Question

In the following figure both $ABCD$ and $PQRS$ are parallelogram find the value of $x.$

Answer 1

In parallelogram $ABCD,$

$\angle A=\angle C={120}^o$ [opposite angles of a parallelogram are equal]

and $\angle DCB+\angle SCO={180}^{\circ }$ [Linear pair]

$\angle SCO=180-120={60}^{\circ }$ $...\left(1\right)$

Now, in parallelogram $PQRS,$

$\angle S+\angle R={180}^{\circ }$ [adjacent angles of parallelogram]

$\angle RSP=180-50={130}^{\circ }$

and $\angle RSP+\angle CSO={180}^{\circ }$ [linear pair]

$\angle CSO=180-130={50}^{\circ }$ $...\left(2\right)$

Now In $\Delta COS,$

$\angle COS+\angle OCS+\angle CSO={180}^{\circ }$ [angle sum proportion]

$\angle COS=180-50-60$ [From equation $\left(1\right)$ & $\left(2\right)$]

$\angle COS={70}^{\circ }$

Hence $\angle COS=\angle POB$ [vertically opposite angles]
$\Rightarrow x={70}^{\circ }$

Therefore the value of $x$ is ${70}^{\circ }$