A Manual of the Protracting Trigonometer: With Its Application to Rectilinear Draughting and Plotting, Trigonometry, and Surveying

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Sheldon, 1862 - Protractors - 77 pages

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Page 14 - The circumference of every circle is supposed to be divided into 360 equal parts, called degrees ; each degree into 60 equal parts, called minutes ; and each minute into 60 equal parts, called seconds.
Page 55 - To find the area of a trapezoid. RULE. Multiply half the sum of the two parallel sides "by the perpendicular distance between them : the product will be the area.
Page 15 - After remarking that the mathematician positively knows that the sum of the three angles of a triangle is equal to two right angles...
Page 68 - AC 16.25 ch., and the distance of a point P in the side AB, from the angle A 8.50 ch. ; it is required to find the distance AG of a point G in the line AC, so that a line drawn from P to G may cut off a triangle APG containing 3 acres. Ans. 9.25 ch. PROBLEM VIII. The area and base of a triangle being given, to cut off a triangle containing a given area by a line running parallel to one of the sides.
Page 13 - The diameter of a circle is a straight line passing through its center and terminated at both ends by the circumference.
Page 73 - B sin 6 2. Given the bearings of three adjacent sides of a tract of land, and the length of the middle side, to cut off, by a line running a given course, a trapezium of a given area.
Page 67 - Аь the rectangle of the given side and sine of the given angle, Is to twice the given area ; So is radius, To the other side adjacent to the given angle. Then having two sides and the included angle given, the other angles and side, if required, may be found by trig...
Page 15 - All the meridians passing through a survey of moderate extent, are considered as straight lines parallel to each other. The bearing or course of a line is the angle which it makes with a meridian passing through one end ; and it is reckoned from the north or south point of the horizon, toward the east or west. Thus, if NS represent a meridian, and the angle NAB is 40°, then the bearing of AB from the point A is 40° to the west of north, and is written N. 40° W., and read north forty degrees west....
Page 73 - To the square of the ordinate add f of the square of the abscissa, and twice the square root of the sum will be the length of the curve required.
Page 15 - В = 36° 43' ..... log. 9.776598 sum. • CASE II. 5. Wben two sides and the included angle are given, the solution may be effected by means of propositions 3 and 4. Thus, take the given angle from 180° ; the remainder will be the sum of the other two angles. Then, by proposition 3, — As the sum of the given sides is to their difference, So is the tangent of half the sum of the remaining angles to the tangent of half their difference. Half the sum of the remaining angles added to half their difference...

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