Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p
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 = \(\frac{12}{13} - \frac{5}{12}\)
=\(\frac{79}{156}\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}