In the figure POQ is the diameter of the circle PQR. If PSR = 145o, find xo
A.
25o
B.
35o
C.
45o
D.
125o
E.
55o
Correct answer is E
< PRQ = \(\frac{1}{2}\) < POQ = 90o
< PSR + < PQR = 180o
< PQR = 180o - 145o = 35o
\(\bigtriangleup\)PQR is a right angled triangle
x = 90 - < PQR
= 90o - 35o
= 55o