Question

In the given figure, PQRS is a rectangle and its diagonals PR and QS intersect at O. If $\angle POQ={110}^o$, find the measure of $\angle PSQ$.

Answer 1

To find $\angle PQO$ and $\angle PSQ$.

PQRS is a rectangle, and O is the intersection point of diagonals PR and SQ.

$PR=SQ$ [Diagonals of rectangle are equal]

$PO=Qo$ [Diagonals of rectangle bisect each other]

$\therefore \angle PQO=\angle OPQ\to \left(1\right)$ [Angles opposite to equal sides]

In $△POQ$ ,

$\angle PQO+\angle POQ+\angle OPQ={180}^{\circ }$

$2\angle PQO+{110}^{\circ }={180}^{\circ }$ [From (1)]

$\angle PQO=\frac{{180}^{\circ }-{110}^{\circ }}{2}={35}^{\circ }$

Now, in $△PQS$

$\angle PQS+\angle QPS+\angle PSQ={180}^{\circ }$

${35}^{\circ }+{90}^{\circ }+\angle PSQ={180}^{\circ }$

$\angle PSQ={180}^{\circ }-{125}^{\circ }={55}^{\circ }$