Trigonometry! #155

In $\Delta ABC$ , $\angle B=60^{\circ}$ and $\angle C=45^{\circ}$ . Let $D$ divide $BC$ internally in the ratio $1:3$ . Then find the value of $\frac{\sin\angle BAD}{\sin\angle CAD}$

Express your answer in the form of $\frac{a}{b}$ , where $a$ and $b$ have nothing in common. Enter the value of $a^{2}+b^{2}$ .

This problem is part of the set Trigonometry .