Quantitative Aptitude Ques 2015
Question: In a $ \Delta ABC, $ $ AB=AC $ and D is a point on AB, such that $ AD=DC=BC. $ Then, $ \angle BAC $ is
Options:
A) $ 40{}^\circ $
B) $ 45{}^\circ $
C) $ 30{}^\circ $
D) $ 36{}^\circ $
Show Answer
Answer:
Correct Answer: D
Solution:
- Given that, $ AB=AC $ and $ AD=CD=AC $ Let $ \angle ABC=\theta $ Then, $ \angle ACB=\theta $ $ [\because AB=AC] $
$ \Rightarrow $ $ \angle BAC=180{}^\circ -2\theta $
$ \Rightarrow $ $ \angle ACD=180{}^\circ -2\theta $ $ [\because AD=CD] $
$ \Rightarrow $ $ \angle BCD=\angle ACB-\angle ACD $
$ \Rightarrow $ $ \angle BCD=\theta -(180{}^\circ -2\theta )=3\theta -180{}^\circ $ and $ \angle BDC=\theta $ $ [\because CD=BC] $ New, in $ \Delta BCD $ $ \angle CBD+\angle BDC+\angle BCD=180{}^\circ $
$ \Rightarrow $ $ \theta +\theta +3\theta -180{}^\circ =180{}^\circ $
$ \Rightarrow $ $ 5\theta =360{}^\circ $
$ \Rightarrow $ $ \theta =72{}^\circ $
$ \therefore $ $ \angle BAC=180{}^\circ -2\theta $ $ =180{}^\circ -144=36{}^\circ $
BETA
