Circumpolar star at elongation: If δ is the declination and φ is the observer’s latitude, the hour angle (H) of the star at elongation satisfies which relation?

Introduction / Context:
Greatest elongation of a circumpolar star occurs when the change of azimuth with respect to hour angle becomes zero. This yields a characteristic relation connecting the star’s hour angle at elongation with the observer’s latitude and the star’s declination.

Given Data / Assumptions:

• Circumpolar star (does not set).
• Observer’s latitude = φ, star’s declination = δ.
• Elongation condition dZ/dH = 0.

Concept / Approach:
Applying spherical trigonometry to the astronomical triangle and enforcing the tangency condition at elongation eliminates azimuth and produces a closed-form relation for H. The standard result used in surveying and navigation is cos H = tan φ * cot δ.

Step-by-Step Solution:
Start from the relation linking azimuth, hour angle, latitude, and declination.Apply the elongation condition to set the derivative of azimuth with respect to hour angle to zero.Solve for H in terms of φ and δ to obtain cos H = tan φ * cot δ.

Verification / Alternative check:
Test φ = 45°, δ = 60°: tan φ = 1, cot δ ≈ 0.577 → cos H ≈ 0.577 → H ≈ 55°, a reasonable hour angle for elongation.

Why Other Options Are Wrong:
sin H = tan φ * cot δ or tan H = tan φ * cot δ: not the canonical elongation relation; they would produce inconsistent values.H = φ − δ: pertains to meridian transit relations, not elongation.

Common Pitfalls:
Using culmination (meridian) formulas instead of elongation relations; forgetting that the star must be circumpolar or the expressions may be invalid.

Final Answer:
cos H = tan φ * cot δ.