In this fiqure, PQ = PR = PS and SRT = 68o o . Find QPS

In this fiqure, PQ = PR = PS and SRT = 68o ${}^o$. Find QPS 

Answer Details

Since PQRS is quadrilateral

2y + 2x + QPS = 360o ${}^o$

i.e. (y + x) + QPS = 360o ${}^o$ 

QPS = 360o ${}^o$ - 2 (y + x)

But x + y + 68o ${}^o$ = 180o ${}^o$

There; x + y = 180o ${}^o$ - 68o ${}^o$ = 112o ${}^o$

QPS = 360 - 2(112o ${}^o$)

= 360o ${}^o$ - 224 = 136o