The elevation of the spaceship, is the altitude of the triangle formed by
the line to the near and far side of the crater and the crater's distance.
Correct response:
The given parameters are;
Distance above the surface of the Moon Apollo 11 is orbiting = 3 miles
Angle of depression to the near side of the crater = 25°
Angle of depression to the far side of the crater = 18°
Required:
The distance across the crater.
Solution:
Angle, θ, formed by the line perpendicular to the Moon and the line
from Apollo to the start of the crater is given as follows;
θ = 90° - 25° = 75°
Therefore;
[tex]\displaystyle tan(75^{\circ}) = \mathbf{\frac{x_1}{3} }[/tex]
x₁ = 3 × tan(65°)
Similarly, the distance from the far side from the point directly under
Apollo 11, x₂, is given as follows;
Horizontal distance of the crater from the Apollo 11 = 3 × tan(90° - 18°)
Which gives;
x₂ = 3 × tan(72°)
Which gives;
[tex]\displaystyle x_2 = \mathbf{3 \times \frac{\sqrt{10 + 2 \cdot \sqrt{5} } }{\sqrt{5} - 1 } }[/tex]
Distance across the crater, D = x₂ - x₁
Therefore;
[tex]\displaystyle D = 3 \times \frac{\sqrt{10 + 2 \cdot \sqrt{5} } }{\sqrt{5} - 1} - (3 \times tan(65^{\circ})) \approx \mathbf{ 2.8}[/tex]
Learn more about trigonometric ratios here:
https://brainly.com/question/22599614
