Trigonometry (11th Edition) Clone

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