The distances are x cot. A, x cot. B, and a cot. C, which, when x has been calculated, may be themselves calculated at once.

=

Note. When DD,, or D + D, for logarithmic computation, being then

2D, the expression becomes better suited

x = D÷√ cot.2 A+ cot.2 C cot.2 B).

In this case, therefore, the rule is as follows:-Double the log. cotangents of the angles of elevation at the extreme stations, find the natural numbers answering thereto, and take half their sum; from which subtract the natural number anwering to twice the log. cotangent of the middle angle of elevation: then half the log. of this remainder subtracted from the log. of the measured distance between the 1st and 2d, or the 2d and 3d stations, will be the log. of the height of the object.

Let the subjoined example be worked out.

Going along a straight and horizontal road which passed by a tower, I wished to find its height, and for this purpose measured two equal distances each of 252 feet, and at the extremities of these distances took three angles of elevation of the top of the tower; viz. 36° 50′, 21° 24′, and 14°. What is the height of the Ans. 161.88 feet.

tower?

EXAMPLE VII.

The side of a hill forms an inclined plane whose angle of inclination is known; required the direction in which a rail or other road must run along the side of the hill, in order that it may have an ascent of 1 in every n feet.

In the annexed figure, let ADB be in the horizontal plane, CD a line on the side or slope of the hill perpendicular to AD, and DB a horizontal line perpendicular to AD, to which, as well as to AB, BC is perpendicular, and AC the rail-road; its projection on the horizontal plane will be AB, and the angle BAC, therefore, will determine the question.

cot. CDB

Therefore, sin. BAD = tan. CAB. cot. CDB.

Suppose, for example, CDB = 10°, and n = 100, or the ascent one foot in a hundred.

8.7537029= log. sin. 3° 15′ 5′′, the horizontal angle required.

EXAM. VIII. From the edge of a ditch, of 36 feet wide, surrounding a fort, having taken the angle of elevation of the top of the wall, it was found to be 62° 40′ required the height of the wall, and the length of a ladder to reach from my station to the top of it?

Ans. {height of wall 69:64, ladder, 78·4 feet.

1 EXAM. IX. Required the length of a shoar, which being to strut 11 feet from the upright of a building, will support a jamb 23 feet 10 inches from the ground? Ans. 26 feet 3 inches.

EXAM. X. A ladder, 40 feet long, can be so placed, that it shall reach a window 33 feet from the ground, on one side of the street; and by turning it over without moving the foot out of its place, it will do the same by a window 21 feet high, on the other side required the breadth of the street? Ans. 56 649 feet. EXAM. XI. A maypole, whose top was broken off by a blast of wind, struck the ground at 15 feet distance from the foot of the pole: what was the height of the whole maypole, supposing the broken piece to measure 39 feet in length? Ans. 75 feet.

EXAM. XII. At 170 feet distance from the bottom of a tower, the angle of its elevation was found to be 52° 30': required the altitude of the tower?

Ans. 221.55 feet. EXAM. XIII. From the top of a tower, by the sea-side, of 143 feet high, it was observed that the angle of depression of a ship's bottom, then at anchor, measured 35°: what was the ship's distance from the bottom of the wall?

Ans. 204 22 feet. EXAM. XIV. What is the perpendicular height of a hill; its angle of elevation, taken at the bottom of it, being 46°, and 200 yards farther off, on a level with the bottom, the angle was 31°? Ans. 286 28 yards.

EXAM. XV. Wanting to know the height of an inaccessible tower: at the least distance from it, on the same horizontal plane, I took its angle of elevation equal to 58°; then going 300 feet directly from it, found the angle there to be only 32°: required its height, and my distance from it at the first station?

Ans. (height 307:53

distance 192.15

EXAM. XVI. Being on a horizontal plane, and wanting to know the height of a tower placed on the top of an inaccessible hill; I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a line directly from it to the distance of 200 feet farther, I found the angle to the top of the tower to be 33° 45'. What is the height of the tower?

Ans. 93.33148 feet. EXAM. XVII. From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple equal 40°; then from another window, 18 feet directly above the former, the like angle was 37° 30′: required the height and distance of the steeple?

Ans. {height 210-44.

distance 250.79.

