Find angle a,b,c ,d and e

The angle $30^\circ$ and angle a are vertically opposite angles. Vertically opposite angles are equal.
Therefore, a = $30^\circ$.

The line segment from the center to the vertex P and the line segment from the center to the vertex R are both radii of the circle. Thus, triangle OPR is an isosceles triangle.
The angle $30^\circ$ is given. The angle at the center subtended by the arc PR is $2 \times 30^\circ = 60^\circ$ (angle at center is twice the angle at circumference subtended by the same arc).
In triangle OPR, the sum of angles is $180^\circ$. So, angle OPR + angle ORP + angle POR = $180^\circ$.
Since triangle OPR is isosceles with OP = OR, angle OPR = angle ORP.
Let angle OPR = angle ORP = x.
x + x + $60^\circ$ = $180^\circ$
2x = $180^\circ - 60^\circ$
2x = $120^\circ$
x = $60^\circ$.
So, angle OPR = $60^\circ$.
Angle b is the same as angle OPR.
Therefore, b = $60^\circ$.

Angle c and angle b are angles on a straight line. The sum of angles on a straight line is $180^\circ$.
c + b = $180^\circ$
c + $60^\circ$ = $180^\circ$
c = $180^\circ - 60^\circ$
c = $120^\circ$.

The line segment from the center to the vertex P and the line segment from the center to the vertex Q are both radii of the circle. Thus, triangle OPQ is an isosceles triangle.
The angle $a^\circ$ is given as $30^\circ$. This is angle OPQ.
Since triangle OPQ is isosceles with OP = OQ, angle OPQ = angle OQP.
Therefore, angle OQP = $30^\circ$.
Angle d is the same as angle OQP.
So, d = $30^\circ$.

Angle e is part of the angle at the center subtended by the arc PQ.
In isosceles triangle OPQ, angle POQ + angle OPQ + angle OQP = $180^\circ$.
Angle POQ + $30^\circ$ + $30^\circ$ = $180^\circ$.
Angle POQ + $60^\circ$ = $180^\circ$.
Angle POQ = $180^\circ - 60^\circ$.
Angle POQ = $120^\circ$.
The angle $120^\circ$ is the angle at the center subtended by arc PQ.
Angle e is the angle at the circumference subtended by the arc PQ.
The angle at the circumference is half the angle at the center subtended by the same arc.
e = (1/2) × Angle POQ
e = (1/2) × $120^\circ$
e = $60^\circ$.

Summary of angles:
a = $30^\circ$
b = $60^\circ$
c = $120^\circ$
d = $30^\circ$
e = $60^\circ$