 | John Bonnycastle - 1806 - عدد الصفحات: 464
...10.0000000 Is to AB 2.2988431 So is sin ZB - - 46° 9.8569314 To height AC-- 143.14 - - - 2/1557772 3. It is required to find the perpendicular height of...vertical plane, were 64° and 35° ; their distance asunder being half a mile, or 880 yards. As sin AB C - - -^ AD C - - ^ DAB - -- DAB - - - 29° G -... | |
 | Charles Hutton - 1807 - عدد الصفحات: 464
...when its angles of elevation were 35° and 64°, as taken by two observers, at the same time, both on the same side of it, and in the same vertical plane ; the distance between them being half a mile or 880 yards. And what was its distance from the said... | |
 | Charles Hutton - 1811 - عدد الصفحات: 494
...when its angles of elevation were 35° and 64°, as taken by two observers, at the same time, both on the same side of it, and in the same vertical plane ; the distance between them being half a mile or 880 yards. And what was its distance fronl the said... | |
 | John Bonnycastle - 1818 - عدد الصفحات: 488
...Is to AB 2.2929272 So is sin / B - - - 46° 10' - - 9.858150-5 ToheightAC -- l4l.6l -- 2.1510777 3. It is required to find the perpendicular height of...side of it, and in the same vertical plane, were 64° 10' and 35° 25'; their distance asunder being half a mile, or 880 yards. ± A. B c 64° 10' / ADC... | |
 | John Farrar - 1822 - عدد الصفحات: 270
...9,71184 to AB 2,29884 Then, as radius = sin 90" 10,00000 is to AB 2.29884 so is sin ABC = 46° 9,85693 to the height AC= 143,14 2,15577 50. It is required...problem may be solved in the same manner as the last. 51. From the top of a tower AC (Jig. 30) 120 feet high, thepig. SO. angles CAB, CAD, formed by the... | |
 | John Farrar - 1822 - عدد الصفحات: 244
...2,29884 Then, as radius = sin 90° ... .. . 10,00000 h| to AB 2.29884 so is sin ABC = 46° 9,85693 to the height AC— 143,14 , 2,15577 50. It is required...that this problem may be solved in the same manner as I he last. 51. From the top of a tower AC (Jig. 30) 120 feet high, thepig. 39. angles CAB, CAD, formed... | |
 | John Farrar - 1833 - عدد الصفحات: 274
...2,29884 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...problem may be solved in the same manner as the last. . so. 51- From the top of a tower AC (fig. 30), 120 feet high the angles CAB, CAD, formed by the perpendicular... | |
 | John Farrar - 1840 - عدد الصفحات: 270
...2,29884 Then, as radius = sin 90° .... 10,00000 is to AB 2,29884 so is sin ABC = 46° . . . 9,85693 to the height AC = 143,14 . . . 2,15577 50. It is...last. being measured and found to be 33°, and 64° 30', what is the distance of the two objects B, D 1 As radius or sin 90° 10,00000 is to AC = 120 feet... | |
 | Oliver Byrne - 1852 - عدد الصفحات: 600
...when its angles of elevation were 35° and 64°, as taken by two observers,. at the same time, both on the same side of it, and in the same vertical plane ; their distance, as under, being half a mile, or 880 yards. And what was its distance from the said... | |
 | John Radford Young - 1855 - عدد الصفحات: 218
...earth ( (6) The angles of elevation of a balloon were taken at the same time by two observers, both on the same side of it and in the same vertical plane as the balloon ; the angles were found to be 35° and 64°, and the observers were 880 yards apart... | |
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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,... 
