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

 

Given: ∠QRS = 110^o

We know that the sum of either pair of opposite angles of a cyclic quadrilateral is 180^o.

So in cyclic quadrilateral PQRS,

$\angle QRS+\angle QPS={180}^o$

$\Rightarrow \angle QPS+{110}^o={180}^o$

$\Rightarrow \angle QPS={70}^o$   ...(i)

We know that the angle subtended by a diameter/semicircle on any point of a circle is 90^o.

$\therefore \angle PSQ={90}^o$   ...(ii)

Now in $\Delta PSQ,$

$\angle PSQ+\angle QPS+\angle SQP={180}^o$    [Angle sum property of a triangle]

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

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

$\Rightarrow \angle PQS={180}^o-{160}^o$

$\Rightarrow \angle PQS={20}^o$

Hence, the measure of $\angle PQS$ is ${\mathbf{20}}^{\mathbf{o}}.$