Question

On the unit circle, the point P(-5/13, 12/13) lies on the terminal arm of an angle in standard position?

Answer 1

The angle is `1.9656` radians or `112.62^@`
(assuming that is what the question intended to ask).

Explanation:

In standard position with the unit length terminal arm at `P(-5/13,12/13)`
implies
`color(white)("XXXX")` `cos(theta) = -5/13`

which, in turn, implies
`color(white)("XXXX")` `theta = arccos(-5/13)`
which can be solved using a calculator to obtain the result above.

Answer 2

I have no one sentence answer. See below.

Explanation:

If the question was to find the values of the six trigonometric functions, use the definition. Note that
`r = sqrt((-5/13)^2 + (12/13)^2) = sqrt ((25+122)/169) = sqrt (169/169) = 1`

(The point P is on the unit circle.)

Call the angle `theta` (I'll use `r` in the definitions even though it is `1`.

`sin theta = y/r = y/1 = 12/13` `color(white)"sssss"` `csc theta = r/y = 1/y = 1/(12/13) = 13/12`

`cos theta = x/r = x/1 = -5/13` `color(white)"sss"` `sec theta = r/x = 1/x = 1/(-5/13) = -13/5`

`tan theta = y/x = (12/13)/(-5/13) = 12/13 * -13/5 = -12/5`

`cot theta = x/y = (-5/13)/(12/13) = -5/13*13/12 = -5/12`