Quantitative Aptitude Ques 1381
Question: In the figure, $ \angle CAB=72{}^\circ , $ $ \angle CBA=74{}^\circ $ and $ \angle CED=112{}^\circ $ Find $ \angle CDE. $
Options:
A) $ 34{}^\circ $
B) $ 33{}^\circ $
C) $ 35{}^\circ $
D) $ 38{}^\circ $
Show Answer
Answer:
Correct Answer: A
Solution:
- In $ \Delta ABC, $ $ \angle CAB+\angle ABC+\angle BCA=180{}^\circ $ $ 72{}^\circ +74{}^\circ +\angle BCA=180{}^\circ $ $ 146{}^\circ +\angle BCA=180{}^\circ $ $ \angle BCA=34{}^\circ $ Now, $ \angle BCA=\angle DCE $ [vertically opposite angle] In $ \Delta CDE, $ $ \angle CED+\angle CDE+\angle DCE=180{}^\circ $ $ 112{}^\circ +\angle CDE+34{}^\circ =180{}^\circ $
$ \Rightarrow $ $ \angle CDE=180{}^\circ -(112+34){}^\circ $
$ \Rightarrow $ $ \angle CDE=180{}^\circ -146{}^\circ =34{}^\circ $
$ \Rightarrow $ $ \angle CDE=34{}^\circ $
BETA
