question_answer

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

A) \[1/\sqrt{3}\]                    

B) \[1/\sqrt{6}\]

C) \[\sqrt{2/3}\]                    

D) \[1/3\]

Correct Answer: B

Solution :

Given,\[\angle B=\pi /3\]and\[\angle C=\pi /4\] Applying sin formula in\[\Delta ABD\]                 \[\frac{\sin \angle BAD}{BD}=\frac{\sin \pi /3}{AD}\] \[\Rightarrow \]               \[\sin \angle BAD=\frac{BD}{AD}\times \frac{\sqrt{3}}{2}\]                          ... (i) Again applying sin formula in\[\Delta ADC\]                 \[\frac{\sin \angle CAD}{DC}=\frac{\sin \pi /4}{AD}\]                       \[\Rightarrow \]               \[\sin \angle CAD=\frac{DC}{AD}\times \frac{1}{\sqrt{2}}\]                          ? (ii) From Eqs. (i) and (ii), we get                 \[\frac{\sin \angle BAD}{\sin \angle CAD}=\frac{BD}{AD}\times \frac{\sqrt{3}}{2}\times \frac{AD}{DC}\times \sqrt{2}\]                 \[=\frac{BD}{DC}\times \frac{\sqrt{3}}{\sqrt{2}}\] Given, \[DC=3BD\] \[\therefore \]  \[\frac{\sin \angle BAD}{\sin \angle CAD}=\frac{BD}{3BD}\times \frac{\sqrt{3}}{\sqrt{2}}=\frac{1}{\sqrt{6}}\]