In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ∠SQR = 68°, find ∠PQT.
A.
34o
B.
48o
C.
56o
D.
68o
E.
73o
Correct answer is C
< STQ = < SQR = 68° (alternate segment)
\(\therefore\) < STQ = 68°
< TQS = \(\frac{180° - 68°}{2}\)
= \(\frac{112}{2} = 56°\)
\(\therefore\) < PQT = 180° - (68° + 56°)
= 180° - 124°
= 56°