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February 16th, 2019, 09:45 AM   #1
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Pythagorean theorem/Lesson Learned

Hello,

I learned something interesting that I want to share.

So, me and my brother were talking about math, and he told me that, "The Pythagorean Theorem only works for right triangles."

I then thought to myself, "How would I find side length of a triangle like an isosceles from the two side lengths?"

I then found this formula that answered the question:

Side * 2cos(angle)

However, I am the kind of guy that likes to find his own way. So, I had the idea. What if I modified the Pythagorean Theorem to fit other triangles types?

Eventually, I got it. This is what I came up with:

(a+b) sin(angle/2)

I checked my work with a simple Isosceles Triangle. Side a = 1. Side b =1. The angles are 72, 54, and 54. According to the equation I said earlier, The answer should be 1.175570505.

Then I tried the equation I made up. 1+1 is 2. The angle divided by 2 is 36. 2Sin(36) equals 1.175570505.

It worked. I tried it with other side lengths, and it worked. Sides a and b = 2. Angles are 40, 40, and 100. Answer should be 3.064177772.

The equation I made up.

2 + 2 is 4. 4sin(50) equals 3.064177772.

It worked.

Then, I tried something harder. A scalene with the following measurements. According to the law of sines, C=1, A=0.732050808, B=0.896575472. Angle AB=75. Angle BC=45. Angle CA=60

I tried to find Side C my own way. Here is what I came up with:

a cos(angle AC) + b cos(angle BC) = c

Checked my work.

0.732050808 cos(60)= 0.366025404

0.896575472 cos(45) =0.633974596.

0.366025404 + 0.633974596 = 1

It worked.

I learned two things from this. First, I learned more about triangles and how to find out side lengths. Second, I learned there is more than one way to find an answer in mathematics. Some people use s*2cos(angle), I use my own way. Both ways work.

I hope you enjoyed reading this, and if anyone has input, feel free to add it.

Jared

Last edited by skipjack; February 16th, 2019 at 05:47 PM.
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February 16th, 2019, 01:46 PM   #2
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The usual extension to the Pythagorean theorem is the law of cosines: $c^2=a^2+b^2-2ab\cos C$, where $C$ is the angle opposite side $c$.

Last edited by skipjack; February 16th, 2019 at 05:20 PM.
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February 16th, 2019, 03:13 PM   #3
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Thanks for clarifying that. To be honest, I have difficulty understanding the law of cosines. I do use it, but sometimes I use it incorrectly. Mainly because it's hard for me to understand.

Again, thanks for telling me that. I appreciate you and others here help me out with understanding mathematics.
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February 16th, 2019, 05:48 PM   #4
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Quote:
Originally Posted by jnicholes View Post
I then found this formula that answered the question:

Side * 2cos(angle)
That works, but the angle isn't the apex angle. If you want to use the apex angle, the
formula becomes side * 2sin(angle/2).

Quote:
Originally Posted by jnicholes View Post
Eventually, I got it. This is what I came up with:

(a+b) sin(angle/2)
That formula works only if a = b, so it can be written as (2a) sin(angle/2), which is equivalent to the formula I gave above.

Quote:
Originally Posted by jnicholes View Post
Then, I tried something harder. . . . Here is what I came up with:

a*cos(angle AC) + b*cos(angle BC) = c
That is usually written as c = a * cos(B) + b * cos(A). It's a standard result, but is often overlooked, perhaps because it doesn't seem to have a handy description or name.
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February 16th, 2019, 06:39 PM   #5
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@skipjack, you are correct with everything you just said. Thank you for telling me that.

You just made me realize something. I went and did days of research and experimenting to try to find my own way, and when I do find my own way, and I share it, it turns out it was already figured out by someone else.

To be honest, I do not know how I am coming up with this stuff without even knowing that it already existed. I guess I just like experimenting and learning how things work my own way. I learn from experience. It's just who I am.

Thanks for the help and the input. I appreciate it.

Jared
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February 17th, 2019, 03:58 PM   #6
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Quote:
Originally Posted by jnicholes View Post
Thanks for clarifying that. To be honest, I have difficulty understanding the law of cosines. I do use it, but sometimes I use it incorrectly. Mainly because it's hard for me to understand.

Again, thanks for telling me that. I appreciate you and others here help me out with understanding mathematics.
What about the cosine law is so hard to understand? I suggest you sketch a triangle and see how it works.
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February 17th, 2019, 05:06 PM   #7
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@mathman, I don't really understand it, because I didn't really learn about it well enough when I went through school. You have to understand I am mentally challenged, math is one of the area's I excel in, however, because of my disability, I was alternating between homeschool and regular school periodically. Because of this, I think I missed learning about certain math things, like the law of cosines. The Law of Sines, I just learned recently, to be honest.

Would you mind explaining to me the law of cosines? It's difficult to understand because I never really learned it well enough. I know a little bit of it, but I don't know all of it.

Jared
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February 17th, 2019, 05:59 PM   #8
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Google it; you'll get helpful sites like:
https://betterexplained.com/articles/law-of-cosines/
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February 17th, 2019, 07:15 PM   #9
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Thanks, @Denis. I just looked it up. It's making more sense now. Thanks for helping out.

Jared
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February 18th, 2019, 07:47 AM   #10
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cosine.jpg
Code:
c² = EB²          + EA²
     = CB² - CE² + (CA - CE)²
     = a²   - CE² + CA² - 2CE*CA + CE²
     = a²            + b²   - 2(a cos(C))*b
     = a² + b² - 2ab cos(C)
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