CASE IV. Fig. 49. The three sides given to find the angles. Fig. 49. 85 75 30 106 The solution of this case depends on the following PROPO. SITION, IN EVERY PLANE TRIANGLE, AS THE LONGEST SIDE IS TO THE SUM OF THE OTHER TWO SIDES, SO IS THE DIFFERENCE BETWEEN THOSE TWO SIDES TO THE DIFFERENCE BETWEEN THE SEGMENTS OF THE LONGEST SIDE, MADE BY A PERPENDICULAR LET FALL FROM THE ANGLE OPPOSITE THAT SIDE. Half the difference between these segments, added to half the sum of the segments, that is, to half the length of the longest side, will give the greatest segment; and this half difference subtracted from the half sum will be the lesser seg. ment. The triangle being thus divided, becomes two right angled triangles, in which the hypothenuse and one leg are given to find the angles. In the triangle ABC, given the side AB 105, the side AC 85, and the side BC 50, to find the angles. Side AC 85 AC 85 BC 50 BC 50 Thus the triangle is divided into two right angled triangles, ADC and BDC; in each of which the hypothenuse and one leg are given to find the angles. To find the angle DCA. To find the angle DCB 1.698970 10.000000 :: seg. AD, 75 1.875061 : : seg. BD, 30 1.477121 The angle DCA 619 56' subtracted from 90° leaves the angle CAD 28° 4'. The angle DCB 36° 52' subtracted from 90° leaves the an. gle CBD 53° 16. The angle DCA 61° 56' added to the angle DCB 36° 52 gives the angle ACB 98° 48'. This case may also be solved according to the following PROPOSITION. IN EVERY PLANE TRIANGLE, AS THE PRODUCT OF ANY TWO SIDES CONTAINING A REQUIRED ANGLE IS TO THE PRODUCT OF HALF THE SUM OF THE THREE SIDES, AND THE DIFFERENCE BETWEEN THAT HALF SUM AND THE SIDE OPPOSITE THE ANGLE REQUIRED, SO IS THE SQUARE OF RADIUS TO THE SQUARE OF THE CO-SINE OF HALF THE ANGLE REQUIRED. Those who make themselves well acquainted with TRIGONOMETRY will find its application easy to many useful purpos. es, particularly to the mensuration of heights and distances; These are here omitted, because, as this work is designed principally to teach the art of common FIELD-SURVEYING, it was thought improper to swell its size, and consequently in crease its price, by inserting any thing not particularly connected with that art. It is recommended to those who design to be surveyors to study TRIGONOMETRY thoroughly; for though a common field may be measured without an acquaintance with that science, yet many cases will occur in practice where a knowledge of it will be found very beneficial; particularly in dividing land, and ascertaining the boundaries of old surveys. Indeed no one who is ignorant of TRIGONOMETRY, can be an accomplished surveyor. D2 SURVEYING. SURVEYING is the art of measuring, laying out, and dividing land. PART I. MEASURING LAND. The most common measure for land is the acre ; which contains 160 square rods, poles or perches; or 4 square roods, each containing 40 square rods. The instrument most in use, for measuring the sides of fields, is GUNTER's chain, which is in length 4 rods or 66 feet; and is divided into 100 equal parts, called links, each containing 7 inches and 92 hundredths. Consequently, 1 square chain contains 16 square rods, and 10 square chains make 1 acre. In small fields, or where the land is uneven, as is the case with a great part of the land in New England, it is better to use a chain of coly two rods in length ; as the survey can be more accurately taken. SECTION I. PRELIMINARY PROBLEMS. PROBLEM I. To reduce two rod chains to four rod chains. Rule. If the number of two rod chains be even, take half the number for four rod chains, and annex the links if any ; thus, 16 two rod chains and 37 links make 8 four rod chains and 37 links. But if the number of chains be odd, take half the greatest even number for chains, and for the remaining number add 50 to the links: Thus, 17 two rod chains and 42 links make 8 four rod chains and 92 links. PROBLEM II. To reduce two rod chains to rods and de. Rule. Multiply the chains by 2, and the links by 4, which will give hundredths of a rod; thus, 