Solution:
Join \[OP,\text{ }OQ,\text{ }OR\text{ }and\text{ }OS\]
Given, \[PQ\text{ }=\text{ }QR\text{ }=\text{ }RS\]
So, \[\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS\]
[Equal chords subtends equal angles at the centre]
\[Arc\text{ }PQRS\]subtends \[\angle POS\] at the centre and \[\angle PTS\]at the remaining part of the circle.
Hence,
\[\angle POS\text{ }=\text{ }2\text{ }x\angle PTS\]
\[=\text{ }2\text{ }x\text{ }{{75}^{o}}~=\text{ }{{150}^{o}}\]
Or,
\[\angle POQ\text{ }+\angle QOR\text{ }+\angle ROS\text{ }=\text{ }{{150}^{o}}\]
\[\angle POQ\text{ }=\angle QOR\text{ }=\angle ROS\]
So,
\[=\text{ }{{150}^{o}}/\text{ }3\text{ }=\text{ }{{50}^{o}}\]
In \[\vartriangle OPQ\] we have,
\[OP\text{ }=\text{ }OQ\][Radii of the same circle]
So, \[\angle OPQ\text{ }=\angle OQP\]
[Angles opposite to equal sides are equal]
By angle sum property of \[\vartriangle OPQ\]
\[\angle OPQ\text{ }+\angle OQP\text{ }+\angle POQ\text{ }=\text{ }{{180}^{o}}\]
\[\angle OPQ\text{ }+\angle OQP\text{ }+\text{ }{{50}^{o}}~=\text{ }{{180}^{o}}\]
\[\angle OPQ\text{ }+\angle OQP\text{ }=\text{ }{{130}^{o}}\]
Or,
\[2\angle OPQ\text{ }=\text{ }{{130}^{o}}\]
\[\angle OPQ\text{ }=\angle OPQ\]
So,
\[=\text{ }{{130}^{o}}/\text{ }2\text{ }=\text{ }{{65}^{o}}\]
Similarly, we can prove that
In \[\vartriangle OQR\]
\[\angle OQR\text{ }=\angle ORQ\text{ }=\text{ }{{65}^{o}}\]
And in \[\vartriangle ORS\]
\[\angle ORS\text{ }=\angle OSR\text{ }=\text{ }{{65}^{o}}\]
So,
(i) \[\angle POS\text{ }=\text{ }{{150}^{o}}\]
(ii) \[\angle QOR\text{ }=\text{ }{{50}^{o}}~\]