2026 TPSS P1 Q6

2026 TPSS P1 Q6

Secondary 4
4 marks

In the diagram, \(PRS\) is a straight line. \(P\) is due north of \(Q\). \(PQ=12\text{ cm}\), \(QR=35\text{ cm}\) and angle \(PQR\) is \(90^\circ\).

  1. Find the value of \(\cos\angle QRS\), leaving your answer as a fraction in its simplest form.[2]
  2. Find the bearing of \(R\) from \(P\).[2]

Solution:

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Answer:(a) \(-\dfrac{35}{37}\) (b) \(108.9^\circ\)

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