EXAM. XVIII. Wanting to know the height of, and my distance from, an object on the other side of a river, which appeared to be on a level with the place where I stood, close by the side of the river; and not having room to measure backward, in the same line, because of the immediate rise of the bank, I placed a mark where I stood, and measured in a direction from the object, up the ascending ground, to the distance of 264 feet, where it was evident that I was above the level of the top of the object; there the angles of depression were found to be, viz. of the mark left at the river's side 42°, of the bottom of the object 27°, and of its top 19°. Required the height of the object, and the distance of the mark from its bottom? 57.26.

Sheight
Ans.{distance 150-50.

EXAM. XIX. If the height of the mountain called the Peak of Teneriffe be

2 miles, as it is very nearly, and the angle taken at the top of it, as formed between a plumb-line and a line conceived to touch the earth in the horizon, or farthest visible point, be 88° 2′; it is required from these measures to determine the magnitude of the whole earth, and the utmost distance that can be seen on its surface from the top of the mountain, supposing the form of the earth to be perfectly globular? dis. 135 943

Ans. {diam. 7918

miles.

EXAM. XX. Two ships of war, intending to cannonade a fort, are, by the shallowness of the water, kept so far from it, that they suspect their guns cannot reach it with effect. In order therefore to measure the distance, they separate from each other a quarter of a mile, or 440 yards; then each ship observes and measures the angle which the other ship and the fort subtends, which angles are 83° 45′ and 85° 15'. What is the distance between each ship and the fort?

Ans. {2293:26} yards.

EXAM. XXI. Wanting to know the breadth of a river, I measured a base of 500 yards in a straight line close by one side of it; and at each end of this line I found the angles subtended by the other end and a tree, close to the bank on the other side of the river, to be 53° and 79° 12′. What was the perpendicular breadth of the river? Ans. 529 48 yards.

EXAM. XXII. Wanting to know the extent of a piece of water, or distance between two headlands, I measured from each of them to a certain point inland, and found the two distances to be 735 yards and 840 yards; also the horizontal angle subtended between these two lines was 55° 40'. What was the distance required? Ans. 741-2 yards.

EXAM. XXIII. A point of land was observed, by a ship at sea, to bear east-bysouth; and after sailing north-east 12 miles, it was found to bear south-eastby-east. It is required to determine the place of that headland, and the ship's distance from it at the last observation? Ans. 26 0728 miles.

EXAM. XXIV. Wanting to know the distance between a house and a mill, which were seen at a distance on the other side of a river, I measured a base line along the side where I was, of 600 yards, and at each end of it took the angles subtended by the other end and the house and mill, which were as follow, viz. at one end the angles were 58° 20′ and 95° 20′, and at the other end the like angles were 53° 30′ and 98° 45′ What was the distance between the house and mill? Ans. 959-5866 yards.

EXAM. XXV. Wanting to know my distance from an inaccessible object O, on the other side of a river; and having no instrument for taking angles, but only a chain or chord for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured in a direct line from the object O 100 yards, viz. AC and BD each equal to 100 yards; also the diagonal AD measured 550 yards, and the diagonal BC 560. What was the distance of ths object O from each station A and B ? AO 536.81. BO 500'47.

Ans. {B

EXAM. XXVI. In a garrison besieged are three remarkable objects, A, B, C, the distances of which from each other are discovered by means of a map of the place, and are as follow, viz. AB 2661, AC 530, BC 327 yards. Now, having to erect a battery against it, at a certain spot without the place, and being desirous to know whether my distances from the three objects be such, as that they may from thence be battered with effect, I took, with an instrument, the horizontal angles subtended by these objects from the station S, and found them

to be as follow, viz. the angle ASB 13° 30′, and the angle BSC 29° 50′. Required the three distances, SA, SB, SC; the object B being situated nearest to me, and between the two others A and C. SA 757-14. Ans. SB 537.10. SC 655.30.

EXAM. XXVII. Required the same as in the last example, when the object B is the farthest from my station, but still seen between the two others as to angular position, and those angles being thus, the angle ASB 33° 45′, and BSC 22° 30'; also the three distances, AB 600, AC 800, BC 400 yards?

SA 710.3. Ans. SB 1041.85. SC 934-14.

