A knotty problem

$\angle AED = 108^{\circ}$ (angle of a regular pentagon)

Therefor $\angle AEH = 180^{\circ}-108^{\circ} = 72^{\circ} = \theta$

$\displaystyle\frac{a}{t} = \sin(\theta) = \sin(72^{\circ}) = \frac{\sqrt{10+2\sqrt{5}}}{4} = 0.951\dots$

Value of $\sin(72^\circ)$

Let, $A = 18^\circ$

Therefore, $5A = 90^\circ$

$2A+3A=90^\circ$

$2A=90^\circ-3A$

$\sin(2A)=\sin(90^\circ-3A)=\cos(3A)$

$2\sin(A)\cos(A)=4\cos^3(A)-3\cos(A)$

$2\sin(A)=4\cos^2(A)-3$

$2\sin(A)=4(1-\sin^2(A))-3$

$4\sin^2(A)+2\sin(A)-1=0$

Therefore, $\displaystyle\sin(A) = \frac{-2\pm \sqrt{2^2-4(4)(-1)}}{2(4)}$

$\displaystyle\sin(A) = \frac{-1+\sqrt{5}}{4}$ (we take plus sign since $\sin(18^\circ)$ is positive.)

Therefore, $\displaystyle\sin(18^\circ) = \frac{\sqrt5-1}{4}$

$\sin(72^\circ)=\sin(90^\circ-18^\circ)=\cos(18^\circ)=\sqrt{1-\sin^2(18^\circ)}$

$\displaystyle\sin(72^\circ) = \sqrt{1-\bigg(\frac{\sqrt5-1}{4}\bigg)^2}$

$\sin(72^\circ) = \displaystyle\frac{\sqrt{10+2\sqrt{5}}}{4}$