Quantitative Aptitude Ques 1430
Question: 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=6cm $ and $ CE=5\sqrt{3}cm. $ Find the value of $ AC:AD. $
Options:
A) $ \frac{3}{4} $
B) $ \frac{4}{5} $
C) $ \frac{2\sqrt{3}}{5} $
D) Cannot be determined
Show Answer
Answer:
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=18cm $
In $ \Delta ACB $ which is right angled triangle,
$ AC=AB\sin 30{}^\circ $
[alternatively apply Pythagoras theorem]
$ \Rightarrow $ $ AC=9cm $ $ [\because \sin 30{}^\circ =1/2] $ and $ AD=AB-BD=18-6=12cm $
$ \therefore $ $ \frac{AC}{AD}=\frac{9}{12}=\frac{3}{4} $
BETA
