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:
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/5hf25yulcjq14vinjmtikb2xvh9bj8t04a.png)
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:
![\begin{gathered} \sin A=\frac{opposite\text{ side}}{hypotenuse} \\ \sin A=(8)/(17) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/trclxx0jlp0hnx4h9zc3m5uk7sg0bjnibx.png)
We do something similar to find the cosine of A:
![\begin{gathered} \cos A=\frac{\text{adjacent side}}{\text{hypotenuse}} \\ \cos A=(15)/(17) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4jawqh3rl3wlop03w7kuv7gn64nau2qtj3.png)
And finally, the tangent of A is:
![\begin{gathered} \tan A=\frac{opposite\text{ side}}{adjacent\text{ side}} \\ \tan A=(8)/(15) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vuvx9krscshv6pc11d1b0q1nuehm6i74gl.png)
Answer:
![\begin{gathered} \sin A=(8)/(17) \\ \cos A=(15)/(17) \\ \tan A=(8)/(15) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4qvsgueoce7j0hxik0d1hx0q3sqr72pc55.png)