and reasoning:

The angles $\theta$ and $\theta’$ are equal as vertically opposite angles:

\begin{equation}
\label{eq:Wed10Oct162927BST2018}
\theta=\theta’
\end{equation}

and they can be both related to the important quantities of the problem as follows:

\begin{align}
\tan\theta=\frac{h}{s-R}\\
\tan\theta’=\frac{h’}{s’-R}\label{eq:Wed10Oct162841BST2018}
\end{align}

On the second line, I note that $h’$ is negative, which is okay as it is an algebraic quantity. Anyway, from the first equation, I can deduce that the magnification $m\equiv\displaystyle\frac{h’}{h}$ is given by

\begin{equation}
m=\frac{s’-R}{s-R}\,.
\end{equation}

However the formula which has been given in class for the magnification is:

\begin{equation}
\label{eq:Wed10Oct163138BST2018}
m=-\frac{s’}{s}\,.
\end{equation}

Can you explain to this student what is going on? (before next tutorial and/or next blog post, where an explanation will be proposed; you are welcome to share your suggestions in the comments below. If you figure it out, it should be clear what is going on).