In the adjoining figure- AB DE - - BAC -53- and - CDE -35- then the value of - CED isADA- 45-B- 88C- 110-D- 53-

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Alternate Angles of Parallel Lines and Its Transversal

In the adjoin...

Question

In the adjoining figure, AB || DE, ∠BAC = 53^o and ∠CDE = 35^o, then the value of ∠CED is

45^o

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110^o

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53^o

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In the given figure, if AB || DE, ∠BAC = 35º and ∠CDE = 53º, find ∠DCE.

A B | | D E, ∠ B A C = 35^{\circ} a n d ∠ C D E = 53^{\circ}, f i n d ∠ D C E .

∠

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∠

°

∠

A B | | D E, ∠ B A C = 35^{\circ} a n d ∠ C D E = 53^{\circ}, f i n d ∠ D C E .

From the adjoining figure AB || DE. Then the value of $x^{\circ}$ is:

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