2. IN DIVISION. Let it be required to find the quotient, which arises by dividing one number by another; suppose 1425 by 57. From the logarithm of the dividend, viz. 1425 Take the logarithm of the divisor, viz. 57 3.15381 = 1.75587 - And the remainder is the logar. of the quotient, viz. 25 1.39794 3. IN THE RULE OF THREE. Three numbers given, to find a fourth, in direct proportion. RULE. From the Tables take the logarithms of each of the proposed numbers, then, add the logarithms of the second and third together, and from the sum take the logarithm of the first, and the remainder will be the logarithm of the fourth number. Let the three proposed numbers be 18, 24, and 33, and the operation will stand thus: 1.38021=the logarithm of 24, the 2d term, 1-51851 the logarithm of 33, the 3d term. = 2-89872 the logarithm of their product. --1·25527 the logarithm of the first term 18. = = 1.64345 the logarithm of the fourth term required, which, by the Table, answers to the natural number 44, the 4th proportional to the three proposed numbers. 4. IN INVOLUTION, OR RAISING POWERS. To find any power of any proposed number, or to involve any number to any proposed power, by logarithms. RULE.-Multiply the logarithm of the given root by the power, viz. by 2 for the square, by 3 for the cube, &c. and the product is the logarithm of the power sought. Required to find the cube of 12? 1.07918 the logarithm of 12. X3 the third power, or cube. 3.237541728 the cube of 12. 5. IN EVOLUTION, OR EXTRACTING ROOTS. To extract any root of any proposed number. RULE. Divide the logarithm of the proposed number, by the index of the required root, viz. by 2 for the square, by 3 for the cube, &c. and the quotient will be the logarithm of the root required. Required to find the cube root of 1728? 3.23754 the logarithm of 1728, and 3·23754÷8=1.07918 is the logarithm of the cube root of 1728, viz. 12. 6. IN COMPOUND INTEREST. To find the amount of any sum for any time, and at any rate, at Compound Interest. RULE.-Multiply the logarithm of the ratio (i. e. the amount of £.1 D.1 for one year) by the number of years, and to the pre duc duct add the logarithm of the principal; the sum will be the loga rithm of the amount, What will .45 amount to, forborne 12 years, at 6 per cent, per annum, compound interest? Log. of 1.06 the ratio, is 02538 12 :30396 Add log. of 45, the principal 1.65321 The sum is 1.95717 which is the logarithm of 90·7= .90 14s. Ans. 7. IN DISCOUNT AT COMPOUND INTEREST. To find the present worth of any sum of money due time hence, at any rate, at Compound Interest. any RULE. From the logarithm of the sum to be discounted, subtract the logarithm of the rate multiplied by the time; and the remainder is the logarithm of the present worth. What present money will pay a debt of £.90 14s. due 12 years hence, discounting at the rate of 6 per cent. per annum ? From the logarithm of £.90 14 = 1.95717 Subt. prod. of the log. of the ratio x by the time = 30396 The remainder 1.65321 is the logarithm of .45 Ans. PLAIN GEOMETRY. Definitions, 1. A POINT in the Mathematicks is considered only as a mark, without any regard to dimensions. 2. A Line is considered as length, without regard to breadth or thickness. 3. A Plain or Surface has two dimensions, length and breadth, but is not considered as having thickness. 4. A Solid has three dimensions, length, breadth and thickness, and is usually called a Body, 5. A line is either straight, which is the nearest distance between two Points; or crooked, called a Curve Line, whose ends may be drawn further asunder. 6. If two Lines are at equal distance from one another in every part, they are called parallel Lines, which, if continued infinitely, will never meet. 7. If two lines incline one towards another, they will, if continued, meet in a point: by which meeting is formed an Angle. 8. If one Line fall directly upon another, so that the Angles on both sides are equal, the Line, so falling, is called a perpendicular, and the Angles, so made, are called right Angles, and are equal to 90 degrees, each. 9. All Angles, except right Angles, are called oblique Angles, whether they are acute, that is, less than a right Angle; or obtuse, that is, greater than a right Angle. GEOMETRICAL PROBlems. PROBLEM I. To divide a Line AB into two equal parts. Set one foot of the compasses in the point A, and, opening them beyond the middle of the line, describe arches above and below the line; with the same extent of the compasses, set one foot in the point B, and describe two arches crossing the former : draw a line from the intersection of the arches above the line, to the intersec B tion below the line, and it will divide the line AB into two equal parts. PROBLEM II. To erect a perpendicular on the point C in a given line. Set one foot of the compasses