Question

$\text{In the given figure,}AB||CDandEF\text{is a transversal.If}\angle AEF={65}^{\circ },\angle DFG={30}^{\circ },\angle EGF={90}^{\circ }and\angle GEF=x^{\circ },findthevalueofx.$

Answer 1

ANSWER:
It is given that, AB || CD and EF is a transversal.
∴ $\angle EFD=\angle AEF$
(Pair of alternate angles)
⇒ $\angle EFD={65}^{\circ }$
⇒ $\angle EFG+\angle GFD={65}^{\circ }$
⇒ $\angle EFG+{30}^{\circ }={65}^{\circ }\Rightarrow \angle EFG={65}^{\circ }-{30}^{\circ }={35}^{\circ }$
In $\angle EFG$,
$\angle EFG+\angle GEF+\angle EGF={180}^{\circ }$
(Angle sum property)
⇒ ${35}^{\circ }+x+{90}^{\circ }={180}^{\circ }$
⇒ ${125}^{\circ }+x={180}^{\circ }$
⇒$x={180}^{\circ }-{125}^{\circ }={55}^{\circ }$
Thus, the value of x is${55}^{\circ }$.