coalt. 41°33′11′′

sin. colat. 79 43 20 arith. comp. 0·0070251

sin. sun's polar dist. 113 22 48 arith. comp. 0.0372078

Hour angle 24 54 35 1h 39m 38.3 in time.

Time at Greenwich

L. in time, reckoning westward.

15 32 33

13 52 54 7.

Or, subtracting this from 24 hours, we have 101 7m 4·3′, for the longitude east, in time, and therefore the longitude in degrees is 151° 46′ '4" E.

2. Given the apparent altitude of the moon's centre 8° 26′ 13′′, the true altitude 9° 20′ 45′′, the apparent altitude of a star 35° 40', the true altitude 35° 38′ 49′′, and their apparent distance 31° 13′ 26′′; to determine the true distance.

The true distance is 30° 23′ 56′′.

Those who are desirous of entering more at large into the problem of the Longitude, and of becoming acquainted with the best methods of shortening the computation by the aid of subsidiary tables, may advantageously consult, besides the works already referred to, the Quarto Tables of J. De Mendoza Rios, Lynn's Navigation Tables, Captain Kater's Treatise on Nautical Astronomy, in the Encyclopædia Metropolitana, Kerrigan's Navigator's Guide and Nautical Tables, and Dr. Myers's translation of Rossel on the Longitude.

Variation of the Compass.

(81.) We shall conclude this part of our subject by briefly considering the methods of finding the variation of the compass, or the quantity by which the north point, as shown by the compass, varies easterly or westerly from the true north point of the horizon.

The solution of this problem merely requires that we find by computation, or by some means independent of the compass, the bearing of a celestial object, that we observe the bearing by the compass, and then take the difference of the two. The problem resolves itself, therefore, into two cases, the object whose bearing is sought being either in the horizon or above it: in the one case we have to compute its amplitude, and in the other its azimuth.

The computation of the amplitude is simply determining the hypotenuse of a right-angled triangle, of which one side is given, viz. the declination of the object, as also the angle opposite to it, viz. the colatitude. The computation of the azimuth requires the solution of an oblique spherical triangle, the three sides being given to find an angle; the three given sides are the colatitude; the zenith distance of the object and its polar distance and the azimuth being measured by the angle at the zenith opposite the polar distance, this is the angle sought. We shall give an example in each of these cases of the problem.

EXAMPLES.

1. In January 1830, at latitude 27° 36′ N., the rising amplitude of Aldebaran was, by the compass* E. 23° 30′ N.; required the variation.

By the Nautical Almanack the declination of Aldebaran is 16° 9' 37" N., therefore since Rad. x sin. dec. sin. amp. x cos. lat., the computation is as follows.

* The compass amplitude must be taken when the apparent altitude of the object is equal to the depression of the horizon.

As the object is farther from the magnetic east than from the true east, the magnetic east has therefore advanced towards the south, and therefore the magnetic north towards the east; hence the variation is 5° 11′ 43′′ E.

2. In latitude 48° 50′ north, the true altitude of the sun's centre was 22° 2′, the declination at the time was 10° 12′ S., and its magnetic bearing 161° 32′ east. Required the variation.

O's polar distance 100° 12'

sin. zenith distance 67 58 arith. comp. 0.0329363

sin. colatitude 41 10 arith. comp. 0.1816080

The variation is west, because the sun's observed distance from the

north, measured easterly, being greater than its true distance, intimates that the north point of the compass has approached towards the west.

3. In latitude 48° 20′ north, the star Rigel was observed to set 9° 50' to the northward of the west point of the compass; required the variation, the declination of Rigel being 8° 25′ S.

Variation 22° 33′ West.

4. In latitude 50° 12′ north, when the sun's declination was 11° 28′ 53′ N., its true altitude was found to be 37° 0′ 16′′, and the observed azimuth S. 31° E.; required the variation of the compass.

Variation 28° 2′ West.

END OF PART III.

PART IV.

MISCELLANEOUS TRIGONOMETRICAL INQUIRIES.

(82.) WE now come to the final part of our subject, in which we propose to bring together several miscellaneous particulars which properly come under consideration in a treatise on Trigonometry. One or two of these, especially those which relate to certain compendious solutions of plane triangles, and to the trigonometrical lines of small arcs, might have been introduced much earlier, although we have preferred to postpone their consideration for a supplementary chapter, agreeing with Woodhouse, that it is better for the student first "to attend solely to the general solutions, and to postpone to a time of leisure and of acquired knowledge the consideration of the methods that are either more expeditious or are adapted to particular exigencies."

CHAPTER I.

ON THE SOLUTIONS OF CERTAIN CASES OF PLANE TRIANGLES, AND ON DETERMINING THE TRIGONOMETRICAL LINES OF SMALL ARCS.

PROBLEM I.

(83.) Given two sides and the included angle of a plane triangle, to determine the third side, without finding the remaining angles.

The general expression for the side c, in terms of the two sides a, b, and the included angle C, is (17),

R