In triangle $ABC,$ $\angle B = 60^\circ$ and $\angle C = 45^\circ.$  The point $D$ divides $\overline{BC}$ in the ratio $1:3$.  Find
\[\frac{\sin \angle BAD}{\sin \angle CAD}.\]
Question:
In triangle $ABC,$ $\angle B = 60^\circ$ and $\angle C = 45^\circ.$  The point $D$ divides $\overline{BC}$ in the ratio $1:3$.  Find
\[\frac{\sin \angle BAD}{\sin \angle CAD}.\]
Answer:
Let's think step by step.

user
\frac{\sqrt{6}}{6}
In triangle $ABC,$ $\angle B = 60^\circ$ and $\angle C = 45^\circ.$  The point $D$ divides $\overline{BC}$ in the ratio $1:3$.  Find
\[\frac{\sin \angle BAD}{\sin \angle CAD}.\]
train