If $$\alpha $$, H, A and $$\delta $$ be the altitude, hour angle, azimuth and declination of a circumpolar star at its elongation, in latitude $$\lambda $$, the following relation holds good A. $$\cos {\text{H}} = \frac{{\tan \lambda }}{{\tan \delta }}$$ B. $$\sin \alpha = \frac{{\sin \lambda }}{{\sin \delta }}$$ C. $$\sin {\text{A}} = \frac{{\cos \delta }}{{\cos \lambda }}$$ D. All the above

The correct answer is D. All the above.

The altitude $\alpha$ is the angle between the horizon and the star. The hour angle $H$ is the angle between the star and the meridian, measured eastward from the observer’s meridian. The azimuth $A$ is the angle between the observer’s north point and the line of sight to the star, measured eastward from north. The declination $\delta$ is the angle between the star’s celestial equator and the observer’s celestial equator.

The

[Diagram of a circumpolar star at its elongation]

The star is at its elongation when it is at its maximum angular distance from the observer’s north celestial pole. This occurs when the hour angle is $90^\circ$ minus the observer’s latitude $\lambda$.

The altitude of the star at its elongation is given by

$$\alpha = \arctan \left( \frac{\sin \delta}{\sin \lambda} \right)$$

The hour angle of the star at its elongation is given by

$$H = 90^\circ – \lambda$$

The azimuth of the star at its elongation is given by

$$A = \arctan \left( \frac{\cos \delta}{\cos \lambda} \right)$$

Therefore, the following relations hold good:

$$\cos H = \frac{\tan \lambda}{\tan \delta}$$

$$\sin \alpha = \frac{\sin \lambda}{\sin \delta}$$

$$\sin A = \frac{\cos \delta}{\cos \lambda}$$