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.

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