| In the given figure, ADEC is a cyclic quadrilateral, CE and AD are extended to meet at B.\[\angle CAD=60{}^\circ \]and \[\angle CBA=30{}^\circ .\]\[BD=6\,\,cm\] and \[CE=5\sqrt{3}\,\,cm.\]Find the value of \[AC:AD.\] |
 |
A) \[\frac{3}{4}\]
B) \[\frac{4}{5}\]
C) \[\frac{2\sqrt{3}}{5}\]
D) Cannot be determined
Correct Answer: A
Solution :
| \[\angle CED=120{}^\circ \] |
| [since, ACED Is a cyclic quadrilateral] |
| \[\angle BED=60{}^\circ \]\[\Rightarrow \]\[\angle EDB=90{}^\circ \] |
| \[\therefore \] \[\frac{BD}{BE}=\cos 30{}^\circ \] |
| \[\Rightarrow \] \[\frac{6}{BE}=\frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[BE=4\sqrt{3}\,\,cm\] |
| \[\therefore \] \[=4\sqrt{3}+5\sqrt{3}=9\sqrt{3}\,\,cm\] |
| Now, since AB and CB are the secants of the circle. |
| \[\therefore \] \[BD\times BA=BE\times BC\] |
| \[\Rightarrow \] \[6\times BA=4\sqrt{3}\times 9\sqrt{3}\]\[\Rightarrow \]\[BA=18\,\,cm\] |
| In \[\Delta ACB\] which is right angled triangle, |
| \[AC=AB\sin 30{}^\circ \] |
| [alternatively apply Pythagoras theorem] |
| \[\Rightarrow \] \[AC=9\,\,cm\] \[[\because \sin 30{}^\circ =1/2]\] |
| and \[AD=AB-BD=18-6=12\,\,cm\] |
| \[\therefore \] \[\frac{AC}{AD}=\frac{9}{12}=\frac{3}{4}\] |
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