If $\angle PQR$ of cyclic quadrilateral $PQRS$ is $110^{\circ}$, then $\angle PSR$ is—

Question

Accepted Answer

$70^{\circ}$

Answer

$110^{\circ}$

Answer

$90^{\circ}$

Answer

$55^{\circ}$

If ∠PQR\angle PQR∠PQR of cyclic quadrilateral PQRSPQRSPQRS is 110∘110^{\circ}110∘, then ∠PSR\angle PSR∠PSR is—