PROBLEM V. From a given point C to let fall a perpendicular CB on a given Line AB, when the perpendicular is to fall so near the end of the given line that it cannot be done as above. Upon any point A of the line AB as a centre, and with the distance AC describe an arch E; choose any other point in the line AB, as D, and with the distance DC describe another arch intersecting the former in E, join CE cutting AB in B, and it is done, for CB will be the perpendicular required. PROBLEM VI. D To make an angle that shall contain any proposed On C erect a perpendicular CD, and it is done; CASE 2. When the angle is acute, as for example 36° 30' let CB be the given line and C the point at which the angle is to be made. CASE 3. When the given angle is obtuse, as for example 127° 20′ let CB With the chord of 600 in your compasses, and one foot on C, as a centre, draw the arch FB, on which set off from B to F, the given angle 3640 taken from the line of chords; through F and the centre C draw the right line AC, and it is done; for the angle ACB will be an angle of 36° 30' as was required. be the given line and C the angular point. H E G Take the chord of 60° in your compasses, and with one foot on C as a centre, describe an arch BGHE, upon which set off the chord of 60° (which you already have in your compasses) from B to G, and from G to H; then set off from G to E, the excess of the given angle above 60°, which is 673° taken from the line of chords; or you may set off from H to E, the excess of the given angle above 1200, which is 740; draw the line CE, and it is done, for the -angle ECB will be an angle of 1270 20'. 127.20' B Were it required to measure a given angle, the process. would have been nearly the same, by sweeping an arch as BE, and measuring it on the line of chords, as is evident. PROBLEM VII. To bisect a given arch of a circle AB, whose centre is C. than the half of AB, and with one foot in A, de- *For a description of the line of Chords see page 20 B D E A PROBLEM VIII. To find the centre of a given Circle. With any radius, and one foot in the circumference as at A, describe an arch of a circle, as CBD, cutting the given circle in B; with the same extent, and one foot in B, describe another arch CAD, cutting the former in C and D; through C and D draw the line CD, which will pass through the centre of the circle; in like manner may another right line be drawn, as EFG, which shall cross the first right line at the centre required. This construction depends upon article 43 of Geometry. PROBLEM IX. A B G G D H To draw a Circle through any three given points not situated in a Right Line. Let A, B and D be the given points. Take in your compasses any distance greater than half AB, and with one foot in A describe an arch E EF; with the same extent, and one foot in B, describe another arch cutting the former in the points E, F, through which draw the indefinite right line EFC; then take in your compasses any extent greater than half BD, and with one foot in B, describe an arch GH; with the same extent, and one foot in D, describe another arch cutting the former in the points G, H, through which draw the right line GHC, cutting the former right line EFC, in the point C; upon the point C as a centre, with an extent equal to CA, CB, or CD, as radius, describe the sought circle. PROBLEM X. To divide a Circle into 2, 4, 8, 16, or 32, equal parts. Draw a diameter through the centre, dividing the circle into two equal parts; bisect this diameter by another drawn perpendicular thereto, and the circle will be divided. into four equal parts or quadrants; bisect each of these quadrants again by right lines drawn through the centre, and the circle will be divided into eight equal parts; and so you may continue the bisections any number of times. This problem is useful in constructing the mariner's compass. · PROBLEM XI. 3 2 1 To divide a given Line into any number of equal parts. Let it be required to divide the line AB into five equal parts. From the point A draw any line AĎ, making an angle with the line AB; then through the point B draw a line BC parallel to AD; and from A, withany small opening in your compasses, set off a number of equal parts on the line AD, less by one than the proposed number (which C, number of equal parts in this example is 4:) 4 then from B set off the same number of the A 3 2 1 same parts on the line BC, then join 4 and 1, 3 and 2, 2 and 3, 1 and 4, and these lines will cut the given line as required. CONSTRUCTION OF THE PLANE SCALE, 1st. WITH the radius you intend for your scale, describe a semicircle ADB, (Plate II. fig. 1.) and from the centre C draw CD perpendicular to AB, which will divide the semicircle into two quadrants, AD, BD; continue CD towards S, draw BT perpendicular to CB, and join BD and AD. 2dly. Divide the quadrant BD into 9 equal parts, then will each of these be 10 degrees; subdivide each of these parts into single degrees, and if your radius will admit of it, into minutes or some aliquot parts of a degree greater than minutes. Sdly. Set one foot of the compasses in B and transfer each of the divisions of the quadrant BD to the right line BD, then will BD be a line of chords. 4thly. From the points 10, 20, 30, &c. in the quadrant BD draw right lines parallel to CD, to cut the radius CB, and they will divide that line into a line of sines which must be numbered from C towards B. 5thly. If the same line of sines be numbered from B towards C, it will become a line of versed sines, which may be continued to 180°, if the same divisions be transferred on the same line on the other side of the centre C. 6thly. From the centre C, through the several divisions of the quadrant BD, draw right lines till they cut the tangent BT, so will the line BT become a line of tangents. 7thly. Setting one foot of the compasses in C, extend the other to the several divisions 10, 20, 30, &c. in the tangent line BT, and transfer these extents severally to the right line CS, then will that line be a line of secants. 8thly. Right lines drawn from A to the several divisions 10, 20, 30, &c. in the quadrant BD, will divide the radius CD into a line of semi-tangents. 9thly. Divide the quadrant AD into eight equal parts, and from A as a centre transfer these divisions severally into the line AD, then will AD be a line of Rhumbs, each division answering to 11° 15' upon the line of chords. The use of this line is for protracting and measuring angles, according to the common division of the mariner's compass. If the radius AC be divided into 100 or 1000, &c. equal parts, and the lengths of the several sines, tangents, and secants, corresponding to the several arches of the quadrant, be measured thereby, and these numbers be set down in a table,* each in its proper column, you will by these means have a collection of numbers by which the several cases in trigonometry may be solved. Right lines graduated as above, being placed severally upon a ruler, form the instrument called the Plane Scale, (see Plate II. fig. 2.) by which the lines and angles of all triangles may be measured. All right lines, as the sides of plain triangles, &c. when they are considered simply as such without having any relation to a circle, are measured by scales of equal parts, one of which is subdivided equally into 10, and this serves as a common division to all the rest. In most scales an inch is taken for a common measure, and what an inch is divided into is generally set at the end of the scale. By any common scale of equal parts, divided in this manner, any number less than 100 may be readily taken; but if the number should consist of three places of figures, the value of the third figure cannot be exactly ascertained, and in this case it is better to use a diagonal scale, by which any number consisting of three places of figures, may be exactly found. The figure of this scale is given in Plate II. fig. 3: its construction is as follows. Having prepared a ruler of convenient breadth for your scale, draw near the edges thereof two right lines, af, cg, parallel to each other; divide one of these lines as af, into equal parts, according to the size of your scale;† and * In table XXIV. is given the sine and co-sine to every minute of the quadrant, to five places of decimals. The length of one of these equal parts at the end of the scale to which this description refers is ab or cd: The length of one of the equal parts of the scale of the other end being the half of ab. |