Values of Angle in Standard Position

I'm having a little bit of difficulty with this question. I think I may have solved it, but I'm not sure about the negative symbols and whether or not they should be included in my calculations.

Determine the values of sin θ, cos θ, tan θ, csc θ, sec θ, and cot θ at

(-3, -4) on the terminal arm of an angle θ in standard position.

I know that if the angle is in standard position the vertex is at (0,0) and that the initial arm is on the positive x-axis. By plotting (-3,-4) on the Cartesian plane I know that the terminal arm is in quadrant III.

Therefore, side a= -4 units and side b= -3 units. I then used the Pythagorean theorem to solve for the hypotenuse and found it to be 5 units.

Solving for each of the trigonometric functions:

sinθ = opp/hyp

sinθ = 4/5

sinθ = 0.8

θ = sin^-1 0.8

θ = 53.1301

** OR, should the 4 be a -4?? **

cosθ = adj/hyp

cosθ = 3/5

cosθ = 0.6

θ = cos^-1 o.6

θ = 53.1301

** OR, should the 3 be a -3?? **

tanθ = opp/adj

tanθ = 4/3

tanθ = 1.333333333

θ = tan^-1 1.333333333

θ = 53.1301

** OR, should the 4 and 3 be a -4 and -3?? **

If I make each of those values negative like they are in the right triangle then the angle measures all come out differently which has me confused.

Then,

cscθ = hyp/opp

cscθ = 5/4

cscθ = 1.25

** Can I solve for θ here??**

secθ = hyp/adj

secθ = 5/3

secθ = 1.666666667

** Can I solve for θ here? **

cotθ = adj/hyp

cotθ = 3/4

cotθ = 0.75

** Can I solve for θ here? **

Therefore, angle θ = 180 + 53.1301 = 233.1301

Am I on the right track here or totally off-base?? Thanks!!