A) \[130{}^\circ \]
B) \[110{}^\circ \]
C) \[70{}^\circ \]
D) \[100{}^\circ \]
Correct Answer: C
Solution :
\[\angle CAB+\angle DBA=180{}^\circ \] \[\angle DBA=180{}^\circ -\angle CAB=180{}^\circ -130{}^\circ =50{}^\circ \] (\[AC|\,\,|BD,\]co-int angles are supplementary)) \[\Rightarrow \] \[\angle DBA+\angle ABF+\angle FBG=180{}^\circ \] \[\Rightarrow \] \[50{}^\circ +\angle ABF+60{}^\circ =180{}^\circ \] \[\Rightarrow \] \[\angle ABF=180{}^\circ -110{}^\circ =70{}^\circ \] Now, \[AE|\,|BF\Rightarrow \angle EAB+\angle ABF=180{}^\circ \] (co-int angles) \[\Rightarrow \] \[x+70{}^\circ =180{}^\circ \] \[\Rightarrow \] \[x=70{}^\circ \]
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