JAMB - Mathematics (1988 - No. 37)
In triangle PQR, PQ = 1cm, QR = 2cm and PQR = 120o Find the longest side of the triangle
\(\sqrt{3}\)cm
\(\sqrt{7}\)cm
3cm
7cm
Обяснение
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
PR2 = 12 + 22 - 2(1)(2) x - cos 60o
= 5 - 2(1)(2) x -\(\frac{1}{2}\)
= 5 + 2 = 7
PR = \(\sqrt{7}\)cm