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 $
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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 $