Note that we must have $\triangle ABD \sim \triangle ABC$. Therefore, $\angle ABD = 15^\circ$. Now let the foot of the altitude from $D$ to $AB$ be $E$. We have $AD=10 \sin {15^\circ}$ and $BD = 10 \sin {75^\circ}$, so $ED=\boxed{10 \sin {15^\circ} \sin {75^\circ}}$.

OlympusHero · 39 minutes ago (view)helo SolutionNote that we must have $\triangle ABD \sim \triangle ABC$. Therefore, $\angle ABD = 15^\circ$. Now let the foot of the altitude from $D$ to $AB$ be $E$. We have $AD=10 \sin {15^\circ}$ and $BD = 10 \sin {75^\circ}$, so $ED=\boxed{10 \sin {15^\circ} \sin {75^\circ}}$. x 4hellloolo Y 1 hellloolo