Quantitative Aptitude Ques 1749
Question: The angles $ x{}^\circ , $ $ a{}^\circ , $ $ c{}^\circ $ and $ (\pi -b{}^\circ ) $ are indicated in the figure given below. Which one of the following is correct?
Options:
A) $ x{}^\circ =a{}^\circ +c{}^\circ -b{}^\circ $
B) $ x{}^\circ =b{}^\circ -a{}^\circ -c{}^\circ $
C) $ x{}^\circ =a{}^\circ +b{}^\circ +c{}^\circ $
D) $ x{}^\circ =a{}^\circ -b{}^\circ +c{}^\circ $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \because $ $ \angle PCT+\angle PCB=\pi $ [linear pair]
$ \angle PCB=\pi -(\pi -b{}^\circ )=b{}^\circ $ … (i) In $ \Delta BPC, $
$ \angle PCB+\angle BPC+\angle PBC=\pi $
$ \Rightarrow $ $ \angle PBC=\pi -\angle PCB+\angle BPC $ $ =\pi -b{}^\circ -a{}^\circ $ (ii) $ \because $ $ \angle ABE+\angle EBC=\pi $ $ [\because \angle PBC=\angle EBC] $ [linear pair]
$ \Rightarrow $ $ \angle ABE=\pi -\angle PBC=\pi -(\pi -b{}^\circ -a{}^\circ ) $ $ =a{}^\circ +b{}^\circ $ (iii) Now, in $ \Delta ABE, $ Sum of two interior angles = Exterior angle $ \angle EAB+\angle ABE=\angle BES $
$ \Rightarrow $ $ c{}^\circ +b{}^\circ +a{}^\circ =x{}^\circ $
$ \therefore $ $ x{}^\circ =a{}^\circ +b{}^\circ +c{}^\circ $
BETA
