In the figure, $O$ is the center of the circle and $SP$ is the tangent.
$\angle TSP = \angle TSR = 90^\circ$ $\;\;\;$ [angle between the radius of a circle and the tangent at the point of contact]
Given: $\;$ $\angle SRT = 65^\circ$
In right $\triangle TSR$,
$\angle TSR + \angle STR + \angle SRT = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $90^\circ + x + 65^\circ = 180^\circ$ $\implies$ $x = 25^\circ$
Now, $\;$ $OT = OQ$ $\;\;\;$ [radii of the same circle]
$\therefore \;$ $\angle OQT = \angle OTQ = 25^\circ$ $\;\;\;$ [angles opposite equal sides are equal]
In $\triangle OTQ$,
$\angle OTQ + \angle OQT + \angle TOQ = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $25^\circ + 25^\circ + \angle TOQ = 180^\circ$ $\implies$ $\angle TOQ = 130^\circ$
Now, $\;$ $\angle TOQ + \angle OSR = 180^\circ$ $\;\;\;$ [linear pair]
i.e. $\;$ $130^\circ + y = 180^\circ$ $\implies$ $y = 50^\circ$
In right $\triangle OSP$,
$\angle OSP + \angle SPO + \angle POS = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $90^\circ + z + 50^\circ = 180^\circ$ $\implies$ $z = 40^\circ$
$\therefore \;$ $x = 25^\circ, \; y = 50^\circ, \; z = 40^\circ$