Question

During a half-moon phase, the Earth-Moon-Sun form a right triangle. If the Moon-Earth-Sun angle at this half-moon phase is measured to be $89.85^\circ$, the ratio of the Earth-Sun and Earth-Moon distances is closest to

• A. 328
• B. 382
• C. 238
• D. 283

Hint:

For small angles like $0.15^\circ$, using the small angle approximation ($\tan(\theta) \approx \sin(\theta) \approx \theta$ in radians) can simplify calculations.

Answer 1

The Correct Option is B

In the half-moon phase, the Earth-Moon-Sun form a right triangle, where the Earth-Sun distance forms the hypotenuse and the Earth-Moon distance forms one of the legs. We are given the Moon-Earth-Sun angle as $89.85^\circ$.
This means the angle $\angle \text{Earth-Moon-Sun} = 90^\circ - 89.85^\circ = 0.15^\circ$.
We are tasked with finding the ratio of the Earth-Sun distance to the Earth-Moon distance, denoted as $\frac{d_{\text{ES}}}{d_{\text{EM}}}$.
Step 1: Apply Trigonometry
In the right triangle, we use the tangent of the angle, where: \[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{d_{\text{EM}}}{d_{\text{ES}}} \] Given that $\theta = 0.15^\circ$, we use the small angle approximation for tangent, where: \[ \tan(0.15^\circ) \approx 0.002617 \] Thus, we can write: \[ 0.002617 = \frac{d_{\text{EM}}}{d_{\text{ES}}} \] Step 2: Calculate the Ratio
Rearranging the equation to solve for $\frac{d_{\text{ES}}}{d_{\text{EM}}}$: \[ \frac{d_{\text{ES}}}{d_{\text{EM}}} = \frac{1}{0.002617} \approx 382 \] Thus, the ratio of the Earth-Sun distance to the Earth-Moon distance is approximately 382. \[ \boxed{382} \]