EXAM. XXVIII. If DB in the figure at p. 383 represent a portion of the earth's surface, and D the point where the levelling instrument is placed, then LB will be the difference between the true and the apparent level; and you are required to demonstrate that, for distances not exceeding 5 or 6 miles measured on the earth's surface, BL, estimated in feet, is equal to of the square of BD, taken in miles.

EXAM. XXIX. On the opposite bank of a river to that on which I stood, is a tower, known to be 216 feet high. With a pocket sextant I ascertained the vertical angle subtended by the tower's height to be 47° 56'. Required the distance, across the river, from the place where I stood, to the bottom of the tower; supposing my eye to be 5 feet above the horizontal plane which passes through it? Ans. 200-3 feet.

EXAM. XXX. In the valley of Chamouni three positions, A, B, C, were selected, in a straight line, such that AB = 800, and BC 750 yards. Three remarkable points, A', B', C', on the side of the Jura, were also chosen to be observed. The angles of elevation of A', as seen from A, B, and C, were 67° 10′, 58° 15′, and 52° 18′; those of B', as seen from the same points, were 72° 18′, 78° 15′, and 70° 10′; and finally, those of C' were 60° 5′, 61° 10, and 58° 5′ respectively. It is required from these observations to find the distances of A', B', C', from one another, and from A, B, C; together with their altitudes above Chamouni, and the inclination of the side of the Jura to the horizon at the place in question.

MENSURATION OF PLANES.

THE Area of any plane figure is the measure of the space contained within its extremes or bounds; without any regard to thickness.

This area, or the content of the plane figure, is estimated by the number of little squares that may be contained in it; the side of those little measuring squares being an inch, or a foot, or a yard, or any other fixed quantity. And hence the area or content is said to be so many square inches, or square feet, or square yards, &c.

Thus, if the figure to be measured be the rectangle ABCD, and the little square E, whose side is one inch, be the measuring unit proposed: then as often as the said little square is contained in the rectangle, so many square inches the rectangle is said to contain, which in the present csae is 12.

3

D

B

E

PROBLEM I.

To find the Area of any Parallelogram; whether it be a Square, a Rectangle, a Rhombus, or a Rhomboid.

MULTIPLY the length by the perpendicular breadth, or height, and the product will be the area *.

EXAMPLES.

Ex. 1. Find the area of a parallelogram, the length being 12:25, and breadth or height 8.5.

12.25 length

8.5 breadth

Ex. 2. Find the area of a square, whose side is 35.25 chains.

Ans. 124 acres, 1 rood, 1 perch. Ex. 3. Find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches.

Ans. 9 feet. Ex. 4. Find the content of a piece of land, in form of a rhombus, its length being 6.20 chains, and perpendicular breadth 5:45.

Ans. 3 acres, 1 rood, 20 perches. Ex. 5. Required the number of square yards of painting in a rhomboid, whose length is 37 feet, and height 5 feet 3 inches. Ans. 21 square yards.

Ex. 6. The two diagonals of a parallelogram are 185.5 and 137·9, and they intersect under an angle of 42° 10′ 18′′. What is the area?

Ex. 7. The side of a rhombus is 18, and its longer diagonal 23: what is its area and its other diagonal?

PROBLEM II.

To find the Area of a Triangle.

RULE I.—MULTIPLY the base by the perpendicular height, and take half the product for the area †. Or, when either of them is an even number, multiply the one of these dimensions by half the other.

* The truth of this rule is proved in the Geom. theor. 81. cor. 2.

The same is otherwise proved thus: Let the foregoing rectangle be the figure proposed; and let the length and breadth be divided into several parts, each equal to the linear measuring unit, being here 4 for the length, and 3 for the breadth; and let the opposite points of division be connected by right lines. Then it is evident that these lines divide the rectangle into a number of little squares, each equal to the square measuring unit E; and further, that the number of these little squares, or the area of the figure, is equal to the number of linear measuring units in the length, repeated as often as there are linear measuring units in the breadth, or height; that is, equal to the length drawn into the height; which here is 4 × 3 or 12.

And it is proved (Geom. theor. 25, cor. 2), that any oblique parallelogram is equal to a rectangle, of equal length and perpendicular breadth. Therefore the rule is general for all parallelograms whatever.

The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base and altitude, by Geom. theor. 26.