In the figure,AB parallel CD and CD parallel | vedclass

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(D) Given that $AB \parallel CD$ and $CD \parallel EF$,therefore $AB \parallel EF$.Since $CD \parallel EF$ and $ED$ is a transversal,the interior angl

Vedclass · Accepted Answer

$x=125^o, y=125^o, z=35^o$

Vedclass · Answer

(D) Given that $AB \parallel CD$ and $CD \parallel EF$,therefore $AB \parallel EF$.Since $CD \parallel EF$ and $ED$ is a transversal,the interior angles on the same side of the transversal are supplementary.Thus,$y + 55^o = 180^o$.$y = 180^o - 55^o = 125^o$.Since $AB \parallel CD$ and $BE$ is a transversal,the corresponding angles are equal.Thus,$x = y = 125^o$.Since $EA \perp AB$ and $AB \parallel EF$,then $EA \perp EF$. Therefore,$\angle AEF = 90^o$.From the figure,$\angle AEF = z + 55^o$.$z + 55^o = 90^o$.$z = 90^o - 55^o = 35^o$.Therefore,$x = 125^o, y = 125^o, z = 35^o$.

In the figure,$AB \parallel CD$ and $CD \parallel EF$. Also,$EA \perp AB$. If $\angle BEF = 55^o$,find the values of $x, y$,and $z$.

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