Explanation:



Since |PR| = |RS| = |SP|
\(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70^o
But < PQR + < PSR = 180^o (Opposite interior angles of a cyclic quadrilateral)
< PQR + 60 = 180^o
< PR = 180 - 60 = 120^o
But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)
< QPR + < PRQ + < PQR = 180^o (Angles in a triangle)
2 < QPR + 120 = 18-
2 < QPR = 180 - 120
QPR = \(\frac{60}{2}\) = 30^o
From the diagram, < QRS = < PRQ + < PRS
30 + 60 = 90^o