Find the declination from meridian altitudes of a circumpolar star A circumpolar star has meridian altitudes of 70° (upper culmination) and 10° (lower culmination), and both culminations occur north of the zenith. What is the declination of the star?

Introduction / Context:
For circumpolar stars (declination same sign and sufficiently large relative to latitude), upper and lower culmination altitudes on the meridian provide direct equations linking site latitude and star declination. This is a common reduction in positional astronomy and astro-surveying.

Given Data / Assumptions:

• Upper culmination altitude h_up = 70°.
• Lower culmination altitude h_low = 10°.
• Both culminations are north of the zenith → star north of zenith implies δ > φ (northern declination larger than latitude).
• Let φ = latitude, δ = declination (both positive in N hemisphere).

Concept / Approach:
For a circumpolar star with δ and φ both north and δ > φ:
h_up = 90° − δ + φ and h_low = 90° − φ − δ. Solve these two linear equations for φ and δ.

Step-by-Step Solution:
Use h_up: 90 − δ + φ = 70 ⇒ −δ + φ = −20 ⇒ φ − δ = −20.Use h_low: 90 − φ − δ = 10 ⇒ −φ − δ = −80 ⇒ φ + δ = 80.Add the two: (φ − δ) + (φ + δ) = −20 + 80 ⇒ 2φ = 60 ⇒ φ = 30°.Substitute: φ + δ = 80 ⇒ 30 + δ = 80 ⇒ δ = 50°.

Verification / Alternative check:
Check both altitudes with φ = 30°, δ = 50°: h_up = 90 − 50 + 30 = 70°; h_low = 90 − 30 − 50 = 10° → matches given data.

Why Other Options Are Wrong:
Options 80°, 70°, 60°, 40° do not satisfy the pair of equations for the given heights; only 50° does.

Common Pitfalls:
Using formulas for the δ < φ case or forgetting the sign convention (north positive); misinterpreting “north of zenith,” which here implies δ > φ.

Final Answer:
50°