#### Answer

$AE = \sqrt{6}-\sqrt{2}$
$sin~15^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$

#### Work Step by Step

We can use the Pythagorean theorem to find the length of $AE$:
$(AE)^2 = (DE)^2-(AD)^2$
$AE = \sqrt{(DE)^2-(AD)^2}$
$AE = \sqrt{(4)^2-(\sqrt{6}+\sqrt{2})^2}$
$AE = \sqrt{16-(6+2\sqrt{12}+2)}$
$AE = \sqrt{6-2\sqrt{12}+2}$
$AE = \sqrt{(\sqrt{6}-\sqrt{2})~(\sqrt{6}-\sqrt{2})}$
$AE = \sqrt{6}-\sqrt{2}$
We can find $sin~15^{\circ}$:
$sin~15^{\circ} = \frac{opposite}{hypotenuse}$
$sin~15^{\circ} = \frac{\sqrt{6}-\sqrt{2}}{4}$