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] . |
\[\Rightarrow \]\[\frac{3}{4}\angle BAC+\angle BAC=112{}^\circ \] |
\[\Rightarrow \] \[\frac{7}{4}\angle BAC=112{}^\circ \] |
\[\Rightarrow \]\[\angle BAC=64{}^\circ \] |
\[\therefore \]\[\angle ABC=\frac{3}{4}\angle BAC\]\[\Rightarrow \]\[\frac{3}{4}\times 64{}^\circ =48{}^\circ \] |
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