WAEC - Mathematics (1993 - No. 19)

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

34^o

48^o

56^o

68^o

73^o

< 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°