# 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 and Company, 1833 - Trigonometry - 155 pages
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Page 6 - Every section of a circular cone made by a plane parallel to the base is a circle.
Page 116 - Knowing therefore the absolute distance A x, and the difference of latitude of these two points, we readily find the length of a degree, and thence the length of 360�, or the whole circumference of the earth considered as a sphere. By performing similar operations in different latitudes, we obtain the length of degrees in those latitudes, and thence the true figure of the earth, on the supposition that it is a spheroid of revolution, or such as would be formed by the revolution of an ellipse about...
Page 135 - ... 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 31 - ... to obtain the logarithm of the fourth. term. Or, adding the arithmetical complement of the logarithm of the first term to the logarithms of the other two, to obtain that of the fourth.
Page 32 - 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.
Page 89 - 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. . . 1,09272 6,78 . . . log. . . . 0,83123 40� 24
Page 88 - ABCD (Jig. 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...