Question

In the given figure, PQ, RS and UT are parallel lines.

If $c={57}^{\circ }anda=\frac{c}{3}$, find the value of d.

Answer 1

Since, $PQ||UTandPT$ is transversal.
$Therefore,\angle QPT=\angle UTP\left[alternateinteriorangles\right]\Rightarrow a+b=c\Rightarrow \frac{c}{3}+b=c[\because a=\frac{c}{3},given]\Rightarrow b=c-\frac{c}{3}\Rightarrow b=\frac{3c-c}{3}\Rightarrow b=\frac{2c}{3}=\frac{2}{3}\times {57}^{\circ }\therefore b={38}^{\circ }$
Again, $PQ||RSandPR$ is transversal.
$\angle QPR+\angle PRS={180}^{\circ }[co-interiorangles]\Rightarrow b+d={180}^{\circ }\Rightarrow d={180}^{\circ }-b\Rightarrow d={180}^{\circ }-{38}^{\circ }[\because b={38}^{\circ }]\Rightarrow d={142}^{\circ }$