How to find the direction cosines and direction angles for a given line Given the line $5x – 12y = 60$    What are the direction angles and direction cosines?
From my notes, the direction cosines as the $x, y$-components of the unit vector having the same slope as the line.
Slope vector is $\langle 12, 5 \rangle$ and converting to a unit vector: $\bigl\langle \frac{12}{13}, \frac{5}{13} \bigr\rangle$.
Using $\arccos$ with these two values, I get $22.6^\circ$ and $67.4^\circ$ respectively.
Since the line has a positive slope, I can picture a positive acute angle that the line would make with the $x$-axis, so $22.6^\circ$ makes sense.
But looking at the angle made with the y axis, it should be obtuse so it makes sense to subtract $67.4$ from $180$, yielding $180^\circ - 67.4^\circ = 112.6^\circ$. This aligns with the correct answer provided.
However, the answer sheet gave the direction $\cos$ for $y$ as $-\frac{5}{13}$ which would get the $112.6^\circ$ answer automatically. But how do I know when to include a negative, when the slope is positive? 
 A: $y=\frac{5}{12}x-5$ is your formula in slope-intercept form, so your slope is indeed positive.
However, there is a gap within your reasoning, and that is the fact that the y-axis direction cosine is not actually about the y-value, but rather about the rotation to the y-axis. Slope is actually about the y-value, rather than rotation to the axis.
A line with a positive slope will form an obtuse angle with the y-axis, and cosines of obtuse angles are always negative. A line with a negative slope will form an acute angle with the y-axis, and cosines of acute angles are always positive. 
Which means:
Lines with positive slope have a negative y value for their unit vector, and 
lines with negative slope have a positive y value for their unit vector.
The result of this is that you will need to use a negative version of the y part of the unit vector. Either your notes are slightly off (the cosines are for the unit vector with the negative of the slope of the line), or your textbook/teacher teaches a different "direction cosine" (which is the rarer case). I suggest going to ask your teacher for help if you don't understand. 
Tl;dr/Quick Summary: The unit vector needs to have the negative version of the slope of the line. The reason is because obtuse angles have negative cosines and acute angles have positive cosines.
Note: Everything I just stated may or may not apply to 3d direction cosines. I assumed from your question that you were only learning 2d direction cosines.
Edit: If the textbook was wrong, then you were indeed looking for the angle made with the positive y-axis, and you were right. It would be clearer if you gave a picture or a transcript of the textbook problem.
