How can trigonometric functions be negative? I cannot understand why $\cos(180-\theta)$  say is $-\cos\theta$. This is probably because my teacher first introduced trigonometry in triangles. I do not understand it for obtuse angles because I cannot think of them in a right triangle. 
I realised that I couldn't feel what I had read "in my spleen" when I was looking at the proof for the law of cosines in an obtuse-angled triangle. I have spent quite some time thinking about how the "$-\cos\theta$" entered the derivation. I cannot fully understand, why the negatives which work in the $XY$-plane work in triangles. For instance, since in a triangle, all the sides are positive while taking the ratio of sides we do not get any negative values but how then does $\cos 120^{\circ}=-0.5$. My brain is in a mess right now. I would appreciate it if someone could help me out or suggest something that I can do.
Let me illustrate what I can't get around.

It is given that in the triangle $\angle BAC=120$ degrees,$|AC|=3$ and that D is the foot of the perpendicular from C to BD. Then $\cos\angle BAC=-0.5=\dfrac{AD}{AC} \implies AD=-1.5$ ?

For right angle triangles, we all know that values of $sin (\theta)$, $cos (\theta)$ and $tan (\theta)$ works when $0\le\theta\le90$.
We do not touch any angles beyond 90 in right angle triangle trigonometry as total angles in a triangle is 180 degrees.
Now consider these triangles (not to scale):

Where triangle A and Triangle B have different angles labelled.
And we name angle c=$\theta$ and angle d=$(180-\theta)$
Now notice that the cosine rule is: $C^2=B^2+A^2-2\cdot A\cdot B\cdot cos(C)$ for any non-right angle triangle like this one:

Now back to the topic. Notice that for those two triangles (Triangle A and Triangle B), the side length D is clearly bigger than C.
Using the cosine rule for those two triangles:
For triangle A: $C^2=A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)$ (as $c=\theta$)
For Triangle B: $D^2=A^2+B^2-2\cdot A\cdot B\cdot cos(180-\theta)$ (as $d=180-\theta)$
Now you know that $cos(180-\theta) =-cos(\theta)$, we can have for Triangle B: $D=\sqrt{A^2+B^2-(2\cdot A\cdot B\cdot -cos(\theta))}=\sqrt{A^2+B^2+2\cdot A\cdot B\cdot cos(\theta)}$
Compared to triangle A for side C, $C=\sqrt{A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)}$
Now we know that the formula for both triangles A and B that the side D is bigger than side C and the formulas derived are:
$C=\sqrt{A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)}$ for angle $c=\theta$ 
$D=\sqrt{A^2+B^2+2\cdot A\cdot B\cdot cos(\theta)}$ for $d=180-\theta$
And we make sense that $\sqrt{A^2+B^2+2\cdot A\cdot B\cdot cos(\theta)}$ is clearly bigger than $\sqrt{A^2+B^2-2\cdot A\cdot B\cdot cos(\theta)}$
That is one of the reasons why $cos(180-\theta)=-cos(\theta)$
