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 problems involving small angles close to \(90^\circ\):
1. Use trigonometric approximations to simplify calculations.
2. Ensure you compute the reciprocal for cosine or sine functions correctly.
3. Verify the result for reasonableness based on the problem context.

Answer 1

The Correct Option is B

Step 1: Geometry of the problem. The given problem involves a right triangle formed by: - The Earth, the Moon, and the Sun. - The Moon-Earth-Sun angle \(\angle \text{MES} = 89.85^\circ\). - The Earth-Moon distance (\(d_\text{EM}\)) as one leg of the triangle. - The Earth-Sun distance (\(d_\text{ES}\)) as the hypotenuse. Step 2: Trigonometric relation. Using the cosine rule in the right triangle: \[ \cos(\angle \text{MES}) = \frac{d_\text{EM}}{d_\text{ES}}. \] Rearranging for the ratio \(\frac{d_\text{ES}}{d_\text{EM}}\): \[ \frac{d_\text{ES}}{d_\text{EM}} = \frac{1}{\cos(89.85^\circ)}. \] Step 3: Calculate \(\cos(89.85^\circ)\). Since \(89.85^\circ\) is very close to \(90^\circ\), \(\cos(89.85^\circ)\) can be approximated using a calculator: \[ \cos(89.85^\circ) \approx 0.002617. \] Step 4: Compute the ratio. Substitute \(\cos(89.85^\circ) \approx 0.002617\): \[ \frac{d_\text{ES}}{d_\text{EM}} = \frac{1}{0.002617} \approx 382. \]