Related Questions

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)
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
If ‘δ’ is the declination of the star and ‘φ’ is the latitude of the observer, then the azimuth of the star at elongation is given by
A sin z = sec φ . cos δ
B cos z = sec φ . cos δ
C tan z = sec φ . cos δ
D None of these
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
There are 23 poles with a constant distance between each pole. A car takes 24 seconds to reach the 12th pole. How much more time will it take to reach the last pole?
A 40 seconds
B 44 seconds
C 48 seconds
D 46 seconds
The latitude (λ) of a place and the altitude (α) of the pole are related by
A λ = α
B λ = 90° - α
C λ = α - 90°
D λ = 180° - α
From the principal point the horizon point lies on the principal line at a distance of
A f tan θ
B f sin θ
C f cot θ
D f cos θ
A hat contains 12 cards marked with a star and 18 unmarked cards. What is the probability that one card slelected at random will be marked with a star?
A 3/5
B 2/5
C 2/3
D 5/2

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