Question

In the given figure AB||CD and CD||RE. If y:z =3:7 then x

$\left(a\right){108}^{\circ }$

$\left(b\right){126}^{\circ }$

$\left(c\right){162}^{\circ }$

$\left(d\right){63}^{\circ }$

Answer 1

ANSWER:
$\left(b\right){126}^{\circ }$
Let y=3a° and z=7a°
Let the transversal intersect AB at P, CD at O and EF at Q.
Then, we have:
$\angle COP=\angle DOF=y$

Vertically-
Opposite Angles∴ $\angle OQF+\angle DOQ={180}^{\circ }ConsecutiveInteriorAngles\Rightarrow 3a+7a={180}^{\circ }\Rightarrow 10a={180}^{\circ }\Rightarrow a={18}^{\circ }\therefore y=3\times {18}^{\circ }={54}^{\circ }$
Also,
$\angle APO+\angle COP={180}^{\circ }\Rightarrow x+{54}^{\circ }={180}^{\circ }\Rightarrow x={126}^{\circ }$