The ancient Greek astronomer and mathematician Aristarchus wanted to know how far away the Sun and Moon were. He reasoned that when the phase of the Moon was exactly half full, the Sun, Moon, and Earth would form a right triangle. He measured the angle between the Sun and half-full Moon in the sky to be $87^\circ$.

Using this measurement, how many times farther away is the Sun than the Moon? If you don’t have a calculator, use the small angle approximation $\sin\theta\approx \theta$ (where $\theta$ is measured in radians). Show answer

Let $m$ and $s$ be the distance to the Sun and Moon, respectively. Then, \begin{align*} \frac{s}{m} &= \frac{1}{\cos 87^\circ}\\ &\approx 19.1. \end{align*} (If you’re not using a calculator, you could say $\cos 87^\circ = \sin3^\circ \approx 3\pi/180$. If you’re not too concerned with accuracy, you could use $\pi\approx 3$ and get $s/m \approx 20$, or you could say $3\pi \approx 10$ and get $18$, or if you want more accuracy, you can use $\pi\approx 22/7$ and get $19\frac{1}{11}$.)

So according to Aristarchus, the Sun is about $19$ times farther away than the Moon, and is therefore $19$ times bigger than the Moon, since they look the same size in the sky.

Aristarchus’ measurement was not very accurate. The real angle is about $89^\circ51’$ (it varies a little over the years). Use this to find the true ratio between the distance of the Sun and Moon. Show answer

\begin{align*} \frac{s}{m} &= \frac{1}{\cos 89^\circ51’}\\ &\approx 382 \end{align*} (Without a calculator, $\cos 89^\circ51’ = \sin 9’ \approx \frac{9}{60} \times \frac{\pi}{180}$, which gives $s/m = 400$ if you use $\pi\approx 3$, or $381\frac{9}{11}$ if you use $\pi\approx 22/7$.)

So the Sun is really about $380$ times farther away than the Moon.