,

marked, and the last three perpendiculars are because they lie towards the east, or minus side of the base.

NOTE 2. The above exercise contains 28 questions like the first three examples, for each of the 8 places may be combined with all the others; those who wish to compare the relative merits of the old method and that given above, may calculate them both ways; the results ought to be the same.

NOTE 3. If the objects observed above were the corners in the boundary of a county or an estate, its area could easily be found from the above, as will be shown in Mensuration or Land Surveying.

97. Given the angular bearings of three objects whose distances from one another are known, to find the distance of the station where the angles are taken from each of the three given objects.

Let the three objects be (A, A', or A"), and B and C, of which B and C are the extreme objects, and A, A', or A", the object which has its angular position between the objects B and B C, as seen from the station S, and let CSA, BSA, be the angles formed by these objects at the station S.

CONSTRUCTION. At the point B, in the line joining the extreme objects, make the angle DBC=CSA, and at C, in the same line, make the

[DCB=BSA; then about the BDC describe a circle, and it will pass through the station S; join (A, A', or A"), and D, and produce the line to meet the circumference in S, and S will be the position of the station sought.

Now in the DBC there are given the Ls at B and C, and the side BC, hence BD and DC can be found, (Art. 39); and since the three sides of the ABC are given, the Ls can be found, (Art. 49), therefore their difference ACD or ABD can be found; then in the ▲s ACD or ABD, there are given two sides and the contained angle, therefore (Art. 40) the Ls CAD or CAS, and BAD or BAS, can be found. Again, in the ACS there are given the Ls CAS and CSA, and the side AC, hence (Art. 39) AS and CS can be found, and from the ▲ ABS, in the same manner, can be found BS.

1. If the three objects B, A', C, lie in the same straight line, having calculated CD as above, we have in the ADCA', two sides, DC, A'C, and the contained angle at C, to find (Art. 40) the LCA'D, or BA'S; then in the ABA'S, there are given the Ls BA'S and A'SB, and the

side A'B, to find BS and A'S, by (Art. 39); and in a similar manner, from the AA'SC the distance CS can be found. If the station were taken in the line BC, or BC produced, the distance could not be determined.

2. If the third object lie between the station S and the line BC, as at A"; to the LDCB add the LBCA", then in the ADCA" there are given two sides and the contained L, hence the LCA"D can be found, and consequently its supplement, CA'S; then in the ACA"S there are given two angles, and the side A"C, to find (Art. 39) A′′S and CS, and in the same manner, from the ABA"S, BS can be found. The solution when the three objects form a triangle fails, when the points D and A coincide, or when the station is in the circumference of a circle described about the three objects. If the station S were in the production of one of the sides of the formed by the objects, or in one of the sides themselves, the solution is easy, without describing a circle.

3. If the station were within the ▲ formed by the three objects, as at S, in the annexed figure, then the Ls ASB and ASC being measured, construct the LBCD the supplement of ASB, and B the angle CBD = the_supplement of ASC, and about the ABCD describe a circle, and join DA, and the point S, where the line AD cuts the circle, will be the station required, for the BSD

will be the BCD; .. BSA, its supplement, will be of the given magnitude, and the same may be shown of the LASC. Now, in the BDC there are given the angles and a side BC, .. BD and DC can be found; and since the sides of the ABC are given, the Ls can be found, .. in the ABD, AB, and BD, are known, and the angle ABD, hence the BAD, or BAS, can be found, .. since ASB is also known, and the side AB, the sides SB and SA can be determined; and if from the [BAC, the BAS be subtracted, the LCAS will be known, and since ASC is also known, and the side AC, the side SC can be found. 4. The station is without the triangle, upon one of its sides, or within it, according as the sum of the two measured angles is less than, equal to, or greater than two right angles.

EXERCISES.

1. In diagram, (Art. 97), let BC=290 yards, AC-195

yards, and AB=240; LASB=30° 5′, and LASC=25° 45'. It is required to find the distances, AS, BS, and C4. Ans. AS 432-07 yards, BS 270-46 yards, and CS 336-34 yards.

2. In the same diagram, let the three objects, B, A', C, in the same straight line, be distant from each other as follows: viz. BA' 490 yards, A'C 300 yards, and consequently BC 790 yards; and let the LA'SC=43°, and BSA' 33° 45'. It is required to find the distances SB, SA', SC. Ans. SB 782-17 yards, SA' 423-93 yards, and SC 390-104 yards.

