Question

In the given figure, if $l_1||l_{2}and{l}_3||l_4$, what is ${}^{\prime }{y}^{\prime }$ in terms of ${}^{\prime }{x}^{\prime }$?

• A. ${90}^{\circ }+x$
• B. ${90}^{\circ }+2x$
• C. ${90}^{\circ }-\frac{x}{2}$
• D. ${90}^{\circ }-2x$

Answer 1

The correct option is C

${90}^{\circ }-\frac{x}{2}$

At point A
$Here,\angle x+\angle BAD={180}^{\circ }\text{(linear pair)}\to \angle BAD=({180}^{\circ }-x)\text{As ABCD is a parallelogram, then}\angle BAD=\angle BCD=({180}^{\circ }-x)$
$\because$ Intersection of two pair of parallel lines forms a parallelogram and opposite angles are equal in a parallelogram

$Also,\angle ECF=y+y=2y\to \left(2y\right)=\angle ECF=\angle BC=({180}^{\circ }-x)\text{[Vertically Opposite angles]}\to y=\frac{({180}^{\circ }-x)}{2}\to y={90}^{\circ }-\left(\frac{x}{2}\right)$