17 two rod chains and 21 links make 34 rods and 84 hundredths ; expressed thus, 34.84 rods. If the links exceed 25, add 1 to the number of rods and multiply the excess by 4: thus, 15 two rod chains and 38 links make 31.52 rods. PROBLEM III. To reduce four rod chains to rods and decimal parts. RULE. Multiply the chains, or chains and links, by 4 ; the product will be rods and hundredths: thus, 8 chains and 64 links make 34.56 rods. Note. The reverse of this rule, that is, dividing by 4, will reduce rods and decimals to chains and links : thus, 105.12 rods make 26 chains and 28 links. PROBLEM IV. To reduce square rods to acres. RULE. Divide the rods by 160, and the remainder by 40, if it exceeds that number, for roods or quarters of an acre: thus 746 square rods make 4 acres, 2 roods, and 26 rods. PROBLEM V. To reduce square chains to acres RULE. Divide by 10; or, which is the same thing, cut off the right hand figure: thus, 1460 square chains make 146 acres; and 846 square chains make 84 acres and 6 tenths. PROBLEM VI. To reduce square links to acres. RULE. Divide by 100000 ; or, which is the same, thing, cut off the 5 right-hand figures: thus, 3845120 square links make 38 acres and 45120 decimals. Note. When the area of a field, by which is meant its superficial contents, is expressed in square chains and links, the whole may be considered as square links, and the number of acres contained in the field, found as above. Then multiply the figures cut off by 4, and again cut off 5 figures, and you have the roods; multiply the figures last cut off by 40, and again cut off 5 figures, and you have the rods. EXAMPLE. How many acres, roods, and rods, are there in 156 square chains and 3274 square links? 15)63274 square links, 4 2)53096 40 21) 23840 Answer. 15 acres 2 roods and 21 rods. PROBLEMs for finding the area of right lined figures, and also % PROBLEM VII. To find the area of a square or rectan, gle. Rule. Multiply the length into the breadth ; the product will be the area. PROBLEM VIII. To find the area of a rhombus or rhomboid. Rule. Drop a perpendicular from one of the angles to its opposite side, and multiply that side into the perpendicular; the product will be the area. PROBLEM IX. To find the area of a triangle. Rule 1. Drop a perpendicular from one of the angles to its opposite side, which may be called the base ; then multi. ply the base by half the perpendicular; or the perpendicular by half the base ; the product will be the area. Or, multiply the whole base by the whole perpendicular, and half the product will be the area. Role 2. If it be a right angled triangle, multiply one of the legs into half the other; the product will be the area. Or, multiply the two legs into each other, and half the pro. duct will be the area. Rule 3. When the three sides of a triangle are known, the area may be found arithmetically, as follows: Add together the three sides; from half their sum subtract each side, noting down the remainders ; multiply the half sum by one of those remainders, and that product by another remainder, and that product by the other remainder; the square root of the last product will the area.* EXAMPLE. Suppose a triangle whose three sides are 24, 20, and 18 chains. Demanded the area. 24x20x18=62, the sum of the three sides, the half of which is 3]. From 31 subtract 24, 20, and 18; the three remainders will be 7, 11, and 13. 31x7=217; 217x11=2387; 2387x13=31031, the square root of which is 176.1, or 17 acres 2 roods and 17 rods. BY LOGARITHMS. As the addition of logarithms is the same as the multiplication of their corresponding numbers ; and as the number an. swering to one half of a logarithm will be the square root of the number corresponding to that logarithm : it follows, that if the logarithm of the half sum of the three sides and the logo arithms of the three remainders be added together, the num. * Better expressed thus. From half the sum of the sides subtract each side separately Multiply the half sum and the several remainders together, |