In the given right triangle,
BC=5
AC=12.
Hypotenuse of the triangle, AB=13.
Now, the ratio of sin A can be expressed as,
![\sin A=\frac{opposite\text{ side}}{hypotenuse}](https://img.qammunity.org/2023/formulas/mathematics/college/y1scz5ybgpq4alfr9qi26le192yz0patpc.png)
The opposite side to angle A is BC.
Hence,
![\begin{gathered} \sin A=(BC)/(AB) \\ \sin A=(5)/(13) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bxtp17z4zwpvabtrjepi61l88deknrkvlq.png)
The ratio of cos A can be expresssed as,
![\cos \text{ A=}\frac{\text{adjacent side}}{hypotenuse}](https://img.qammunity.org/2023/formulas/mathematics/college/p7z9kamhqe2hsjveuzzk0oa0z2gx0i6lj6.png)
The side adjacent to angle A is AC.
Hence,
![\begin{gathered} \cos \text{ A=}(AC)/(AB) \\ \cos \text{ A=}(12)/(13) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/r59294b4r9xejkvxov6g2064hl7n16lgj4.png)
The ratio tan A can be expressed as,
![\begin{gathered} \tan \text{ A=}\frac{\text{opposite side}}{adjacent\text{ side}} \\ \tan \text{ A=}(BC)/(AC) \\ \tan \text{ A=}(5)/(12) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/95ib0ifptd68knbrdiegzie9gbu1m0s360.png)
Therefore, sin A=5/13, cos A=12/13 and tan A=5/12.