$AB$ is the diameter of a circle. $\angle PAB = 35^\circ$, $\angle PQB = 25^\circ$.
- Given: $AB$ is the diameter of the circle.
$\therefore \;$ $\angle APB = 90^\circ$ $\;\;\;$ [angle in a semi-circle]
$\angle APB + \angle BPQ = 180^\circ$ $\;\;\;$ [linear pair]
$\therefore \;$ $\angle BPQ = 180^\circ - \angle APB$
i.e. $\;$ $\angle BPQ = 180^\circ - 90^\circ = 90^\circ$
In $\triangle BPQ$,
$\angle BPQ + \angle PQB + \angle QBP = 180^\circ$ $\;\;\;$ [sum of angles of a triangle equal $180^\circ$]
i.e. $\;$ $\angle QBP = 180^\circ - \left(\angle BPQ + \angle PQB\right)$
Given: $\angle PQB = 25^\circ$
$\therefore \;$ $\angle QBP = 180^\circ - \left(90^\circ + 25^\circ\right) = 65^\circ$
$ARBP$ is a cyclic quadrilateral.
$\therefore \;$ $\angle PAR = \angle QBP$
[external angle of a cyclic quadrilateral is equal to internal opposite angle]
$\therefore \;$ $\angle PAR = 65^\circ$
But $\;$ $\angle PAR = \angle PAB + \angle RAB$
$\therefore \;$ $\angle RAB = \angle PAR - \angle PAB$
Given: $\;$ $\angle PAB = 35^\circ$
i.e. $\;$ $\angle RAB = 65^\circ - 35^\circ = 30^\circ$ - $\angle PRB = \angle PAB$ $\;\;\;$ [angles in the same segment of a circle are equal]
$\therefore \;$ $\angle PRB = 35^\circ$ - $\because \;$ $AB$ is the diameter of the circle.
$\therefore \;$ $\angle ARB = 90^\circ$ $\;\;\;$ [angle in a semi-circle] - $\angle ARB = \angle PRA + \angle PRB$
$\therefore \;$ $\angle PRA = \angle ARB - \angle PRB$
i.e. $\;$ $\angle PRA = 90^\circ - 35^\circ = 55^\circ$