Quantitative Aptitude Ques 1029

Question: The side BC of the $ \Delta ABC $ is produced to D. If $ \angle ACD=112{}^\circ $ and $ \angle ABC=\frac{3}{4}\angle BAC, $ then $ \angle ABC $ is equal to

Options:

A) $ 68{}^\circ $

B) $ 58{}^\circ $

C) $ 48{}^\circ $

D) $ 38{}^\circ $

Show Answer

Answer:

Correct Answer: C

Solution:

  • Given, $ \angle ACD=112{}^\circ $ and $ \angle ABC=\frac{3}{4}\angle BAC $ Now, $ \angle ABC+\angle BAC=\angle ACD $ [since, exterior angle of triangle is equal to the sum of two interior opposite angles] .

$ \Rightarrow $ $ \frac{3}{4}\angle BAC+\angle BAC=112{}^\circ $

$ \Rightarrow $ $ \frac{7}{4}\angle BAC=112{}^\circ $

$ \Rightarrow $ $ \angle BAC=64{}^\circ $

$ \therefore $ $ \angle ABC=\frac{3}{4}\angle BAC $
$ \Rightarrow $ $ \frac{3}{4}\times 64{}^\circ =48{}^\circ $