In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS
150o
120o
90o
60o
Correct answer is C
Since |PR| = |RS| = |SP|
\(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70o
But < PQR + < PSR = 180o (Opposite interior angles of a cyclic quadrilateral)
< PQR + 60 = 180o
< PR = 180 - 60 = 120o
But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)
< QPR + < PRQ + < PQR = 180o (Angles in a triangle)
2 < QPR + 120 = 18-
2 < QPR = 180 - 120
QPR = \(\frac{60}{2}\) = 30o
From the diagram, < QRS = < PRQ + < PRS
30 + 60 = 90o
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