Assuming uncorrelated errors (which is not necessarily the case here, but is fair enough for high school), the error $\delta f$ in the function $f(\theta_i,\theta_r) = \frac{\sin(\theta_i)}{\sin(\theta_r)}$ due to uncertanties $\delta \theta_i$ and $\delta \theta_r$ is
$$\delta f = \sqrt{\left(\frac{\partial f}{\partial \theta_i}\delta\theta_i\right)^2+\left(\frac{\partial f}{\partial \theta_r}\delta\theta_r\right)^2}$$
in this case,
$$\frac{\partial f}{\partial \theta_i} = \frac{\cos(\theta_i)}{\sin(\theta_r)}$$
and
$$\frac{\partial f}{\partial \theta_r} = -\frac{\sin(\theta_i)}{\sin^2(\theta_r)} \cos(\theta_r)$$
This can be simplified somewhat if you note that
$$ \frac{\partial f}{\partial \theta_i} = f(\theta_i,\theta_r) \cdot \cot(\theta_i)$$
and
$$\frac{\partial f}{\partial \theta_r} = -f(\theta_i,\theta_r) \cdot \cot(\theta_r)$$
Putting all of that together and doing a bit of algebra,
$$\frac{\delta f}{f} = \sqrt{\delta\theta_i^2 \cdot \cot^2(\theta_i) + \delta\theta_r^2 \cdot \cot^2(\theta_r)}$$
This quantity is the relative uncertainty in $f$, and can be interpreted as a percent. If $\frac{\delta f}{f} = 0.1$, then your uncertainty in $f$ is 10% of its calculated value. That is, if you measure $f$ to be 1.5, then you would report your measurement as $f = 1.5 \pm 0.15$.