The problem asks for the ratio of the distance from the Earth to the Sun (ES) to the distance from the Earth to the Moon (EM) during a half-moon phase, given a specific angle measurement.

Geometric Setup

We can model the positions of the Sun (S), Earth (E), and Moon (M) as forming a triangle, $\triangle EMS$.

• The problem states it's a half-moon phase, which implies the angle subtended by the Moon at the Earth, $\angle MES$, is approximately $90^{\circ}$.
• The problem specifies that the "Moon-Earth-Sun angle" is measured to be $89.85^{\circ}$. This corresponds to the angle at the Moon, $\angle EMS = 89.85^{\circ}$.
• We need to find the ratio $\frac{ES}{EM}$.

Calculating the Angle at the Moon

The sum of angles in a triangle is $180^{\circ}$. Using the angles identified:

$ \angle MES + \angle ESM + \angle EMS = 180^{\circ} $

Substituting the known values:

$ 90^{\circ} + \angle ESM + 89.85^{\circ} = 180^{\circ} $

Solving for $\angle ESM$ (the angle subtended by the Earth at the Moon):

$ \angle ESM = 180^{\circ} - 90^{\circ} - 89.85^{\circ} $

$ \angle ESM = 0.15^{\circ} $

Applying the Law of Sines

The Law of Sines relates the sides of a triangle to the sines of its opposite angles:

$ \frac{EM}{\sin(\angle ESM)} = \frac{ES}{\sin(\angle EMS)} $

Rearranging the formula to find the desired ratio $\frac{ES}{EM}$:

$ \frac{ES}{EM} = \frac{\sin(\angle EMS)}{\sin(\angle ESM)} $

Calculating the Distance Ratio

Substitute the angle values into the equation:

$ \frac{ES}{EM} = \frac{\sin(89.85^{\circ})}{\sin(0.15^{\circ})} $

Using trigonometric values:

• $ \sin(89.85^{\circ}) = \sin(90^{\circ} - 0.15^{\circ}) = \cos(0.15^{\circ}) \approx 0.9999966 $
• $ \sin(0.15^{\circ}) \approx 0.0026177 $

Now, calculate the ratio:

$ \frac{ES}{EM} \approx \frac{0.9999966}{0.0026177} \approx 381.975 $

Conclusion

The calculated ratio of the Earth-Sun distance to the Earth-Moon distance is approximately 381.975. This value is closest to 382.