# In Fig. 6.30, if $\mathrm{AB} \| \mathrm{CD}$, EF $\perp \mathrm{CD}$ and $\angle \mathrm{GED}=126^{\circ}$, find $\angle \mathrm{AGE}, \angle \mathrm{GEF}$ and $\angle \mathrm{FGE}$."

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Given :

$AB \parallel CD$ and $EF$ is perpendicular to $CD$.

$\angle GED = 120^o$.

To find :

We have to find  $\angle GEC, \angle EGF, \angle GEF$

Solution :

$\angle GEF + \angle CEG = 120^o$

$120^o = \angle GEF + 90^o$

$\angle GEF = 120^o-90^o$

$\angle GEF = 30^o$

$CD$ is a straight line.

Therefore,

$\angle GED + \angle CEG = 180^o$

$120^o+\angle CEG = 180^o$

$\angle CEG = 180^o-120^o$

$\angle GEC = 60^o$

In Triangle $GFE$,

$\angle GFE + \angle GEF+ \angle EGF= 180^o$

$90^o+30^o+\angle EGF= 180^o$

$\angle EGF= 180^o-120^o$

$\angle EGF= 60^o$.

$\angle GEF = 30^o$ , $\angle GEC = 60^o$

$\angle EGF= 60^o$.

Updated on 10-Oct-2022 13:40:32