A diagram of the right triangle ABC with right angle at C is shown in the following image:
We are going to assume that side a is opposite to angle A, side b is opposite to angle B, and that side c is opposite to angle C:
We are asked to find sinA, cosA, and tanA.
First, let's remember the definitions for sine, cosine, and tangent for any angle X:
[tex]\begin{gathered} \sin X=\frac{opposite\text{ side}}{hypotenuse} \\ \cos X=\frac{\text{adjacent side}}{\text{hypotenuse}} \\ \tan X=\frac{opposite\text{ side}}{adjacent\text{ side}} \end{gathered}[/tex]In this case, for angle A:
The opposite side to angle A is a=8
The adjacent side to angle A is b=15
The hypotenuse is c=17.
Substituting these values to find the sine of A:
[tex]\begin{gathered} \sin A=\frac{opposite\text{ side}}{hypotenuse} \\ \sin A=\frac{8}{17} \end{gathered}[/tex]We do something similar to find the cosine of A:
[tex]\begin{gathered} \cos A=\frac{\text{adjacent side}}{\text{hypotenuse}} \\ \cos A=\frac{15}{17} \end{gathered}[/tex]And finally, the tangent of A is:
[tex]\begin{gathered} \tan A=\frac{opposite\text{ side}}{adjacent\text{ side}} \\ \tan A=\frac{8}{15} \end{gathered}[/tex]Answer:
[tex]\begin{gathered} \sin A=\frac{8}{17} \\ \cos A=\frac{15}{17} \\ \tan A=\frac{8}{15} \end{gathered}[/tex]
