And in the triangle ABC, find BC thus, 2. S. ACB : AB :: S. CAB: BC. 22° 30' 113 36° 30' 175. 6. In the triangle DBC, you have DBC=ABCABD=72°; likewise the sides BD, BC, as before found, given to find DC. 3. BD+BC : BD-BC :: T. of DCB+CDB : 391.6 40.4 54° T. of 1 DCB-CDB. 8° 05'. 54° +8° 05' = 62° 051 = DCB. 4. S. CDB : BC :: S. DBC: DC 45° 55' 175.6 72232.5 If from a point C, of a triangle ABC, inscribed in a circle, there be a perpendicular CD, let full upon the opposite side AB ; that perpendicular is to one of the sides, including the angle, as the other side, including the angle, is to the diameter of the circle, i. e. DC : AC :: CB: CE. Let the diameter CE be drawn and join EB; it is plain the angle CEB=CAB (by cor. 2. theo. 7. sect. 1.) and CBE is a right angle (by cor. 5. theo. 7. sec. 1.) and = ADC: whence ECB = ACD. The triangles CEB, CAD, are therefore mutually equiangular, and (by theo. 16. sect. 1.) DC : AC :: CB : CE, or DC : CB : : AC: CE. Q. E. D. PROBLEM V. Plate VI. fig. 5. Let three gentlemen's seats, A, B, C, be situate in a triangular form : there is given, AB 2.5 miles, AC 2.3, and BC 2. It is required to build a church at E, that shall be, equi-distant from the seats A, B, C. What distance must it be from each-seat, and by what angle may the place of it be found ? Geometrically. By prob. 15. sect. 1. Find the centre of a circle that will pass through the points A, B, C: and that will be the place of the church; the measure of which, to any of these points, is the answer for the distance : draw a line from any of the three points to the centre, and the angle it makes with either of the sides that contain the angle it was drawn to; that angle laid off by the direction of an instrument, on the ground, and the distance before found, being ranged thereon, will give the place of the church required. By Calculation. 1: AB : AC + BC :: ACBC : AD-DB. 2.5 4.9 .3 .516 1.25 +.2581.508=AD. By cor. 2. theo. 14. sect. 1. The square root of the difference of the squares of the hypothenuse AC, and given leg AD, will give DC. i. e. 5.29-2.274064=3.015936. Its square root is 1.736=CD. Then by the preceding lemma, 2. CD : AC :: CB : the diameter. the half of which, viz. 1.325 is the semi-diameter, or distance of the church from each seat, that is, AE. CE. BE. From the centre E let fall a perpendicular upon any of the sides, as EF, and it will bisect in E: (by theo. 8. sect. 1.) Wherefore, AF=CF=1 AC=1.15. In the right angled triangle AFE, you have AF 1.15, and AE the radius 1.325 given, to find FAE, thus; S. AF : R. : : AE : Sec. FAE. 1.15 90 1.325 29° 47' Wherefore directing an instrument to make an angle of 29° 47', with the line AC; and measuring 1.325 on that line of direction, will give the place of the church, or the centre of a circle that will pass through A, B, and C. The above angle FAE, may be had without a secant, as before, thus : AE: R. : : AF:S. AEF. 1.325 90° 115 60°. 13 Its complement 29•. 47', will give FAE, as before. : The questions that may be proposed on this head, being innumerable, we have chosen to give only a few of the most useful. SECTION III. Containing a particular description of the several Instruments used in surveying, with their respective uses. And first, OF THE CHAIN. THE stationary distance, or mearings of ground, of four poles or perches, which consist of 100 links ; (and this is the most natural division) or by one of 50 links, which contains two poles or perches : but because the length of a perch differs in many places, therefore the length of chains and their respective links will differ also. The English statute-perch is 57 yards, the twopole chain is 11 yards, and the four-pole one is 22 yards ; hence the length of a link in a statute-chain is 7.92 inches. There are other perches used in different parts of England, as the perch of woodland measure, which is 6 yards; that of church-land measure, which is 7 yards (or the same with the plantation perch) and the forest measure perch which is 8 yards. S |