JEE Advance - Mathematics (1995 - No. 15)
In a triangle $$ABC$$, $$\angle B = {\pi \over 3}$$ and $$\angle C = {\pi \over 4}$$. Let $$D$$ divide $$BC$$ internally in the ratio $$1:3$$ then $${{\sin \angle BAD} \over {\sin \angle CAD}}$$ is equal to
$${1 \over {\sqrt 6 }}$$
$${1 \over 3}$$
$${1 \over {\sqrt 3 }}$$
$$\sqrt {{2 \over 3}} $$