EXAM. VI.—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. Sheight of wall 69-64, ladder 78.4 feet. EXAM. VII. 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. VIII.-A ladder, 40 feet long, can be so planted, 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. IX.-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. X.-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 feet. EXAM. XI. From the top of a tower, by the sea-side, of 143 feet height, it was observed that the angle of depression of a ship's bottom, then at anchor, measured 35°: what then was the ship's distance from the bottom of the wall? Ans. 204-22 feet. EXAM. XII.-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 of it, the angle was 31°? Ans. 286 28 yards. EXAM. XIII.-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 320: required its height, and my distance from it at the first station? Ans. {Distance, 192-15 Height, 307.53 EXAM. XIV. 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 equal 40°, and of the top of the tower equal 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 then is the height of the tower? Ans. 93.33148 feet. EXAM, XV. 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': what then is the height and distance of the steeple ? Ans. Height, 210-44 Distance, 250-79 EXAM. XVI.1.—Wanting to know the height of, and my distance from, an object on the other side of a river, which seemed to be on a level with the place where I stood, close by the side of the river; and not having room to measure backwards, on the same plane, because of the immediate rise of the bank, I placed a mark where I stood, and measured in a direction from the object, un 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 42o, of the bottom of the object 27o, and of its top 19°. Required then the height of the object, and the distance of the mark from its bottom? 57.26 Ans. Height, EXAM. XVII.—If the height of the mountain called the Peak of Teneriffe be 4 miles, 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 87° 25′ 55′′; it is required from hence 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 round? 178.458 miles. Ans. {Dist. {Diam, 7957-818 EXAM. XVIII.— -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 angles which the other ship and the fort subtends, which angles were 83° 45′ and 85° 15'. What then was the distance between each ship and the fort? $2292.26 yards. Ans. 2298.05 EXAM, XIX.-Being on the side of a river, and wanting to know the distance to a house which was seen at a distance on the other side; I measured out for a base 400 yards in a right line by the side of the river, and found that the two angles, one at each end of this line, subtended by the other end and the house, were 68° 2′ and 73° 15'. What then was the distance between each sta EXAM. XX..—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 a tree close on the bank on What then was the perpenAns. 529-48 yards. I found the angles subtended by the other end and the other side of the river, to be 53° and 79° 12'. dicular breadth of the river? EXAM. XXI.—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 then was the distance required? Ans. 741-2 yards. EXAM. XXII.-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-east-byeast. 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. XXIII.-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 then was the distance between the house and mill ? Ans. 959-5866 yards. EXAM. XXIV.— v.—Wanting to know my distance from an inaccessible object 0, on the other side of a river; and having no instrument for taking angles, but only a chain or cord 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 100 yards, viz. AC and BD each equal to 100 yards; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object O from each station A and B? Ans. AO 536.25 BO 500-09 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, a foot, 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 case is 12. 3 D 4 C A B E PROBLEM L 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. * • The truth of this rule is proved in the Geometry, 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 equal parts, each equal to the lineal 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 farther, that the number of these little squares, or the area of the figure, is equal to the number of lineal measuring units in the length, repeated as often as there are lineal measuring units in the breadth, or height; that is, equal to the length drawn into the height; which here is 4 X 3 or 12. And it is proved (Geometry, Theor. 25, Cor. 2), that a rectangle is equal to any oblique parallelogram, of equal length and perpendicular breadth. Therefore, the rule is general for all parallelograms whatever. EXAMPLES. Ex. 1.—To find the area of a parallelogram, whose length is 12-25, and height 8.5. 12.25 length 8.5 breadth 6125 9800 104.125 area Ex. 2. To find the area of a square, whose side is 35.25 chains. Ans. 124 acres, 1 rood, 1 perch. Ex. 3.—To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 93 feet. Ex. 4.-To find the content of a piece of land, in form of a rhombus, its length being 6.20 chains, and perpendicular height 5'45. Ans. 3 acres, I rood, 20 perches. Ex. 5.-To find the number of square yards of painting in a rhomboid, whose length is 37 feet, and breadth 5 feet 3 inches. PROBLEM II. Ans. 21 square yards. To find the area of a triangle. RULE 1-Multiply the base by the perpendicular height, and half the product will be the area.* Or, multiply the one of these dimensions by half the other. EXAMPLES. Ex. 1.-To find the area of a triangle, whose base is 625, and perpendicular height 520 links? Here 625 X 260 = 162500 square links, or equal 1 acre, 2 roods, 20 perches, the answer. Ex. 2.—How many square yards contains the triangle, whose base is 40, and perpendicular 30 feet? Ans. 66 square yards. Ex. 3.—To find the number of square yards in a triangle, whose base is 49 feet, and height 254 feet. Ans. 68, or 68 7361. Ex. 4. To find the area of a triangle, whose base is 18 feet 4 inches, and height 11 feet 10 inches. Ans. 108 feet, 5 inches. RULE II. -When two sides and their contained angle are given: Multiply the two given sides together, and take half their product: Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle. Or, multiply that half product by the natural sine of the said angle. † * The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base and altitude, by Geometry, Theor. 26. + For, let AB, AC, be the two given sides, including the given angle A. Now AB X CP is the area, by the first rule, CP being perpendicular. But, by Trigonometry, as sine angle P, or radius, is to sine angle A :: AC: CP sine angle A X AC, taking radius 1. Therefore, the area AB X CP is X sin, angle A, to radius 1; or, AB X AC A as radius sin. angle A :: AB X AC: the area. Ex. 1.—What is the area of a triangle, whose two sides are 30 and 40, and their contained angle 28° 57′ 18" ? HereX 40 X 30 = 600, Therefore, 1:4841226 nat. sin. 28° 57′ 18′′ 600 290 47356, the answer. Ex. 2.—How many square yards contains the triangle, of which one angle is 45o, and its containing sides 25 and 21 feet? Ans. 20-86947. RULE III.-When the three sides are given: Add all the three sides together, and take half that sum. Next, subtract each side severally from the said half sum, obtaining three remainders. Lastly, multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle. * Ex. 1.-To find the area of the triangle whose three sides are 20, 30, 40. The root of which is 290 4737, the area. Ex. 2.-How many square yards of plastering are in a triangle, whose sides are 30, 40, 50? Ans. 663. Ex. 3-How many acres, &c. contains the triangle, whose sides are 2569, 4900, 5025 links? Ans. 61 acres, 1 rood, 39 perches. PROBLEM III. To find the area of a trapezoid. Add together the two parallel sides; then multiply their sum by the perpendicular breadth or distance between them; and half the product will be the area: by Geometry, theorem 29. Ex. 1.-In a trapezoid, the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links: to find the area. 1225 1975 × 770 = 152075 square links = 15 acres, 33 perches, * For, let a, b, c, denote the sides opposite respectively to A, B, C, the angles of the triangle A B C (see last fig.); then by Theor. 37, Geom. we have BC2=AB2+AC2-2AB.AP, or a2= b2+c2-2c.AP b2+c2-α2 .. AP= ; hence we have .. 4c2.CP2= · AB.CP = { c. CP = √ √ { a+b+c. -atote, a_btc a+b-c} . = √s(s—a) (s—b) (s—c), ... 2 2 -a+b+c 2 = (a+b+c)=half the sum of the three sides. 2 |