In triangle ABC, \(\angle B = 60^\circ\) and \(\angle C = 45^\circ.\) The point D divides BC in the ratio 1:3. Find \(\frac{\sin \angle BAD}{\sin \angle CAD}.\)
By the Law of Sines
sin BAD / BD = sin 60/AD sin CAD / DC = sin 45 / AD
sin BAD / 1 = sin 60 / AD and sin CAD / 3 = sin 45 / AD
sin BAD = sin 60 / AD sin CAD = 3 sin 45 / AD
So
sin BAD sin 60 / AD √3/2 √3/2
_______ = ___________ = _______ = _____ = √3/2 * √2/3 = √6/6 = 1/√6
sin CAD 3 sin 45 / AD 3 * 1/√2 3/√2