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: Option A
Explanation
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)
There is an explanation video available below.