Question

In the given figure, AB || CD. Find the value of x, y and z.

• A. $x={75}^{\circ },y={35}^{\circ }andz={70}^{\circ }$
• B. $x={35}^{\circ },y={70}^{\circ }andz={75}^{\circ }$
• C. $x={70}^{\circ },y={75}^{\circ }andz={70}^{\circ }$
• D. $x={35}^{\circ },y={80}^{\circ }andz={35}^{\circ }$

Answer 1

The correct option is B $x={35}^{\circ },y={70}^{\circ }andz={75}^{\circ }$


Since AB||CD, and BC and AD are the transversals, alternate interior angles will be equal.
$\Rightarrow \angle ABO=\angle OCD$
$\Rightarrow x={35}^{\circ }$

Similarly, $\angle ODC=\angle OAB$
$\Rightarrow z={75}^{\circ }$

We know that sum of all the interior angles in a triangle is ${180}^{\circ }$.
$\Rightarrow In△AOB$,
$\angle ABO+\angle AOB+\angle OAB={180}^{\circ }$
$\Rightarrow x+y+{75}^{\circ }={180}^{\circ }$
$\Rightarrow {35}^{\circ }+y+{75}^{\circ }={180}^{\circ }$
$\Rightarrow y={180}^{\circ }-{35}^{\circ }-{75}^{\circ }$
$\Rightarrow y={180}^{\circ }-{110}^{\circ }={70}^{\circ }$
$\Rightarrow y={70}^{\circ }$
$So,x={35}^{\circ },y={70}^{\circ }andz={75}^{\circ }.$