A trigonometric question

There is this question in the test I did today. It goes something like this:

Given that tan x = 1/p and that x is not acute, p > 0, find

(i)sin(-x)

(ii)sin(90-x)

What I did was I drew out the quadrant that x was in (in this case, the 3rd quadrant) Next I find the Hypotenuse using sqrt (1 + p^2)

So that makes sin(-x) = -1/ sqrt(1 +p^2)... is this correct or is it wrong?

For (ii) sin(90-x) = cos x

= - p / sqrt(1 +p^2)

Please tell me whether there are any mistakes. I ain't sure how to go about doing this question.

don't see anything wrong.

for tan x = 1/p, you can use your triangle method to find that the magnitude of sin x = 1 / sqrt(1 +p^2)

but because x is not acute, and 1/p > 0, x lies in the 3rd quadrant, 180deg <x<270 deg

So for that quadrant, sin x is negative. This gives us sin x = -1 / sqrt(1 +p^2)

Since sin (-x) = -sin x,
sin (-x) = 1 / sqrt(1 +p^2)

Note: It is positive

sin (90 - x) = cos x => correct
From your same triangle, find that the magnitude of cos x = p / sqrt(1 +p^2)

However, since x is in the 3rd quadrant,
cos x = -p / sqrt(1 +p^2)

Ganbatte,
Eagle (Strategic Tuition)