speed of light - Special relativity and missing factors of $c$

I am doing problems in a textbook called 'Introduction to Classical Mechanics' by David Morin. In one of the questions it says the following:

In the lab frame, two particles move with speed $v$ along the path shown

The angle between the trajectories is $2\theta$. What is the speed of one particle, as viewed by the other?

I know how to do this question but what has confused me in the answers he has left out all of the c's. Here is his method

Consider the frame S' that travels along the point P midway between the particles. S' moves at speed $v cos\theta$, so the $\gamma$ factor relating it to the lab frame is: $$\gamma =\frac{1}{\sqrt{1-v^2cos^2\theta}}$$

From what I have learned there should be a $v^2/c^2$ instead of just $v^2$. He has repeated a similar method 4 times in the book (from what I can see) all without the $c^2$, which makes me think that I have got something wrong thinking their should be a c. Please explain thanks.

He then goes onto use the velocity addition formula like this:

$$V=\frac{2u'_y}{1+u^{'2}_{y}}$$

Answer

The beginning of the chapter, the author does use $c$ for the Lorentz transformations (cf., Equation (13.1)). $$\begin{align} A_0&=\gamma(A_0'+(v/c)A_1') \\ A_1&=\gamma(A_1'+(v/c)A_0') \\ A_2&=A_2'\\ A_3&=A_3' \end{align}$$

Shortly after Equation (13.1), the author lists several enumerated remarks. In particular is #4:

4. Lest we get tired of writing the $c$'s over and over, we'll work in units where $c=1$ from now on.

And then he re-writes the Lorentz transformations in the new units in Equation (13.2), $$\begin{align} A_0&=\gamma(A_0'+vA_1') \\ A_1&=\gamma(A_1'+vA_0') \\ A_2&=A_2'\\ A_3&=A_3' \end{align}$$