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
- A. \(\frac{79}{156}\)
- B. \(\frac{85}{156}\)
- C. \(\frac{7}{13}\)
- D. \(\frac{8}{1}\)
Correct Answer: A. \(\frac{79}{156}\)
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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