I am trying to find exact instances of a pentagonal tiling, but I am struggling to solve all the angles for a given type.
Taking type 8 as an example:
where:
$a = b = c = d$
$C = 360 - 2B$
$E = 180 - D/2 $
After choosing a length for $a$ , and an angle for $B$ , what techniques could you use to find a closed form solution for the rest of the angles?
My understanding is that the only property we can use of the pentagon is that all internal angles sum to $540°$.
I have also tried breaking the pentagon into three triangles, using properties of isosceles triangles to find a system of equation so solve, but this was also unsuccessful.
My feeling is that the sine and/or cosine rule would be handy, but I can't see where to apply them. I feel as though I am missing something.
I know we could use an iterative approach (e.g. bisection), but I'd like to avoid that if possible.
From the given angle relations and $A+B+C+D+E=540$ you can easily deduce that $A+D/2=B$. For simplicity I'll let $b=c=d=e=1$.
From the relation between $B$ and $C$ you can prove that $BD$ is perpendicular to $AB$.
The isosceles triangle $\triangle BCD$ gives you that $|BD|=2\sin(B)$.
Using the sine rule in $\triangle ADE$ we find that
$$\frac{\sin(E)}{|AD|} = \frac{\sin(\angle EAD) }{1}$$ $$\frac{\sin(E)}{|AD|} = \sin(A-\angle DAB)$$ $$\frac{\sin(180-D/2)}{|AD|} = \sin(A) \cos(\angle DAB) - \cos(A) \sin(\angle DAB)$$ $$\frac{\sin(D/2)}{|AD|} = \sin(A) \frac{|AB|}{|AD|} - \cos(A) \frac{|BD|}{|AD|}$$ $$\sin(B-A) = \sin(A) |AB| - \cos(A) |BD|$$ $$\sin(B)\cos(A)-\cos(B)\sin(A) = \sin(A) - \cos(A) 2\sin(B)$$ $$\sin(B) -\cos(B) \tan(A) = \tan(A) - 2\sin(B)$$ and then solve for $\tan(A)$ in terms of $B$.
I have not verified whether this is correct, so it is quite likely to have mistakes in it.