In the diagram, PQR is a tangent to the circle QST at Q. ...
In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ∠SQR = 68°, find ∠PQT.
34o
48o
56o
68o
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°
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