Question

In the figure, $POQ$ is a diameter abd $OQRS$ is a cyclic quadrilateral. If $\angle PSR={150}^o$, find $\angle RPQ$.

Answer 1

We know that $PQRS$ is a cyclic quadrilateral

It can be written as

$\angle PSR+\angle PQR={180}^o$

By substituting the values

${150}^o+\angle PQR={180}^o$

On further calculation

$\angle PQR={180}^o-{150}^o$

By subtraction

$\angle PQR={30}^o$

We know that the angle in semi-circle is a right angles

$\angle PRQ={90}^o$

Consider $△PRQ$

Using the angle sum property

$\angle PQR+\angle PRQ+\angle RPQ={180}^o$

By substituting the values

${30}^o+{90}^o+\angle RPQ={180}^o$

On further calculation

$\angle RPQ={180}^o-{30}^o-{90}^o$

By subtraction

$\angle RPQ={180}^o-{120}^o$

So we get

$\angle RPQ={60}^o$

Therefore, $\angle RPQ={60}^o$.