Question

In the adjoining figure $AB||CD$ and $EF||GH$ and $KL{\perp }^r$ FN.
Find the measure of a if $\angle x={42}^{o}and\angle y={30}^o$.

• A. ${72}^o$
• B. ${78}^o$
• C. ${64}^o$
• D. ${60}^o$

Answer 1

The correct option is B ${78}^o$

In $△TFL$,

$\angle TFL+\angle FTL+\angle TLF={180}^{\circ }$

${30}^{\circ }+\angle FTL+{90}^{\circ }={180}^{\circ }$

$\angle FTL={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

In $△QTR$,

$\angle QTR+\angle QRT+\angle TQR={180}^{\circ }$

${60}^{\circ }+\angle QRT+{42}^{\circ }={180}^{\circ }$ ---------($\angle FTL=\angle QTR={60}^{\circ }$ - vertically opposite angles)

$\angle QRT={180}^{\circ }-{102}^{\circ }={78}^{\circ }$

$QRS$ is a straight line, hence

$\therefore \angle QRT+\angle TRS={180}^{\circ }$

$\angle TRS={180}^{\circ }-{78}^{\circ }={102}^{\circ }$

$\angle TRS=\angle USB={102}^{\circ }$ --------(corresponding angles)

$GSU$ is a straight line, hence

$\therefore \angle GSB+\angle USB={180}^{\circ }$

$\angle GSB={180}^{\circ }-{102}^{\circ }={78}^{\circ }$

Option B.