Why is this true? $\frac{x}{b}=cos(180 -\theta) \implies x=-b\ cos\ \theta$

Question

So in my textbook, this is part of a proof for the law of cosines, where two triangles share a side b, and are aligned on a straight line c + x. One triangle has the angle $\theta $ and another has the angle $\alpha =180-\theta$ between b and c, and b and x respectively. Which supposedly leads to the following conclusion: $$\frac{x}{b}=cos(180 -\theta) \implies x=-b\ cos\ \theta$$

I don't understand the logic behind this however, since $180-\theta$ must result into a positive number, as$\theta $ cannot be greater than 180. So why does $\frac{x}{b}= cos(180-\theta)$ imply $x=-b\ cos\ \theta$? What am I missing here?

Answer 1

Note that $90^\circ<\theta<180^\circ\implies\cos(\theta)<0$. On the other hand,$$\cos(180^\circ-\theta)=-\cos(-\theta)=-\cos(\theta).$$

Answer 2

Note that since $\cos(x)$ is an odd-numbered harmonics function (i.e. $f(x+{T\over 2})=-f(t)$ where $T$ is the fundamental period) then we have $$\cos(x+\pi)=\cos(x-\pi)=-\cos x$$The above relation could also be checked in two other different ways. The first is:$$\cos(a+b)=\cos a\cos b-\sin a\sin b$$the second, is the position of $\cos x$ and $\cos(\pi-x)$ on the unit circle. Finally since $\cos(\cdot)$ is an even function, we conclude$$\cos(x-\pi)=-\cos x\implies \cos(\pi-x)=-\cos x$$