Construct the angle whose sine is $\frac{3}{2+\sqrt{5}}$

Question

I saw this question in an old trig text book:

Construct the angle whose sine is $\frac{3}{2+\sqrt{5}}$.

I ask:

1. What solution can people here give ?

2. Is there a solution that does not (like mine) require the construction of $\sqrt[4]{5}$ ? In other words, a solution without assuming the circle intersection property (See Hartshore: Euclid and Beyond pg. 144)

Answer 1

Outline. Let $A=(0,2)$ and $B=(1,0)$. Consider the circle around $A$ with radius $AB=\sqrt{5}$, which intersects the $y$-Axis at $C=(0, 2+\sqrt{5})$. Let now $\Gamma$ be the circle centered at $O=(0,0)$ through $C$ which intersects the line $l$ through $D=(0,3)$ and parallel to the $x$-Axis at $E$. Then

$$\sin\angle DEO=\frac{OD}{OE}=\frac3{2+\sqrt{5}}$$

Answer 2

$$|OB'|= \sqrt{5} \quad\to\quad |C'D|=3(\sqrt{5}-2) \quad\to\quad \sin A'' = \frac{3(\sqrt{5}-2)}{1}=\frac{3}{\sqrt{5}+2}$$