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, \[DC=3BD\] \[\therefore \] \[\frac{\sin \angle BAD}{\sin \angle CAD}=\frac{BD}{3BD}\times \frac{\sqrt{3}}{\sqrt{2}}=\frac{1}{\sqrt{6}}\]You need to login to perform this action.
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