Question

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

If $c={75}^o$ and $a=\frac{2}{5}b$, then find the value of ${}^{\prime }7b^{\prime }$ in degrees.

Answer 1

$\because PQandUTareparallellines$

$⟹\angle QPT=\angle PTU\left(alternateangles\right)$

$⟹a+b=c-\left(1\right)$

$and\because PQ\&RSareparalleltoo$

$⟹\angle SRP=\angle QP{R}^{\prime }\left(correspondingangles\right)$

$⟹d=180-b$

$⟹b+d={180}^{\circ }-\left(2\right)$

$\because c={75}^{\circ }$

$solving\left(1\right)$

$⟹a+b={75}^{\circ }$

$anda=\frac{2}{5}b$

$⟹\frac{2}{5}b+b={75}^{\circ }$

$⟹\frac{7b}{5}={75}^{\circ }$

$⟹7b=75\times 5$

$⟹7b={375}^{\circ }$.