(143a) Given the sun's meridian altitude (a) and the hour angle when rising (h) to find the latitude and declination.

Ex. Meridian altitude 56°, hour angle h=7h.

Ans. Lat. 47° 23′ 30′′. (144) In latitude 45° N. the meridian altitude of the sun = 30°: shew that the tangent of quarter the length

(145a) At a certain place the sun rose at 7h A. M., and its meridian zenith distance was twice the latitude: required the latitude. Ans. Lat. 26° 58'. (146a) In latitude 45° N., the sun rose at 4h A. M.: shew that the tangent of the meridian altitude = 3.

(147a) In latitude 50° N. when the sun's declination is 5° 38' N. required the time it will take the body of the sun to rise out of the horizon, its semidiameter being 16'. Ans. 3" 198.

(148a) Required the time the sun's semidiameter will take to pass the meridian, the declination being 23° 4 and semidiameter 16′ 17′′.3. Ans. Im 118.

(149a) A ship in latitude 59° 6' N. and long. 6° 15'E. observed a point of land bearing N. E., and after sailing E. N. E. 6 miles, the point bore N. W.: required the

latitude and longitude of the point.

Ans. Lat. 59° 11′ 15′′, long. 6° 25′ 13′′ E.

(150) In latitude 50° 48′ N. when the sun's declination=12° 29′ N. and hour angle =2h 53m 1s A. M.: required the azimuth.

Ans. N. 121° 47′. E.

(151a) In latitude 45°N., required the difference in the lengths of the longest and shortest days, (decl. 23° 28′.) Ans. 6h 51m 40s. (152a) In what latitude N will the difference between the longest and shortest days be just 6 hours.

Ans. Lat. 46° 46' N.

(153a) At a certain place, when the sun's declination was 10° N., it rose an hour later than when it was 20° N. required the latitude. Ans. Lat. 52° 27′ N.

:

(154a) In what latitude north will the shortest day be just one third the longest (decl. 23° 29′).

Ans. 58° 27′ N.

(155a) And in what latitude will the shortest day be just ths the longest. Ans. 41° 24' N. (156a) At what time in latitudes 50° N. and 60° N., will the snn have the same altitude, its declination being 22° 57′ 15′′ N. Ans. 4h 51m.

(157a) If a ship sail from a certain place a miles due east, then a miles due south, and then 6 miles due west, and reach the same longitude: required the latitude of the place arrived at.

Ex. a 100' and b = 150'

Ans. 85° 0′ 30′′, (158a) Given the apparent and true altitude of a heavenly body, its declination, and observed distance from a terrestial object: to find the true bearing of the object from a given station.

(1) When the object is in the horizon. Ex. In latitude 7° 51' N., the observed altitude of the sun's lower limb was 10° 30′, and observed distance of his nearest limb

from a well defined point of land on the same level with the eye and to the right of the sun was 95° 16'; index correction of the altitude sextant was 0' 50"- and that of the other was l'10'+; the correction for height of the eye (14 feet) in taking the sun's altitude was 3' 41′′ required the true bearing of the point of land; the sun's declination being 22° 24′ S., and semidiameter 15′ 45′′. Ans. Bearing N. 19° 0 30" W.

- :

(159a) (2) When the object is elevated above the level of the eye, it is necessary to observe its altitude : Ex. 2. In the preceding example suppose the object observed to be a mountain the altitude of whose summit is 10° required the true bearing of that point.

Ans. N. 17° 0' 0" W.

The system of rules and operations by which the relative position of any number of points in a tract of country may be determined, so that it may be delineated on a plane surface, is called Trigonometrical Surveying. When the extent of country is not great, the subject involves little difficulty, but when a kingdom such as Britain or France is to be surveyed, the aid of astronomy, and other branches of natural philosophy is required. In a survey, the most remarkable objects such as the summit of hills, spires, towers &c, must be chosen as stations, and, if necessary, marked by signals. These must be considered as joined by straight lines forming a

chain of triangles, connecting each with all the others. The sides of the triangles should be as long as possible, so as to admit of the stations at any two of the angles being seen from the third: the nearer each triangle approaches to the equilateral form, the better. Supposing a proper disposition of the triangles to have been made, when their angles are known, if a side of any one of them were also known, then the sides of all the others might be found by calculation, and a plan of the country constructed.

This

A small extent of the earth's surface may be regarded as a plane, and lines perpendicular to it as parallel to one another. In an extensive survey, however, such be that of England, the curvature of the earth must be taken into the account, and then its figure and magnitude enter as elements into all the calculations. connection between the figure of the earth and the magnitude and position of lines traced on its surface affords conversely, the means of determining the former when the latter are known: so that such surveys, besides their immediate object, are applicable to the solution of the still more important problem, of finding the magnitude and figure of the earth itself.

When a tract to be surveyed has been covered with a series of triangles, so as to connect the principal points; and all the angles of each, and a side of one are known, the sides of all the triangles may be found by calculation, and a plan made by constructing the triangles on the sides of each other: but in a plan so constructed any error made in the value of one of the sides or angles,

will produce corresponding errors in all the others To avoid as much as possible this source of error, we may determine the position of a side of one of the triangles with respect to the meridian, by compass, or more accurately by astronomical observations, (see problem 158); and then calculate the distances of all the stations from such meridian, and also the meridional distance of one station from the other: from these distances the position of each station may be laid down in the plan independently of the others, and also the direct distance between any two points may be easily found. The manner of proceeding will appear from the following Problem.

(160). fig. (59) Let A B C D E F be six stations connected by four triangles ABC, BCD, BDE, EDF:

a side AB of one of the triangles is 4213 yards, and it makes with the meridian NS an angle SAB=62° 52′ at the point A and the station F makes at the point S an angle ASF with the meridian=52° 40'. It is required to find the points in which perpendiculars from the stations will cut the meridian, and the length of each perpendicular.

L