In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120°. Find the longest side of the triangle

  • A \(\sqrt{3}\)cm
  • B \(\sqrt{7}\)cm
  • C 3cm
  • D 7cm
jamb 1988mathematics

The correct answer is B. \(\sqrt{7}\)cm

\(PR^2 = PQ^2 + QR^2 - 2(QR)(PQ) \cos 120^\circ\)

\(PR^2 = 1^2 + 2^2 - 2(1)(2) \times -\cos 60^\circ\)

\(= 5 - 2(1)(2) \times -\left(\frac{1}{2}\right)\)

\(= 5 + 2 =7\)

\(PR = \sqrt{7} \, cm\)

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