Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos…

Mathematics WAEC 2001

Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p – tan p

  • \(\frac{79}{156}\) checkmark
  • \(\frac{85}{156}\)
  • \(\frac{7}{13}\)
  • \(\frac{8}{1}\)

The correct answer is: A

Explanation

If \(sin P = \frac{5}{13}\) from right angled triangle from pythagoras theorem
\(BC^2 = 13^2 - 5^2\\
=169-25\\
BC = \sqrt{144} = 12\\
∴ cos P - tan P = \frac{12}{13} - \frac{5}{12}\\
=\frac{79}{156}\)
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