A. They have negative reciprocal slopes
![\bf cos\left[tan^{-1}\left(\frac{12}{5} \right)+ tan^{-1}\left(\frac{-8}{15} \right) \right]\\ \left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\ \left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta \\\\\\ \textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5} \\\\\\ \textit{so, we're really looking for }cos(\alpha+\beta)](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5Btan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B12%7D%7B5%7D%20%20%5Cright%29%2B%20tan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B-8%7D%7B15%7D%20%20%5Cright%29%20%5Cright%5D%5C%5C%0A%5Cleft.%20%5Cqquad%20%20%5Cqquad%20%20%5Cquad%20%20%20%5Cright.%5Cuparrow%20%5Cqquad%20%5Cqquad%20%20%5Cqquad%20%20%5Cuparrow%20%5C%5C%0A%5Cleft.%20%5Cqquad%20%20%5Cqquad%20%20%5Cquad%20%20%20%5Cright.%5Calpha%20%5Cqquad%20%5Cqquad%20%20%5Cqquad%20%20%5Cbeta%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bthat%20simply%20means%20%7Dtan%28%5Calpha%29%3D%5Ccfrac%7B12%7D%7B5%7D%5Cqquad%20and%5Cqquad%20tan%28%5Cbeta%29%3D%5Ccfrac%7B-8%7D%7B5%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%27re%20really%20looking%20for%20%7Dcos%28%5Calpha%2B%5Cbeta%29)
now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15
ok, well hmm so, the issue boils down to

now, let's take a peek at the second angle, angle β

now, with that in mind, let's use the angle sum identity for cosine
Answer:
1/625, or 0.0016 is the answer.