Quantitative Aptitude Ques 493
Question: In the given figure, $ AM\bot BC $ and AN is the bisector of $ \angle A. $ If $ \angle ABC=70{}^\circ $ and $ \angle ACB=20{}^\circ , $ then $ \angle MAN $ is equal to
Options:
A) $ 20{}^\circ $
B) $ 25{}^\circ $
C) $ 15{}^\circ $
D) $ 30{}^\circ $
Show Answer
Answer:
Correct Answer: B
Solution:
- In $ \Delta ABC, $ $ \angle B+\angle C+\angle A=180{}^\circ $
$ \Rightarrow $ $ \angle A=180{}^\circ -90{}^\circ =90{}^\circ $ But is bisector of $ \angle A. $
$ \therefore $ $ \angle NAC=\angle NAB=45{}^\circ $ In $ \Delta ANC, $ $ \angle ANC=180{}^\circ -(20{}^\circ +\angle NAC) $ $ =180{}^\circ -(20{}^\circ +45{}^\circ ) $ $ =180{}^\circ -65{}^\circ =115{}^\circ $
$ \therefore $ $ \angle ANM=180{}^\circ -115{}^\circ =65{}^\circ $ $ \angle MAN=180{}^\circ -(90{}^\circ +65{}^\circ ) $ $ =180{}^\circ -155{}^\circ =25{}^\circ $
BETA
