In Fig $\displaystyle AB\ \parallel \ CD$ , and $EF \perp CD$ , $\angle GED = 120°$. Find $\angle GEC , \angle EGF , \angle GEF$
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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 $GFE$,

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

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

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

$\angle EGF= 60^o$.

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

$\angle EGF= 60^o$.

Tutorialspoint

Updated on: 10-Oct-2022

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