Quantitative Aptitude Ques 1821
Question: The bisectors BI and CI of $ \angle B $ and $ \angle C $ of a $ \Delta ABC $ meet in I. What is $ \angle BIC $ equal to?
Options:
A) $ 90{}^\circ -\frac{A}{4} $
B) $ 90{}^\circ +\frac{A}{4} $
C) $ 90{}^\circ -\frac{A}{2} $
D) $ 90{}^\circ +\frac{A}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
- Given that, $ BI $ and $ CI $ are angle bisectors of $ \angle B $ and $ \angle C $ respectively. Now, in $ \Delta BIC $ $ x{}^\circ +\frac{B}{2}+\frac{C}{2}=180{}^\circ $ [let $ \angle BIC=x{}^\circ $ ]
$ \Rightarrow $ $ x{}^\circ =180{}^\circ -\frac{1}{2}(B+C) $
$ \Rightarrow $ $ x{}^\circ =180{}^\circ -\frac{1}{2}(180{}^\circ -A) $ $ [\because \text{in }\Delta ABC,A+B+C=180{}^\circ ] $
$ \Rightarrow $ $ x{}^\circ =180{}^\circ -90{}^\circ +\frac{A}{2} $
$ \therefore $ $ \angle BIC=x{}^\circ =90{}^\circ +\frac{A}{2} $
BETA