3. In the same diagram, let A"B=500 yards, BC=630 yards, A"C-540 yards, BSA"-31°, and LA"SC-28° 24. It is required to determine the distances SB, SA", and SC. Ans. SB 612-523 yards, SA" 137-118 yards, and SC 656-67 yards.

4. In the diagram, (Note 3), let A, B, and C, represent three objects in the same horizontal plane, whose distances are as follows: viz. AC 460 yards, AB 570 yards, and BC 620 yards. At a point S within the triangle formed by these objects, the LASB measured by a circle was found to be 125° 15', the BSC 124° 15', and the LASC 110°30'. Required the distances AS, BS, and CS. Ans. AS 247-855 yards, BS 289-804 yards, and CS 310-323 yards.

5. In the side AB of the triangle ABC, whose sides were AB 1200 yards, BC 1000 yards, and AC 974 yards, I took the angle BSC 74° 12'. Required the distances of the station S from each of the three points, A, B, and C. Ans. SC 814-274 yards, SB 843.094 yards, and SA 356 906 yards.

6. In the side AC, produced beyond C of the same triangle, the LBSC was 37° 25′ 53′′. It is required to find the distance of the station S from each of the objects, A, B, and C. Ans. SC 1000 yards, AS 1974 yards, and SB 1588-16 yards.

PROMISCUOUS EXERCISES.

1. After observing the elevation of a tower, which is 100 feet high, to be 60°, how far must an observer measure back on the level plane before its elevation becomes 30°? Ans. 115-47, or 663/3 feet.

2. From the top of a tower, whose height is 108 feet, the angles of depression of the top and bottom of a vertical column in the horizontal plane, are found to be

30° and 60° respectively. Determine the height of the column. Ans. 72 feet.

3. The sum of the three sides of a triangle is 2539-058, and the angles are to one another as the numbers 3, 4, and 5; it is required to find the sides and angles. Ans. The sides are 707.107, 866-025, 965·926; and the angles are 45°, 60°, and 75°.

4. A person on the top of a tower, whose height is 100 feet, observes the angles of depression of two objects on the horizontal plane, which are in the same straight line with the tower, to be 30° and 45°. Determine their distance from each other, and from the tower on which the observer is situated. Ans. 73-205, 100, and 173.205 feet.

5. From the top of a tower 120 feet high, the angles of depression of two trees on the same horizontal plane with the base, was of the nearest 53° 24', and of the other 20° 16'. Required the distance of each of the trees from the base of the tower.

Ans. Nearest 89.12, and the other 324-982 feet distant. 6. Wishing to know my distance from two distant objects, and their distance from one another, I measured their angular bearings 38° 24', then advancing in a direct line towards the right hand object 300 feet, their angular bearing was 40° 13'; returning to the original station, and then measuring 350 feet towards the left hand object, their angular bearing was then 39° 58'. Find the distance of the first station from each of the objects, and their distance from one another. Ans. From right hand object 8223-08, from the other 6110.24, and their distance 5118.66 feet.

NAVIGATION.

98. NAVIGATION is the art of conducting a ship through the wide and pathless ocean, from one part of the world to another. Or, it is the method of finding the latitude and longitude of a ship's place at sea, and thence determining her course and distance from that place to any other place.

The Equator is a great circle circumscribing the earth, every point of which is equally distant from the poles, thus dividing the globe into two equal parts, called hemispheres; that towards the North Pole is called the northern, and that towards the south the Southern hemisphere.

E

The Meridian of any place on the earth is a great circle passing through that place and the poles, and cutting the equator at right angles. The first meridian with us is that which passes through the Royal Observatory, at Greenwich.

The Latitude of a place is that portion of its meridian which is intercepted between the equator and the given place, and therefore never exceeds 90°.

The Difference of Latitude between two places on the earth is an arc of the meridian, intercepted between their corresponding parallels of latitude, showing how far one of them is to the northward or southward of the other.

The Longitude of any place on the earth is that arc or portion of the equator which is contained between the first meridian and the meridian of the given place, and is called east or west, according as it may be situated with respect to the first meridian. Longitude is reckoned east and west half round the equator, and consequently may be as large as 180°.

The Difference of Longitude between two places is an arc of the equator intercepted between the meridians of those places, showing how far one of them is to the eastward or westward of the other.

The Mariner's Compass is an artificial representation of

the horizon. It is divided into 32 equal parts, called points, each point consisting of 11° 15'. Hence the num