Irregular Pentagon Formula

[Guest]

irr poly.png (10.38 KiB) Viewed 2625 times

Hello, I am having trouble trying to figure out a formula to calculate the angle on this shape.
It's a bit out of my league.
The resulting answer should be 16.0625
In the attached image I need a formula that would calculate the length of the angle.
All of the measurements are variable depending on input.
This is what I have but stuck at this point. It was started by a colleague of mine whom has since moved on.

[tex]((((cabinet_width-cabinet_depth)^2)+((cabinet_width-cabinet_depth)^2 ))^.5)-1.03075[/tex]

Any help would be great.
Mike
Ontario, Canada

[Guest]

I think I got it working.
Anyone see any issues with this version?
[tex]((((cabinet_width-box_material.thickness-cabinet_depth )^2)+((cabinet_width-box_material.thickness-cabinet_depth)^2 ))^.5 )[/tex]

[Guest]

You can use Pythagorean theorem.

You have one right triangle where both sides are 12-5/8 = 12-0.625 = 11.375

Now to find the hypotenuse we use the formula:
[tex]\text{hypotenuse}^2 = (11.375)^2 + (11.375)^2[/tex]
[tex]\text{hypotenuse}^2 = 129.390625 + 129.390625[/tex]
[tex]\text{hypotenuse}^2 = 258.78125[/tex]
[tex]\text{hypotenuse} = \sqrt{258.78125}[/tex]
[tex]\text{hypotenuse} = 16.086679[/tex]

[Guest]

Excellent.
Thanks

Mike

[Guest]

To solve this, apply the Pythagorean theorem. In this scenario, you're dealing with a right triangle where both sides measure 12-5/8, which is equal to 11.375 units.
Now, for the hypotenuse, we use the formula:

hypotenuse^2=(11.375)^2+(11.375)^2

This simplifies to:

Hypotenuse^2=129.390625+129.390625

Thus,

Hypotenuse^2=258.78125,
and taking the square root of both sides gives:
hypotenuse=√258.78125, which approximately equals 16.086679.

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