question_answer

In a \[\Delta \,ABC,\,\,\angle B=\pi /3\,\,and\,\,\angle \,C\,=\,\pi /4\]. If D divides BC internally in ratio \[1:3\], then \[\frac{\sin \,\angle BAD}{\sin \angle CAD}\]

A) \[1/\sqrt{6}\]     

B)        1/3

C) \[1/\sqrt{3}\]     

D)        \[\sqrt{2/3}\]

Correct Answer: A

Solution :

By sine rule, the sine of the angles are proportional to the lengths of opposite sides. i.e. \[\frac{\sin \,A}{a}=\frac{\sin \,B}{b}=\frac{\sin \,C}{c}\] Given a \[\Delta ABC\] in which \[\angle ABD=60{}^\circ \] and \[\angle ACD = 45{}^\circ \], also \[\frac{BD}{DC}=\frac{1}{3}\]. From \[\Delta \,ABD\], (By sine Rule) \[\frac{\sin \angle \,BAD}{BD}=\frac{\sin \,60{}^\circ }{AD}\] Now in \[\Delta \,ADC\]; \[\frac{\sin \angle CAD}{DC}\,\,=\,\,\frac{sin45{}^\circ }{AD}\] \[E{{q}^{n}}\] (i) divided by (ii) \[\frac{\sin \angle BAD}{\sin \angle CAD}=\frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{1}\times \frac{1}{3}=\frac{\sqrt{3}}{\sqrt{2}\times 3}=\frac{1}{\sqrt{6}}\]