JAMB - Mathematics (1983 - No. 6)
In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS
2\(\sqrt{3}\)cm
4\(\sqrt{6}\)cm
2\(\sqrt{6}\)cm
8\(\sqrt{6}\)cm
8cm
Explanation
From the diagram, sin 60o = \(\frac{PR}{8}\)
PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\)
= 4\(\sqrt{3}\)
Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\)
PS Cos45o = 4\(\sqrt{3}\)
PS = 4\(\sqrt{3}\) x 2
= 4\(\sqrt{6}\)
PR = 8 sin 60 = \(\frac{8\sqrt{3}}{2}\)
= 4\(\sqrt{3}\)
Cos 45o = \(\frac{PR}{PS}\) = \(\frac{4 \sqrt{3}}{PS}\)
PS Cos45o = 4\(\sqrt{3}\)
PS = 4\(\sqrt{3}\) x 2
= 4\(\sqrt{6}\)
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