.... No, you need to know the values of at least two segments.

[ssjskipp]

(0, 10) and (10, 0):
would be like
a:10
b:10
c:14.14-
righT?

[PONGpaddle]

There are many different possible answers when given one side. You cannot however get an exact answer without two sides.

[ssjskipp]

yeah, I know, but the point is, I'm trying to find the 'a' and 'b' of the line (given 'c' and two points [the points of 'c']), so if there's a point on (0, 10), and (10, 0), then I can find the 'a' and 'b' of that line (as if there were a triangle there), right?

[PONGpaddle]

At 1/30/06 03:07 PM, ssjskipp wrote: yeah, I know, but the point is, I'm trying to find the 'a' and 'b' of the line (given 'c' and two points [the points of 'c']), so if there's a point on (0, 10), and (10, 0), then I can find the 'a' and 'b' of that line (as if there were a triangle there), right?

You can find possible triangles, yes, but any one wouldn't necessarily be the 'right' answer.

[Toast]

You can't really do that...

Niet.

Nien.

Non.

ìà.

Null point!

[ssjskipp]

@TheDrunkMonkey, thank's that's what I was referring to
@-Toast- ^read

[Rustygames]

At 1/30/06 02:45 PM, ssjskipp wrote: righ,t so given two points (both ends of a line segment), being the hypotenuse, is there then a way to get A and B?

lol just subtract the x values and y values to get a and b you silly boy

At 1/30/06 02:49 PM, -Toast- wrote: .... No, you need to know the values of at least two segments.

ha ha ha. I though you were smart toast ; )

[Rustygames]

(to make this clearer)

if you have 2 points and you want to find the distance between them (the hypotenuse) then just subtract point A's x from point B's x. And that will give you side A. Then do the same for y to get side B. Then apply pythagoras to get side C^2. Then either sqrt or square the distance you are checking against.

Example:
point A (20,30)
point B (100, 50)
Side A = 20-100 = -80
Side B = 30-50 = -20
Side C = Sqrt(-80^2 + -20^2) = 82.something
Side C = 82.blah

Simple

[PONGpaddle]

At 1/30/06 03:15 PM, -Toast- wrote: You can't really do that...

I can prove that one c can have many triangles.

RAR file containing Python executable.

Try a few numbers. c = 100 seems enough to be enough proof.

[PONGpaddle]

I forgot to say that the program can be refined to search within decimal values, therefore finding far more triangles.

[Rustygames]

At 1/30/06 04:39 PM, TheDrunkMonkey wrote:

At 1/30/06 03:15 PM, -Toast- wrote: You can't really do that...

I can prove that one c can have many triangles.

Depends what you mean. If you mean just describing C as a length then yeah it could of course have many different triangles. However if you mean describing C as a line (with gradient and such) then it can only have 2 :P

[Ravens-Grin]

At 1/30/06 04:45 PM, Ninja-Chicken wrote: Depends what you mean. If you mean just describing C as a length then yeah it could of course have many different triangles. However if you mean describing C as a line (with gradient and such) then it can only have 2 :P

Can you put that in mathematical lingo, I have no clue what you're trying to say.

Anyways, there is a possibility of finding a or b if you're given an additional angle( If your just using the base a^2+b^2=c^2 then you are automatically assuming there is one angle of 90 degrees) Then from there you can use the law of sines to determine all of the other sides. Just a simple reminder, the law of cosines is a^2+b^2-2ab(cosC)=c^2, where C is the angle across from line c(or in other words the only angle that is not touching line segment c). From there you can see how they came about with pythagoreans triangle ( which is when C is equal to 90 which is a right angle).

[White-Rhyno]

Do you fuckers not know basic trig or something?

T = O/A, S = O/H, C = A/H

Where;
T = Tan˜
S = Sin˜
C = Cos˜

O = Opposite Length relative to angle
A = Adjacent Length relative to angle
H = Hypotenuse

So if you have the length of the Hypotenuse, and one of the angles(other than 90), you can take the inverse of Cos˜, multiply by the Hypotenuse, to give you the length of Opposite then apply the pythagorean theorem to get the remaining side length.

[White-Rhyno]

At 1/31/06 12:38 AM, White_Rhyno wrote: stuff

GOD DAMNIT.

That ˜ was showing up on my screen as the Greek symbol Theta, which looks like an O with a horizontal line through it.

So just substitute ˜ for the Theta symbol in your heads.

[Toast]

At 1/30/06 04:32 PM, Ninja-Chicken wrote:

At 1/30/06 02:49 PM, -Toast- wrote: .... No, you need to know the values of at least two segments.

ha ha ha. I though you were smart toast ; )

Yeah but he can't do that if he only knows c.
I might be not smart but did you ever try answering those questions when you were 12 and didn't even learn sin,tan,cos,pythagorean theorem, hypotenuse, etc. at school?
Everything I learnt at school so far is how to find the values of a triangle if two values are known. :P

[Rustygames]

I didnt use any trig other then pythagoras. All he was technicaly asking was how does he get the other 2 sides if he has 2 points. Which quite frankly a 5 year old could figure out

[Toast]

At 1/31/06 12:01 PM, Ninja-Chicken wrote: Which quite frankly a 5 year old could figure out

... Stop being so silly! >:(

[Begoner]

Given 2 points of a triangle, each an endpoint of the hypotenuse, you can find the length of the hypotenuse, but there are infinately many possible coordinates for the third point of the triangle, so you could not find the length of sides A and B. Only C.