Question

In Figure, if $PQRS$ and $ABCD$ are parallelogram, find the value of $x$.

Answer 1

In parallelogram $ABCD$,

$\angle B=\angle D=60°(\because \text{Opposite angles in a parallelogram are equal})$

Now, in parallelogram $PQRS$,

$\angle P+\angle S=180°(\because \text{Adjacent angles in a parallelogram are supplementary})$

$\therefore \angle S=180°-110°=70°$

Now,

$\angle Q=\angle S=70°(\because \text{Opposite angles in a parallelogram are equal})$

Now, in $△BOQ$,

$\angle OBQ+\angle OQB+\angle BOQ=180°\left(\text{Angle sum property of a triangle}\right)$

$\therefore 60°+70°+x=180°$

$\Rightarrow x=180°-130°=50°$

Hence the value of $x$ is $50°$.