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 observed at F) 63.25 d. And now (if you please) 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 Objet DE, 361.77 F. So is Radius, Sine 90 d. To the Length of the Visual Line DG 456.00 F. And also, the Distance EF, (by Cafe II. of R, A, P, T,) thus: As the Sine of the Angle observed at F. 63.25 d. Is to the Height of the Object, DE, 361.77 F. So is the Co-fine of the Angle at F, 63.75 d. : Unto which, if you add the Distance FG, 95.25 Foot, their Sum will be 277.6 Foot. And that is the whole Distance from G to E, the Foot of the Object. III. Of the Altitude of an Object standing upon a Hill, Un-acceffible. Suppose M O to be such an Object; and you standing at L, were required to tell the Height thereof. First, 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 standing, as L; where, with your Instrument 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 Pleasure: And also b, an Angle of 22.25 d. and draw the Line Lcat Pleafure. Secondly, Go forwards, in a Right-line towards the Objelt, some confiderable Distance, as to N, 212.5 Foot; and there, by your Instrument 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, crossing 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 MP, that Line shall be equal to the A'ti tude, of the Object, and the Hill together. Now, by the Interfections of these Four Visual Lines, LM, LG, NM, and NO, there are conftituted Four Right-lined Triangles, viz. LMP, and NOP, both Right-angled at P: And LMN, and NMO, Oblique-angled: By the refolving of which, from the Distance Measured, LN, and the several Angles obferved, at L and N, the required Altitude may be obtained. For, 1. In the Oblique-angled Triangle LMN, there is given, (1.) The Angle MLN, 40.52 d. (2.) The Angle LNM, 118.18 d. (it being the Complement of the Angle O NP 61.82d. to 180 d.) (3.) The Side Measured LN, 2125 Foot: And having the Angles at L and N, the Sam of them 158.70 d. takeń from 180 d. there will remain 21.30 d. for the Angle LMN. From which Things given, the Two other Sides, LM and MN, may be found, by Axiom II. thus : As the Sine of L M N, 21.30 d. Is to the Side LN, 212.5 Foot; And fo is the Sine of 61.82 d. (the Complement of MN L, To the Side L M. 515.66 In the Right-angled Triangle NMP, there is given, (1.) The Hypotenuse MN, 380.08 Foot. (2.) The Angle MNP, 61.82d. whereby you may find NP, (by Cafe IV. of R, A, P, T,) thus: As the Radius, Sine sod. Is to the Hypotenuse MN, 380.08 Foot; So is the Sine of the Angle MNP, 61.82 d. To the Side MP, 335.03 Foot: Which is the Height of the Objet and the Hill together. And fo is the Co-fine of MNP, viz. NMP, 28.18 d. To which, if you add the measured Distance L N, 212.05, their Fig. III. Then, 3. In the Triangle LOP, (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 MP (the whole. Height) 335.93, IV. How 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, suppose 60.00 Inches, or 5 Foot) as CD; and the Sun shining, suppose it cafts the Shadore thereof upon the plain Ground to E; which measured, suppose to contain 108 Inches, or 9 Foot. Draw a Line at Pleasure, 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. CD, 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.co Inches, Is to the Radius, Tang. 45 d. So is the Length of the Staff CD, 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. Suppose that the Sun should caft the Shadow of some upright Object, as F G, 84.5 Foot, from Gto 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 Orjed FG. 7 The 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 constitute 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, Is to the Height of the Staff HL, 5.00 Foot; So is the Length of the Shadow of the Objeld G L 84.5 Foot, CHAP. II. Of LONGIMETRIA. I. How (Standing upon an Object of a known Height) to find the Distance from thence, to some other remote Object. Suppofe CA to be the Side of a and being upon the Platform at C, you fee a Tree, or other Objell at B, whose Distance you would know, from the Foot of the Wall at A. Fort or Bulwark 22.5 Foot high, Fig. VI. The Lines A B and A C being drawn, and the Height of the Wall 22.5 Foot, set from A to C, where by your Instrument directed to B, you find the Degrees cut to be 71.25, which Angle lay down, so have you the Right-angled Triangle CAB, in which there is given, (1.) CA, the Height of the Wall 22.5 Foot. (2.) The Angle observed at C, 71.25 Deg. by which you may tind the Distance AB, (by Cafel. of R, A, P, T.) thus: As Radius, Tangent 45 Deg. To C A, the Height of the Wall 22.5 Foot; So is the Tangent of ACB, the Angle observed, 71.25 Deg. And if you would find the Length of the Visual Line CB, As the Sine of the Angle observed at C, 71.25 Deg. Is to the Distance B A, 66.28 Foot; So is the Radius, Sine 90 Deg. To the Visual Line C B; 69.98 Foot. II. To Fig. VII. Fig. VIII. II. To take a Distance, (Acceffible or In-acceffible) at Two Sta tions. Being in a Field at E, there is a Windmill (or other Object, in another Field at F, (feparated by a River or other Impediment) whose Distance is required. In any other Part of the Field, remote from E, cause a Mark to be fet up, as at D; and measure the Distance between E and D, which let be 115.00 Foot. -Then by your Inftrument at E, find the Quantity of the Angle FED, which suppose to be 106.50 Deg. And going from E to D, make Obfervation of the Angle FDE, which suppose to be 57.10 Deg. By these Two Angles, and the Measured Distance, you have conftitated an Oblique Triangle DE F, in which there is given, (1.) The Measured Distance E D, 11500 Foot. (2.) The Two Angles at D and E, 106.50 Deg. and 57.10 Deg. And, (3.) The Angle at F, (the Complements of the other Two to 180 Deg.) 16.40 Deg. Whereby you may find the Sides E F and DF, (by Axiom. II.) thus: As the Sine of the Angle at F, 16.40 Deg. Is to the Measured Distance DE, 115.00 Foot; So is the Sine of the Angle at F, 57.10 Deg. And fo is the Angle at E, 105.50 Deg. (cr 73.50.) To the Distance DF, 390.54 Deg. III. If there were Three (or more) Ships on the Sea, and you be ing upon the Land, defire to know how far those Ships are from you, and alfo, how far they are distant one from the other. ET the Three Ships be A, B, C; and you being upon the L First, Being at M, make choice of fome other Place upon the Secondly, Being at M, observe the Quantity of the Angle AMO, which we will suppose to contain 104.50 Deg. and then removing to O, and observing the Angle MO A, you find it to be 37.82 Deg. through which Degrees, Lines being drawn from M and O, will cross each other in A, which is the Place of the first Ship, which, with the Line of Diftarce MO, will form the Oblique |