Answer:
[tex]\sin (P)=\frac{5}{13}[/tex][tex]\tan (T)=\frac{12}{5}[/tex][tex]\cos (T)=\frac{5}{13}[/tex]
Explanation:
Given the below right triangle;
We can go ahead and determine the below trig ratios as seen below;
For sin(P), we'll have;
[tex]\begin{gathered} \sin (P)=\frac{opposite\text{ side to angle P}}{\text{Hypotenuse}}=\frac{5}{13} \\ \therefore\sin (P)=\frac{5}{13} \end{gathered}[/tex]
For tan(T), we'll have;
[tex]\begin{gathered} \tan (T)=\frac{opposit\text{e side to angle T}}{\text{adjacent side to angle T}}=\frac{12}{5} \\ \therefore\tan (T)=\frac{12}{5} \end{gathered}[/tex]
For cos(T), we'll have;
[tex]\begin{gathered} \cos (T)=\frac{adjacent\text{ side to angle T}}{\text{hypotenuse}}=\frac{5}{13} \\ \therefore\cos (T)=\frac{5}{13} \end{gathered}[/tex]
