In the diagram, |SP| = |SR| and < PRS = 50o. Calculate < PQR
A.
120o
B.
110o
C.
100o
D.
80o
Correct answer is C
< PRS = RPS = 50o (base of isosceles)
RPS = 50o; < RSP + < PRS + < RPS = 180(sum of < s in a triangle)
< RPS + 50 + 50 = 180
< RSP = 180 - 100 = 80
then < PQR + < RSp = 180 (opp. < S of cyclic quad.)
< PQR + 80 = 180o
< PQR = 180 - 80
= 100