In the figure PQRS is a circle with ST RQ Find ... - JAMB Mathematics 1999 Question

< PST = < RQT = x(corresponding angle)

< PST = < PTS = < PTS x (PS = PT0

< SRQ = < SPT = 180^o (sum of < on straight line)

< SPT = 180^o - 110^o = 70^o

in < SPT, < PST = PTS = < PSt = 180^o

2x + 70 = 180^o

2x = 180^o - 70^o = 110^o

x = \(\frac{110^o}{2}\) = 55^o