WAEC - Mathematics (2001 - No. 26)
Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
\(\frac{79}{156}\)
\(\frac{85}{156}\)
\(\frac{7}{13}\)
\(\frac{8}{1}\)
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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