Question
In the given figure, BAD || EF, $\angle\text{AEF}=55^\circ$ and $\angle\text{ACB}=25^\circ,$ find $\angle\text{ABC}.$
✓
Answer
BAD || EF and EC is the transversal.$\Rightarrow\angle\text{AEF}=\angle\text{CAD}$ (corresponding angles)
$\Rightarrow\angle\text{CAD}=55^\circ$
Now, $\angle\text{CAD}+\angle\text{CAB}=180^\circ$ (linear pair)$\Rightarrow55^\circ+\angle\text{CAB}=180^\circ$
$\Rightarrow\angle\text{CAB}=125^\circ$
In $\triangle\text{ABC},$ by angle sum property,$\angle\text{ABC}+\angle\text{CAB}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ABC}+125^\circ+25^\circ=180^\circ$
$ \Rightarrow\angle\text{ABC}=30^\circ$
$\Rightarrow\angle\text{CAD}=55^\circ$
Now, $\angle\text{CAD}+\angle\text{CAB}=180^\circ$ (linear pair)$\Rightarrow55^\circ+\angle\text{CAB}=180^\circ$
$\Rightarrow\angle\text{CAB}=125^\circ$
In $\triangle\text{ABC},$ by angle sum property,$\angle\text{ABC}+\angle\text{CAB}+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ABC}+125^\circ+25^\circ=180^\circ$
$ \Rightarrow\angle\text{ABC}=30^\circ$
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