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In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.

3.  In Fig. 6.30, if AB || CD, EF \bot CD and \angle GED = 126°, find \angle AGE, \angle GEF and \angle FGE.

                

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Given AB || CD, EF\perpCD and \angleGED = 126^0
In the above figure, 
GE is transversal. So, that \angleAGE = \angleGED =  126^0 [Alternate interior angles]
Also, \angleGEF = \angleGED - \angleFED 
                     = 126^0-90^0 = 36^0
        \angle GEF = 36^0 

Since AB is a straight line 
Therefore, \angleAGE  + \angleFGE =  180^0
So, \angleFGE = 180^0-\angle AGE = 180^0 - 126^0
\Rightarrow \angle FGE =54^0

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