Answer:
a = 125°, b = 20°, c = 35°, d = 90°, e = 55°
Step-by-step explanation:
Quadrilateral PQRS is inscribed in a circle, therefore it is a cyclic quadrilateral.
PQ is diameter, SR is chord and PR is transversal such that:
PQ || SR... (given)
[tex] m\angle PRQ = 90°..(\angle \: inscribed \: in\: semicircle) \\
\huge \red {\boxed {\therefore d = 90°}} \\\\
m\angle RPQ= m\angle PRS .. (alternate \: \angle s) \\
\huge \purple {\boxed {\therefore c = 35°}} \\\\
In\: \triangle PQR, \\
c + d + e = 180°\\
35° + 90° + e = 180°\\
125° + e = 180°\\
e = 180° - 125°\\
\huge \orange {\boxed {\therefore e = 55°}} \\\\
a + e = 180°...(opposite \:\angle 's \: of \: cyclic \: quadrilateral) \\
a + 55°= 180°\\
a = 180°- 55°\\
\huge \blue {\boxed {a = 125°}} \\\\
In\: \triangle PSR, \\
a + b + 35°= 180°\\
125° + b + 35° = 180°\\
160° + b = 180°\\
b = 180° - 160°\\
\huge \pink {\boxed {b = 20°}}
[/tex]
