Plate V. fig. 23. S.ACB: AB :: S.A: BC. 18° 87 37° 169.4 Having found BC, we have in the triangle BCD; the angle CBD 55°, consequently BCD 35°, and BC 169.4; to find DC. This is performed by case the first, of right-angled trigonometry, three several ways; thus; 1. R: BC:: S. CBD : DC. 2. Sec. CBD : BC :: T. CBD : DC. 55° 169.4 The height required. 55° 138.8. 3. Sec. BCD: BC :: R: CD. 35° 169.4 90° 138.8 The height required. If BD, the breadth of the moat, were required, it may also be found, by three different statings; as in the first case of right-angled plane trigonometry. Plate VI. fig. 24. PROBLEM VI. Let BC, a may-pole, whose height is 100 feet, be broken at D; the upper part of which, DC, falls upon an horizontal plane, so that its extremity, C, is 34 feet from the bottom or foot of the pole. Required, the segments BD and DC. Geometrically. Lay 34 feet from A to B; on B erect the perpendicular BC of 100 feet; and draw AC; bisect AC (by prob. 4. sect. 1.) with the perpendicular line, EF, and from D, where it cuts the perpendicular BC, draw AD, which will be the upper segment; and DB will be the lower. By cor. to lemma, preceding theo. 7. sect. 1. AD-DC; and (by the lemma) the angle C-CAD. In the triangle ABC, find C as in case 6, of right-angled trigonometry, thus; 1. BC: R:: AB: T. C-GAD. 100 90° 34 18° 47′ By theo. 4. sect. ABD-37° 34' or to C and GAD. 1. The external angle twice the angle C, i. e. to Then in the triangle ABD, there is ABD 37. 34, therefore also its complement DAB 52° 26' and AB 34, given, to find AD and BD. By the second case of rectangular trigonometry. 2. S. ADB AB:: R: AD or DC. 37° 34' 34.90° 55.77 BC-DC- BD. 100-55.77-44.23 required. These may be had from other statings, as in the second case aforesaid. PROBLEM VII. To take the altitude of a perpendicular object on a hill, from a plane beneath it. Plate V. fig. 25. This is done at two stations, thus ; Let the height DC, of a wind-mill on a hill be required. From any part of the plane whence the foot of the object can be seen, let angles be taken to the foot and top; measure thence any convenient distance towards the object, and at the end thereof take another angle to the top and you have the proper requisites, thus: First station. Angle to the foot DAB 21° 00' Angle to the top CAB 35 00 Stationary distance AB 104 feet. Second station Angle to the top 48° 30′. DC required. Geometrically. On an infinite blank line, lay the stationary distance AB 104 feet; from A, set off the second and from B, the third given angle; and from the Plate V. fig. 25 intersecting point C of the line formed by them, let fall the perpendicular CE; from A set off the first angle, and the line formed by it will determine the point D. Thus we have the height of the hill, as well as that of the wind-mill. S The angle CBE-A-ACB, as in the last prob. In the triangle ABC, find AC thus, $ ACB ABS. ACB (or sup. of CBE) AC 13° 30′ : 104 :: 131° 30' The angle CAE-DAE-CAD. The angle ACD=AED+EAD, by theo. 4. In the triangle CAD, find CD thus, S. ADC: AC :: S. CAD : DC : 333.6 111° : 333.6 :: 14 86.46 required. CE, BE, or DE, may be found by other various statings, as set forth in the first and second cases of rectangular trigonometry. Plate V. fig. 26. PROBLEM VIII. To find the length of an object, that stands obliquely on the top of a hill, from a plane beneath. Let CD be a tree whose length is required. This is done at two stations. Make a station at B, from whence take an angle to the foot, and another to the top of the tree; measure any convenient distance backward to A, from whence also let an angle be taken to the foot, and another to the top; and you have the requisites given. Thus, First station. Angle to the foot EBD=36°. 30'. Angle to the top EBC=44°. 30'. Stationary distance AB-104 feet. Second station. Angle to the foot EAD=24°. 30'. Angle to the top EAC-32°. 00'. Let DC and DE be required. The geometrical constructions of this and the next problem are omitted; as what has been already said, and the figures are looked upon as sufficient helps. EBC-A-ACB, or 44°. 30-32°-12°. 30', as before. |