Question

In the figure, AB and CD are parallel lines and transversal EF intersects them at P and Q respectively. If $\angle APR={25}^{\circ },\angle RQC={30}^{\circ }$ and $\angle CQF={65}^{\circ },$ then

Answer 1

In the figure,

$\because AB\left|\right|CD$,EF inetersects them at P and Q respectively,

$\angle APR={25}^{\circ },\angle RQC={30}^{\circ },\angle CQF={65}^{\circ }$

$\because AB\left|\right|CD$

$\therefore \angle APQ=\angle CQF$ (Corresponding angles)

$\Rightarrow y+{25}^{\circ }={65}^{\circ }$

$\Rightarrow y={65}^{\circ }-{25}^{\circ }={40}^{\circ }$

and APQ+PQC=${180}^{\circ }$

(Co -interior angles)

y+${25}^{\circ }+\angle 1+{30}^{\circ }={180}^{\circ }$

${40}^{\circ }+{25}^{\circ }+\angle 1+{30}^{\circ }={180}^{\circ }$

$\Rightarrow \angle 1+{95}^{\circ }={180}^{\circ }$

$\therefore \angle 1+{180}^{\circ }-{95}^{\circ }={85}^{\circ }$

Now ,

$\Delta PQR,$

$\angle RPQ+\angle PQR+\angle PRQ={180}^{\circ }$

(Sum of angles of a triangle )

$\Rightarrow {40}^{\circ }+x+{85}^{\circ }={180}^{\circ }$

$\Rightarrow {125}^{\circ }+x={180}^{\circ }\Rightarrow x={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

$\therefore x={55}^{\circ },y={40}^{\circ }$