## An Elementary Treatise on the Application of Trigonometry: To Orthographic and Stereographic Projection, Dialling, Mensuration of Heights and Distances, Navigation, Nautical Astronomy, Surveying and Levelling : Together with Logorithmic and Other Tables : Designed for the Use of the Students of the University at Cambridge, New EnglandPrinted at the University Press by Hilliard & Metcalf, 1822 - 153 pages |

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altitude azimuth centre chart circle Co-secant Secant Co-sine Co-tang compasses construction correction corresponding decimal Degs departure dial diameter diff difference of latitude difference of level difference of longitude divided divisions ecliptic equal equator feet figure formula Geom given log half height horizon hour angle HourA.M HourA.M.HourP.M inclination intersection latitude and longitude length line of chords logarithms manner measure meridian method miles multiplied N.sine nautical miles number of degrees oblique observed obtain parallax perpendicular plane of projection plane triangle pole pole star primitive proportion radii radius refraction represent respectively right angle right ascension sides Sine sphere spherical excess spherical triangle star stereographic projection straight lines subtract sun's declination supposed tang VBA Tangent tion Trig trigonometry tude vertical whence zenith distance

### Popular passages

Page 26 - ... for half an hour, &c., insert a rod at the centre perpendicularly to the dial plane, and place the dial so that this rod, representing the axis of the earth, shall be situated like cp, that is, in the plane of the meridian, and making an angle with the horizon equal to the latitude of the place. This would be an equinoctial dial, since its plane is parallel to that of the equator. It would moreover be a horizontal dial at the pole, and a vertical south dial at the equator. 42. To construct a...

Page 13 - If two triangles have two sides of the one respectively equal to two sides of the other, and the contained angles supplemental, the two triangles are equal.

Page 127 - E, and the several divisions of the arc AD, and the points of intersection with the radius CA, being numbered with the corresponding figures of the arc AD, will be a line of semi-tangents. This line is generally continued as far as the length of the scale will admit. The divisions beyond 90° are found by dividing the arc AE like the arc AD, and placing a ruler on E and these divisions of AE, and the line of semi-tangents above 90° will be obtained on CA continued. 8. For the line of longitude,...

Page 55 - ... 28". 88. Moreover, if we draw the parallel of declination rs 23° 27' 57" south of EQ, we shall have BQ°f equal to AQ T ; and °fB, equal °PA, converted into time, shows how long it is after 6 o'clock before the sun rises, and how long before 6h the sun sets. The longest night and shortest day, therefore, become equal respectively to the longest day and shortest night, as before found. It will be perceived from what is above shown, that when the latitude and declination are both north or both...

Page 133 - ... polygon, upon a given right line, set off the extent of the given line, as a transverse distance between the points upon the line of polygons, answering to the number of sides of which the polygon is to consist ; as for a pentagon between 5 and 5 ; or for an octagon between 8 and 8 ; then the transverse distance between 6 and 6 will be the radius of a circle whose circumference would be divided by the given line into the number of sides required. The line of polygons may likewise be used in describing,...

Page 93 - It may be remarked that, when the several operations are performed with perfect accuracy, the sum of the northings will be equal to that of the southings, and the, sum of the eastings to that of the westings. This necessarily follows from the circumstance of the surveyor's returning to the place from which he set out ; and it affords a means of judging of the correctness of the work. But it is not to be expected that the measurements and calculations in ordinary surveying will strictly bear this...

Page 33 - Then, as radius = sin 90° .... 10,00000 istoAB 2,29884 so is sin ABC = 46° . . . . 9,85693 to the height AC = 143,14 . . . 2,15577 50. It is required to find the perpendicular height of a cloud or other object, when its angles of elevation, as taken by two observers at the same time, on the same side of it, and in the same vertical plane, were 64° and 35°, their distance apart being half a mile, or 880 yards. It is evident from figure 28, that this problem may be solved in the same manner as...

Page 86 - ABCD (fig- 64), the breadth or perpendicular distance of either two opposite sides, as CP, is equal to the product of the corresponding oblique side CB by the sine of the angle of the parallelogram, radius being unity ((Trig. 30). Hence, the area of a parallelogram is equal to the product of any two contiguous sides multiplied by the sine of the ( contained angle, radius being unity. Given AB = 59 chains 80 links, or...