/sci/ - Science & Math » Thread #13248882

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Anonymous Mon 07 Jun 20:11:39 2021 No.13248882 View Reply Original Report

Quoted By: >>13248944 >>13249015 >>13249021

neat formula i found for the height of any traingle. it doesn't matter how you label b and c. that's all I wanted to say really

Anonymous

Anonymous Mon 07 Jun 2021 20:30:50 No.13248944 Report

Quoted By:

>>13248882
.

Anonymous

Anonymous Mon 07 Jun 2021 20:42:04 No.13248984 Report

Quoted By:

That's really interesting. Will look into that a bit later, but nice job, anon.

Anonymous

Anonymous Mon 07 Jun 2021 20:43:42 No.13248991 Report

Quoted By:

How does the little yugoslavian stick figure factor into it?

Anonymous

Anonymous Mon 07 Jun 2021 20:52:36 No.13249015 Report

Quoted By:

>>13248882
Neat
Tested it for a=2, b=c=sqrt(2)
And a=6, b=c=5

Worked for both

Anonymous

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Anonymous Mon 07 Jun 2021 20:56:26 No.13249021 Report

Quoted By: >>13249132

>>13248882
oh and if you're wondering how I got it
Divide the triangle into 2 right-angled triangles, do the Pythagorean theorem for b and c, put h^2 on one side and now c^2 - x^2 = b^2... bla bla algebra and now you have the formula for x. now just use the Pythagorean theorem in one of the 2 triangles

Anonymous

Anonymous Mon 07 Jun 2021 20:56:42 No.13249024 Report

Quoted By:

Heron's formula says:
$A = \frac{1}{2} \sqrt{ a^2 b^2 - \left( \frac{a^2 + b^2 - c^2}{2} \right)^2}.$
Plug this in the height-area formula $A = \frac{a h_a}{2}$ to get
$h_a = \frac{2A}{a} = \sqrt{b^2 - \left( \frac{a^2 + b^2 - c^2}{2 a} \right)^2}.$

Anonymous

Anonymous Mon 07 Jun 2021 21:25:31 No.13249132 Report

Quoted By:

>>13249021
Good shit anon!
A variant of the same proof can be used to show the law of cosines, which says that in fact $a^2+b^2-c^2=2ab\cos C$ , where C is the angle opposite c.
Substituting the cosine theorem into your formula and $A=\frac{ah}2$ gives $A=\frac12ab\sin C$ .

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