you find the Degrees of your Quadrant to be 52.50 d. and Ob- Fig. II. Jerving at F, you find the Degrees to be 63.25. Now from this Distance Meafured, and the Two Obfervations by the Inftrument made, the Altitude DE may be attained unto by Trigonometrical Calculation, thus: First, Upon a Sheet of Paper, draw a Line at Pleafure, as HK, upon which affume a Point for the Place of your firft ftanding, as at G. and upon G, protract the Angle obferved 52.50 d. drawing a Line through thofe Degrees at Liberty. Secondly, By help of a Scale, fet your Measured Distance 95.25 F. from G to F, and upon F, by a Scale of Chards, protract an Angle of 63.25 d. as you obferved them to be; and through them draw another Line at Pleafure, which will cut the former Line drawn, in the Point D, which will reprefent the top of the Object to be measured; and a Perpendicular let fall from D, upon the Ground-line H K, will fall in the Point E, and fo will the Line DE reprefent the Object it felf. Thirdly, By thefe Lines thus drawn, you will have constituted Two Triangles; one D E F, Right-angled at E; and the other DFG, Obtufe-angled at F; by the refolving of which, the height of the Inacceffible Object D E, will be found. For, 1. In the Oblique-angled Triangle DFG, you have given, the Side F G (which was the Meafured Diftance) 95.25 F. and the obferved Angle DGF, 52.50 d. —And the Angle obferved at F being 63.25 d. the Complement thereof to 180 d. viz. 116.75 d. is the Quantity of the Obtufe Angle D F G, fo have you in the Oblique-angled Triangle DF G, Two Angles given; the Sum of which, viz. 169.25 d. taken from 180 d. there remains 10.75 d. for the Angle FDG: And now in the Triangle DFG you have given Two Angles FDG, and DGF, with the Side F G, oppofite to F DG; whereby, you may find the Side DF, (by Ax. H.) For, As the Sine of FDG, 10.75 d. Is to the Side F G, 95.25 Foot. So is the Sine of FGD, 52.50 d. To the Side D F, 405.14. 2. In the Right-angled Triangle DEF, (having found the Hypotenuse as betore) you have given, (1.) The Obferved Angle at F, 63.25 d. (2.) The Hypotenuse, laft found, DF 405.14 Foot; by which you may find DE, the Height of the Object, (by Cafe IV. of R, A, P, T,) thus: As Fig. II. As the Radius, Sine 90 d. Fig. III. Is to the Hypotenuse DF, 405.14 F. So is the Sine of DF E, (the Angle obferved at F) 63.25 d. And now (if you pleafe) you may find the Visual Line GD, (by Cafe V. of R, A, P, T,) thus: As the Sine of the Angle at G, 52.50 d. Is to the height of the Object DE, 361.77 F. So is Radius, Sine 90 d. To the Length of the Vifual Line DG 456.00 F. And alfo, the Distance EF, (by Cafe II. of R, A, P, T,) thus: As the Sine of the Angle obferved at F. 63.25 d. Unto which, if you add the Distance F G, 95.25 Foot, their III. Of the Altitude of an Object ftanding upon a Hill, Un-acceffible. Suppofe M O to be fuch an Object; and you ftanding at L, were required to tell the Height thereof. Firft, Upon Paper, or the like, draw a Right-line at Pleasure, as QR, and therein, affume any Point, at Pleasure, for the Place of your ftanding, as L; where, with your Inftrument directed to the top of the Object, you find the Degrees cut, to be 40.52 d. and directed to the bottom of the Object at O, the Degrees cut, to be 22.25 d. Wherefore upon L, protract an Angle of 40.52 d. and draw a Line L, at Pleafure: And alfo b, an Angle of 22.25 d. and draw the Line Lcat Pleasure. Secondly, Go forwards, in a Right-line towards the Object, fome confiderable Distance, as to N, 212.5 Foot; and there, by your Inftrument directed to the top of the Object at M, you find the Degrees cut, to be 61.82 d. through which draw a Line at Pleafure, as Na, croffing the Line Lb in the Point M, which is the top of the Object: From whence, a Perpendicular let fall upon the Ground. Ground-line QR, as M P, that Line fhall be equal to the Ati tude, of the Object, and the Hill together. Now, by the Interfections of thefe Four Visual Lines, LM, LG, NM, and NO, there are conftituted Four Right-lined Triangles, viz. LM P, and NO P, both Right-angled at P: And LMN, and N M O, Oblique-angled: By the refolving of which, from the Distance Meafured, LN, and the feveral Angles obferved, at L and N, the required Altitude may be obtained. For, 1. In