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I was designing this model with glasses. The front triangle sides are $a, b$ and $c$. And the slant height is $h$. The top length is $l$ and bottom full length is $L$. enter image description here

enter image description here

One value that is fixed is the back side triangle angle i.e. $\,45\unicode{176}\,$ (forty-five degrees).

enter image description here

Now I need help to find relation between these values so I can design it for different value? Anyone can help me ?

Edited: $x=45\unicode{176}$ (forty-five degrees) which is the front triangle slanted angle. enter image description here

Angelo
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Just doin Gods work
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  • a and b are different? – Vincenzo Tibullo Dec 26 '22 at 11:01
  • Also, which angle are you looking for? – Sassatelli Giulio Dec 26 '22 at 11:10
  • @enzotibes One I'm making its is isosceles triangle, yo yeah a=b – Just doin Gods work Dec 26 '22 at 11:30
  • @SassatelliGiulio First I was thinking to find the angle of the front slant triangle with the ground when I had all the lengths (a, b, c, l, L). But It'd be better if I could get a relation between these values. So I can make changes as per the change in angle i.e. 45 degrees currently (Front slant triangle). – Just doin Gods work Dec 26 '22 at 11:34
  • FYI: When three edges meet at a "trihedral vertex" to create face-angles $a$, $b$, $c$ and dihedral angles $A$, $B$, $C$ (with $a$ opposite $A$, etc) these relations hold $$\cos A = \frac{\cos a -\cos b\cos c}{\sin b\sin c} \qquad \cos a = \frac{\cos A + \cos B \cos C}{\sin B\sin C} \qquad \frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}$$ (These are known as the laws of cosines and sines for spherical trigonometry (ie, triangles drawn on the surface of a sphere), but they apply directly to trihedral vertices.) – Blue Dec 26 '22 at 11:47
  • @JustdoinGodswork, is the base of the back triangle igual to $c$ too ? – Angelo Dec 26 '22 at 11:51
  • @Blue I'm trying to understand it, and how to apply it for my use. Give me some time. – Just doin Gods work Dec 26 '22 at 11:54
  • @Angelo Yes, base of back triangle is equal to c. – Just doin Gods work Dec 26 '22 at 11:55

1 Answers1

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From the values of $c,h,l,L$ you can obtain the values of $a,b$, the height $H$ of the solid and the lengths of the sides of the back triangle.

The height of the solid is

$H=\sqrt{h^2-(L-l)^2}\;.$

The values of $a,b$ are

$a=\sqrt{h^2+(c-H)^2}\;,$

$b=\sqrt{2h^2-(L-l)^2}\;.$

The length of the side of the back triangle which forms the $45\unicode{176}$-angle with the base is $H\sqrt2$.

The length of the base is $c$.

The lenght of the other side is $\sqrt{H^2+(c-H)^2}$.

Addendum :

If we know that $x=45\unicode{176}$ (forty-five degrees) which is the front triangle slanted angle, then we can get all the lengths from $c,l,L$ in the following way :

$H=L-l$

$h=\sqrt2\big(L-l\big)$

$a=\sqrt{2(L-l)^2+(c+l-L)^2}\;,$

$b=\sqrt3\big(L-l\big)$

The length of the side of the back triangle which forms the $45\unicode{176}$-angle with the base is $\sqrt2\big(L-l\big)$.

The length of the base is $c$.

The length of the other side is

$\sqrt{(L-l)^2+(c+l-L)^2}\;.$

Angelo
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