Question on: WAEC Mathematics - 2008

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