In this fiqure, PQ = PR = PS and SRT = 68\(^o\). Find QPS
A.
136\(^o\)
B.
124\(^o\)
C.
112\(^o\)
D.
68\(^o\)
Correct answer is A
Since PQRS is quadrilateral
2y + 2x + QPS = 360\(^o\)
i.e. (y + x) + QPS = 360\(^o\)
QPS = 360\(^o\) - 2 (y + x)
But x + y + 68\(^o\) = 180\(^o\)
There; x + y = 180\(^o\) - 68\(^o\) = 112\(^o\)
QPS = 360 - 2(112\(^o\))
= 360\(^o\) - 224 = 136\(^o\)