in the given point C, extend the ether foot to any distance at pleasure, as to D, and with that extent make the marks D, and E. With the compasses, one foot in D, at any extent above half the distance of D and E, describe an arch above the line, and with the same extent, and one foot in E, describe an arch crossing the former; draw a line from the intersection of the arches to the given point C, which will be perpendicular to the given line in the point C. PROBLEM III. To erect a perpendicular upon the end of a line. Set one foot of the compasses in the given point B, open them to any convenient distance, and describe the arch C D E; set one foot in C, and with the same extent, cross the arch at D: with the same extent cross the arch again from D to E; then with one foot of the compasses in D, and A with any extent above the half of C B ED, describe an arch a; take the compasses from D, and, keeping them at the same extent with one foot in E, intersect the former arch a ina; from thence draw a line to the point B, which will be a perpendicular to AB. PROBLEM PROBLEM IV. From a given point, a, to let fall a perpendicular to a givin line A B. Set one foot of the compasses in the point a, extend the other so as to reach beyond the line A B, and describe an arch to cut the line A B in C and D; put one foot of the compasses in C, and, with any extent above half CD, describe an arch b; keeping the compasses at the same extent, put one foot in D,A and intersect the arch bin b; through which intersection, and the point a, draw a E, the perpendicular required. PROBLEM V. To draw a Line parallel to a given Line A B. Set one foot of the com passes in any part of the line, E F B B part of the line AB, as at e, describe the arch a; lay a ruler to the extremities of the arches, and draw the line E F, which will be paral. lel to the line A B. PROBLEM VI. To make an Angle equal to any number of Degrees. D E It is required to lay off an acute Angle of 35° on a given line AB. Take 60 degrees from the line of chords in the compasses, set one foot of the compasses in the point A, describe an arch C D, at pleasure; then set one foot of the compasses in the brass centre, in the beginning of the line of chords, and bring the other to 35 on the line; with this extent set one foot in C, with the other intersect the arch CD, in a, and through a draw the line AE, so will EAB be an angle of 35 degrees. ୯ If the angle had been obtuse, suppose 125°, then take 90° from the line of chords; set one foot in C, and intersect the arch in b; then take 35° from the same line of chords, and set them from btod: a line drawn from A through d to F will make an angle, FAB, of 125°. To measure an angle by the line of chords, is only to take the distance on the arch between the lines AB and AE, or AB and AF, and laying it on the line of chords. PROBLEM PROBLEM VII. To make a Triangle, whose sides shall be equal to three given lines, provided any two of them be longer than the third. Let A, B, C, be the three given lines; draw a line AB, at pleasure; take the line C in the compasses, set one foot in A, and with the other make a mark at B; then take the given line B in the compasses, and setting one foot in A, draw the arch C; then take the line A in the compasses, and intersect the arch C in C; lastly, draw the lines AC and BC, and the triangle will be completed. B C A A PROBLEM VIII. To make a Square, having equal sides, equal to any given line. Let A be the given line; draw a line A B equal to the given line; from B raise a perpendicular to C equal to AB, with the same extent, set one foot in C and describe the arch D; also with the same extent, set one foot in A and intersect the arch D; lastly, draw the lines AD and CD, and the square will be completed. In like manner may a Parallelogram be constructed, only attending to the difference between the length and breadth. D A A PROBLEM IX. To describe a Circle, which shall pass through any three given Points, which are not in a straight line. Let the three given points be A,B,C, through which the circle is to pass. Join the points AB and BC with right lines, and bisect these lines; the point D, where the bisecting lines cross each other, will be the centre of the circle required. Therefore, place one point of the compasses in D, extending the other to either of the given points, and the circle, described by that radius, will pass through all the points. Hence, it will be easy to find the centre of any given circle; for, if any three points. are taken in the circumference of the given circle, the centre will be readily found as B above. The same may also be observed, when only a part of the circumference is given. A PROBLEM X. To describe an Ellipsis or Oval mechanically. Draw two parallel lines, as L and M, at a mederate distance, by |