In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the...

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

The correct answer is: B

PR² = PQ² + QR² - 2(QR)(PQ) COS 120^o

PR² = 1² + 2² - 2(1)(2) x - cos 60^o

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

= 5 + 2 = 7

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