Related Questions

If ‘δ’ is the declination of the star and ‘φ’ is the latitude of the observer then the hour angle of the star at elongation is given by
A sin H = tan φ . cot δ
B cos H = tan φ . cot δ
C tan H = tan φ . cot δ
D None of these
The latitude of a place was obtained by subtracting the zenith distance of a star from its declination, the observed star was between
A Horizon and equator
B Equator and zenith
C Zenith and pole
D Pole and horizon
The latitude of a place was obtained by subtracting the declination of a star from its zenith distance, the observed star was between
A Horizon and equator
B Zenith and pole
C Equator and zenith
D Pole and horizon
If θ and δ be the latitude of an observer and declination of a heavenly body respectively, the upper culmination of the body will be south of zenith if its zenith distance, is
A δ - θ
B θ - δ
C θ + δ
D $$\frac{1}{2}$$ (θ - δ)
When a star is between the pole and the horizon, the relationship between latitude (λ), zenith distance (z) and declination δ, is
A θ = z + δ
B θ = δ - z
C θ = 180° - (z + δ)
D θ = (z + δ) - 180°
The angle between the observer's meridian and declination circle of a heavenly body, is known as
A Hour angle
B Azimuth
C Right ascension
D Declination
Latitude of the observer's position is equal to altitude of
A North pole
B Pole star
C Celestial pole
D All the above
The latitude of the observer's position, is
A Elevation of the elevated pole
B Declination of the observer's zenith
C Angular distance along the observer's meridian between equator and the observer
D All the above
A star may culminate at zenith if its declination is
A Greater than the longitude of the place
B Less than the latitude of the place
C Equal to the latitude of the place
D None of these
At eastern elongation, the pole star moves
A Eastward
B Westward
C Northward
D Southward

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