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 - 155 من الصفحات
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according altitude angle applied bearing becomes called centre chords circle Co-secant Co-sine Co-tang consequently considered construction contained correction corresponding course declination Degs departure described determined dial difference distance divided divisions draw earth equal equator example extent feet field figure formed formula Geom give given greater half height horizon hour Hourp.m inclination instance known latitude length less logarithms longitude manner means measure meridian method middle miles minutes multiplying N.sine namely object observed obtain opposite parallel pass perpendicular plane pole primitive projection proportion radius refraction remainder represent respectively right angle right ascension rules scale Secant ship sides similar Sine sphere spherical star straight lines subtract sun's suppose taken tang Tangent triangle Trig true whence zenith distance
الصفحة 6 - Every section of a circular cone made by a plane parallel to the base is a circle.
الصفحة 83 - Method of correcting the apparent distance of the Moon from the Sun, or a Star, for the effects of Parallax and Refraction.
الصفحة 56 - ... 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...
الصفحة 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.
الصفحة 33 - To prove that the exterior angle of a triangle is equal to the sum of the two interior opposite angles (see fig.
الصفحة 88 - 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...
الصفحة 89 - ... 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. ....