$in the given figure, A B | | C D a n d E F ⊥ A B . i f E G is a transversal, such that ∠ G E D = 130^{\circ}, f i n d ∠ E G F .$

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Solution

ANSWER:
In the given figure, AB || CD and GE is the transversal.
∴ $∠ G E D + ∠ E G F = 180^{\circ}$
supplementary)
(Sum of adjacent interior angles on the same side of the transversal is
⇒ $130^{\circ} + ∠ E G F = 180^{\circ}$
⇒ $∠ E G F = 180^{\circ} - 130^{\circ} = 50^{\circ}$
Thus, the measure of $∠ E G F i s 50^{\circ}$ .

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in the given figure,AB||CD and EF⊥AB.if EG is a transversal, such that ∠GED=130∘,find ∠EGF.

ANSWER: In the given figure, AB || CD and GE is the transversal. ∴ ∠GED+∠EGF=180∘ supplementary) (Sum of adjacent interior angles on the same side of the transversal is ⇒ 130∘+∠EGF=180∘ ⇒ ∠EGF=180∘−130∘=50∘ Thus, the measure of ∠EGFis50∘.

$in the given figure, A B | | C D a n d E F ⊥ A B . i f E G is a transversal, such that ∠ G E D = 130^{\circ}, f i n d ∠ E G F .$