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In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm....

Mathematics
JAMB 1983

In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS

  • A. 2\(\sqrt{3}\)cm
  • B. 4\(\sqrt{6}\)cm
  • C. 2\(\sqrt{6}\)cm
  • D. 8\(\sqrt{6}\)cm
  • E. 8cm
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Correct Answer: Option B
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\(\sqrt{6}\)


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