If  PQRS is a cyclic quadrilateral and  PQ  is diameter, find the value of $\angle PQS$

Concept: Every angle of a cyclic quadrilateral is on a circle (circumscribed circle), and the sum of any two opposite angles of a cyclic quadrilateral is 180˚.

Here, PQRS is a cyclic quadrilateral.

Then, sum of any pair of opposite angles of cyclic quadrilateral is  $\Rightarrow \angle QRS+\angle QPS={180}^{°}$

$\Rightarrow \angle QPS ....\left(I\right)$

Also we have angle in a semicircle, 

$\angle PSQ={90}^{°} \dots .\left(ii\right)$

In $\Delta PSQ,$  by angle sum property of triangle,

$\angle PSQ+\angle SPQ+\angle SQP={180}^{°}$

From equation (i) and (ii), we get

$\Rightarrow {90}^{°}+{70}^{°}+\angle PQS={180}^{°}$

$\Rightarrow \angle PQS={20}^{°}$

Hence, the value of  $\angle PQS$ is ${20}^{°}$  .