the Oblique-angled Triangle L M N, there is given, (1.) The Angle ML N, 40.52 d. (2.) The Angle L N M, 118.18 d. (it being the Complement of the Angle ONP 61.82d.. to 180 d.) (3.) The Side Meafured L N, 212 5 Foot: And having the Angles at L and N, the Sum of them 158.70 d. taken from 180 d. there will remain 21.30 d. for the Angle L M N. From which Things given, the Two other Sides, L M and MN, may be found, by Axiom II. thus: As the Sine of LM N, 21.30 d. Is to the Side L N, 212.5 Foot; And fo is the Sine of 61.82 d. (the Complement of M N L, To the Side L M. 515.66 In the Right-angled Triangle N M P, there is given, (1.) The Hypotenuse M N, 380.08 Foot. (2.) The Angle MNP, 61.82 d. whereby you may find NP, (by Cafe IV. of R, A, P, F) thus: As the Radius, Sine sod. Is to the Hypotenufe MN, 380.08 Foot; So is the Sine of the Angle M N P, 61.82 d. Which is the Height of the Object and the Hill together. And fo is the Co-fine of M N P, viz. NM P, 28.18 d. To which, if you add the measured Distance L N, 212.05, their Then, 3. In the Triangle LO P, (Right-angled at P) you have given, (1.) The Side L P, 391.54 Foot. (2.) The Angle OL P, 22.25 d. whereby Fig. III. whereby you may find O P, (By Cafe II. of R, A, P, T.) Fig. IV. Fig. V. thus: As the Radius, Tang. 45 d. Is to the Side LP, 391.54 Foot; So is the Tangent of the Angle OL P, 22.25 d. Which substracted from M P (the whole. Height) 335-93, M O. IV. Horo the Altitude of the Sun may be taken, by the Shadow of a Staff, or other Object, of a known Length. Upon A B, being a level Plain, let there be erected a streight Staff, or the like, of any Length, fuppofe 60.00 Inches, or 5 Foot) as CD; and the Sun fhining, fuppofe it cafts the Shadoro thereof upon the plain Ground to E; which meafured, fuppofe to contain 108 Inches, or 9 Foot. Draw a Line at Pleafure, as A B; upon any Point thereof, as D, erect a Perpendicular, upon which fet 60 Inches, the length of your Staff, from D to C; and the length of the Shadow thereof 108 Inches, from D to E; drawing the Line of Umbrago C E, and thus have you conftituted a Right-angled Plain Triangle CDE, in which there is given, (1.) The Leg. C D, 60 Inches. (2) The Leg. DE 108 Inches, by which you may find the Angle CED, (by Cafe I. of R, A, S, T,) thus: As the Length of the Shadow CD 108.00 Inches, Is to the Radius, Tang. 45 d. So is the Length of the Staff C D, 60.00 Inches, And fuch is the Sun's Altitude at that time. V. How the Height of an Acceffible Object may be obtained, by the Length of the Shadow of it. Suppofe that the Sun fhould caft the Shadow of fome upright Object, as FG, 84.5 Foot, from G to L. and at the fame time, your Staff of 5 Foot, does caft its Shadow from L to K, 6.32 Foot: And from hence I would compute the Altitude of the Orje FG. The 7 The Two Lines of the Object FG, and Staff H L, together with Fig. V. the Two Lines of Shadow, GL and LK, being laid down, do conftitute Two Right-angled Plain Triangles, viz. FGL, and HLR, Equiangled, and their Sides Parallel, and therefore Proportional (by Theorem VIII Lib. I.) And therefore, As the Length of the Shadow of the Staff LK, 632 Foot, So is the Length of the Shadow of the Obje&G L 84.5 Foot, CHA P. II. Of LONGIMETRIA. I. How (tanding upon an Object of a known Height) to find the Suppofe CA to be the Side of a fort or Bulwark 22.5 Foot high, Fig. VI. and being upon the Platform at C, you fee a Tree, or other Object at B, whofe Distance you would know, from the Foot of the Wall at A. The Lines A B and A C being drawn, and the Height of the Wall 22.5 Foot, fet from A to C, where by your Inftrument directed to B, you find the Degrees cut to be 71.25, which Angle lay down, fo have you the Right-angled Triangle CA B, in which there is given, (1.) CA, the Height of the Wall 22.5 Foot. (2.) The Angle obferved at C, 71.25 Deg. by which you may find the Distance A B, (by Cafel. of R, A, P, T.) thus: As Radius, Tangent 45 Deg. To CA, the Height of the Wall 22.5 Foot; So is the Tangent of A CB, the Angle obferved, 71.25 Deg. And if you would find the Length of the Vifual Line CB, As the Sine of the Angle obferved at C, 71.25 Deg.. Is to the Distance B A, 66.28 Foot; So is the Radius, Sine 90 Deg. To the Vifual Line C B, 69.98 Foot. B b II. To |