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

A.

\(\sqrt{3}\)cm

B.

\(\sqrt{7}\)cm

C.

3cm

D.

7cm

View answer & explanation

Correct answer is B

PR2 = PQ2 + QR2 - 2(QR)(PQ) COS 120o

PR2 = 12 + 22 - 2(1)(2) x - cos 60o

= 5 - 2(1)(2) x -\(\frac{1}{2}\)

= 5 + 2 = 7

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

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