question_answer

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 [SSC (10+2) 2013]

A) \[68{}^\circ \]                          

B) \[58{}^\circ \]

C) \[48{}^\circ \]  

D) \[38{}^\circ \]

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] .

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

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

\[\Rightarrow \]\[\angle BAC=64{}^\circ \]

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