An Elementary Treatise on the Application of Trigonomentry to Orthographic and Stereographic Projection, Dialling, Mensuration of Heights and Distances, Navigation, Nautical Astronomy, Surveying and Levelling: Together with Logarithmic and Other Tables : Designed for the Use of the Students of the University at Cambridge, New England
Hilliard, Gray, 1833 - 155 páginas
Comentarios de la gente - Escribir un comentario
No encontramos ningún comentario en los lugares habituales.
Otras ediciones - Ver todas
altitude azimuth centre chart circle Co-secant Secant Co-sine Co-tang compasses construction correction corresponding decimal degrees and minutes 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 height horizon hour angle Houram HourP.M inclination latitude and longitude length line of chords logarithms measure meridian miles multiplying N.sine nautical miles Nsine.N number of degrees oblique observed obtain parallax parallel perpendicular plane of projection plane triangle pole pole star primitive proportion radii radius refraction represent respectively right angle right ascension sides Sine ſº sphere spherical excess spherical triangle star stereographic projection straight lines subtract sun's declination sun’s suppose tang Tangent Trig trigonometry tude vertical whence zenith distance
Página 2 - Every section of a circular cone made by a plane parallel to the base is a circle.
Página 79 - Method of correcting the apparent distance of the Moon from the Sun, or a Star, for the effects of Parallax and Refraction.
Página 52 - ... 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 south, the sun rises before and sets after 6 o'clock ; but when one is north and the other south, the sun rises after and sets before 6. 89. We have seen that nm and...
Página 28 - AB 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 the last.
Página 29 - To prove that the exterior angle of a triangle is equal to the sum of the two interior opposite angles (see fig.
Página 84 - 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...
Página 85 - ... the same thing, the product of one side by half the other. Moreover, since any triangle whatever is equal to a right-angled triangle of the same base and altitude (Geom. 170), we can make use of the following simple rule, where the known parts admit of it, as equivalent to the foregoing ; namely, the area of a triangle is equal to the product of the base by half its altitude. Given AB (fig. 65) = 12,38 ch., AC = 6,78 ch., and the Fig. 65. angle A = 46° 24' to find the area. 12,38 . . log. ....