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I'd like to find the distance at which a 2.4 cm coin is subtended by an angle of 10". I've drawn my diagram and found that $D=\dfrac{d}{\alpha}$ using the small angle approximation. Since $10" = 4.85\times10^{-5}$ rad then $ D = \dfrac{2.4 \times 10^{-2} m}{4.85 \times 10^{-5}rad} = 495$
Are the units of distance in this calculation meters? In other words why would the units of radians be 'dropped'?
The units of radians are 'dropped' because unlike most units, they are dimensionless. Recall that the definition of radian angle measure is the ratio of the length of a circular arc to its radius. Thus, radian angle measures have units of $[\mathrm{Length}]/[\mathrm{Length}] \equiv 1$, i.e. dimensionless.
Radians are units in the sense that they give information about what standard the angle quantity is measured by. Thus, the convention of writing $\mathrm{rad}$ is useful in that it distinguishes it from other ways of measuring angles (e.g., in your question, arc-seconds), but in terms of dimensional analysis, $\mathrm{rad} \equiv 1$, so $\mathrm{m}/\mathrm{rad} = \mathrm{m}$.
The equation you are using is:
$$ \tan(\theta) = \frac{d}{D} $$
which in the small angle approximation $\tan(\theta) \sim \theta$. But you have to take into account that $\theta$ has units of radians, where:
$$ 1 \: radian = 3600 \times \frac{180}{\pi} \sim 206265 \: arcsecs. $$
Then, the small angle approximation formula is:
$$ \frac{\theta}{206265} = \frac{d}{D} \: \Rightarrow \: \theta_{arcsecs} = 206265 \times \frac{d}{D}, $$
provided you use the same units for $d$ and $D$.
