Quadrilateral PQRS is a general quadrilateral. The diagonal QS is drawn. Find the sum of the interior angles of PQRS. Also find the sum of the interior angles of triangles PQS and QSR. Express the rel

Here's the solution to question 4:

a) The sum of the interior angles of any quadrilateral is $360^\circ$.
The interior angles of quadrilateral PQRS are $\angle P$, $\angle Q$, $\angle R$, and $\angle S$.

$$\angle P + \angle Q + \angle R + \angle S = 360^\circ$$

The measure of all the interior angles of the given quadrilateral is $\boxed{\text{360 degrees}}$.

b) The sum of the interior angles of any triangle is $180^\circ$.
The interior angles of triangle PQS are $\angle P$, $\angle PQS$, and $\angle PSQ$.

$$\angle P + \angle PQS + \angle PSQ = 180^\circ$$

The measure of all the interior angles of triangle PQS is $\boxed{\text{180 degrees}}$.

c) The sum of the interior angles of any triangle is $180^\circ$.
The interior angles of triangle QSR are $\angle R$, $\angle RQS$, and $\angle QSR$.

$$\angle R + \angle RQS + \angle QSR = 180^\circ$$

The measure of all the interior angles of triangle QSR is $\boxed{\text{180 degrees}}$.

d) From the diagram, the angle $\angle PQR$ is formed by the sum of angles $\angle PQS$ and $\angle RQS$.
The relation is:

$$\angle PQR = \angle PQS + \angle RQS$$

e) Assuming the typo "$\angle PSE$" should be "$\angle PSQ$", the angle $\angle PSR$ is formed by the sum of angles $\angle PSQ$ and $\angle QSR$.
The relation is:

$$\angle PSR = \angle PSQ + \angle QSR$$