Question

Question #89e00

Answer 1

`cos A = 12/13` and `tan A = 5/12`.

Explanation:

Let's say that we have `triangleABC` such that segments `AB` and `BC` are the legs and segment `AC` is the hypotenuse. Thus, if `sin A = 5/13`, we know that the length of the opposite leg divided by the length of the hypotenuse, or `(BC)/(AC)`, is `5/13`. Next, we can use the Pythagorean theorem to find the equation `AB^2 + BC^2 = AC^2`. Since we know `BC` and `AC` are in the ratio `5/13`, we have `AB^2 + (5x)^2 = (13x)^2`. Solving (note the `5:12:13` special right triangle) gives us `AB = 12x`.

Thus, using the definitions of cosine and tangent, we have `cos A = (12x)/(13x) = 12/13` and `tan A = (5x)/(12x) = 5/12`.

Answer 2

`sin A = 5/13`
Using trig identity, we get -->
`cos^2 A = 1 - sin^2 A = 1 - 25/169 = 144/169`
`cos A = +- 12/13`.
Since A is acute, therefor, cos A is positive.
`cos A = 12/13`
`tan A = sin A/(cos A) = (5/13)(13/12) = 5/12`