ent height, by drawing parallel lines; divide the altitude of these rectangles into ten equal parts, and, through each of these parts, draw parallel lines the whole length of the scale. Divide the first division AB into ten equal parts, also CD into as many, and connect these points of division by diagonal lines, and the scale is finished. In taking measures from the diagonal scale--- If the large divisions be reckoned units, the small divisions from A to B will be decimals. If the great divisions be 10, each of the small divisions is an unit; and if the great divisions be 100, then each of the small divisions is 10, and each division in the altitude is. an unit. If it were required to take off 456 from the scale ; with one foot of the compasses on 4, extend the compafles till you have 4 of the great divifions and 5 of the lefler ; then flide up your compafles with a parallel motion till you come to 6 on the pas rallel lines, and you have the extent required. PROBLEM XXVIII. The construction of the line of chords, fines, tangents, and secants, About the centre C, with any convenient radius *, describe the semicircle ADB; erect the perpendicular CF, which will divide the semicircle into two quadrants, viz. AD, BD : divide the quadrant DB into nine equal parts, and upon the point B erect a perpendicular BT, then draw AD and BD. On B as centre, transfer each of these divisions in the quadrant DB, to the straight line BD; then is BD a line of chords. From the points 10, 20, 30, &c. in the quadrant BD, drop perpendiculars upon the diameter AB; transfer the perpendi cụlais The degrees are numbered from B to D. eulars to DC; fo fhall DC be a line of fines, and CB a line of versed sines. From the centre C, draw straight lines through each division in the quadrant BD, to meet the tangent BT; so fhall BT be a line of tangents. From the centre C, with the distances of each of the lines which meet the tangent, sweep arches to cut CF; then shall CF be a line of fecants, If from the point A straight lines be drawn to the several divisions in the quadrant DB, they will divide the radius CD into a line of femitangents. Again---Divide the quadrant AD into eight equal parts, and from A, transfer the divisions to the line AD; then shall AD become a line of rhumbs, each divition answering to a point of the mari ner's compass. PROBLEM XXIX. The angles, and one leg of a right-ang!ed triangle being given, to, construct the figure, and find the other leg. Angle A=30° 40' AB=320 From the diagonal scale make AB 300; upon B erect a per pendicular of an indefinite length; and at the point A make an angle of 30° 40'; then draw the line AC, and it is done. If the angle at C be measured, it will be 59° 20'; and if the Jeg B C be applied to the same diagonal scale from which AB was taken, it will measure 177.9. Plate 3. fig. 43. PRO PROBLEM XXX. The hypothenuse and all the angles being given,, to find the legin AC=568 ? PROBLEM XXXI. The iwo legs of a right-angled triangle beeng given, to find the acute angles, and the hypothenuse. Given AB=150 Req. angle A, angle C, From any diagonal scale, draw AB=150, and from the same fcale draw BC perpendicular to the former=160; join AC, and the triangle is constructed: for if angle A be measured from a line of chords, it will be 46° 51'; also angle C 43° 9'; and AC will be 219.3 equal parts. Plate 3. fig. 45. PROBLEM XXXII. The hypothenuse and one of the legs being given, to find the acute.. angles and the other leg. Given {* AC=150 o} Required ang. C, ang. A, and AB. Draw Draw the base AB, upon B erect the perpendicular BCC qual 69 ; take 150 from the same scale, and with the centre C, and radius 150, describe an arch to cut the base in A; join AC, and it is done : For angle A will measure 27° 23', and angle C 62° 37', and the basc BC 133 equal parts. Plate 3. fig. 46. PROBLEM XXXIII. Given two angles of an oblique angled triangle, and the side opposite to one of them ; to find the other fides. (AngleC 52° 15' Given Angle A 590 A B 276.5 Required AC and BC. Find the supplement of the sum of the two given angles, thus : 59° + 52° 15=111° 15'. And from 180° subtract 11° 15', the remainder will be, 68° 45'; then draw AB equal 276.5: Draw AC, making angle A 59°, and from B draw BC, making angle B 68° 45', and meeting AC in the point C, and it is done : then shall AC meafure 325.9, and BC 299.7. Plate 3. fig. 47. PROBLEM XXXIV. Two fides of an oblique angled triangle, and the angle opposite to one of them being given, to find the otber angles and the third side AB=20 Required ang. A ang. C and BC. Draw the base AB equal 26, and at the point B make an angle of 91° 15' by BC; then on A as centre, with the ra dius 39.42, describe an arch cutting BC in C, and join AC, and it is done. So shall angle A measure 47° 30', and angle C 41° 15'; also BC 29.07 equal parts. Plate 3. fig. 48. PROBLEM XXXV. Two sides, and the contained angle of any triangle being given, to find the remaining angles, and the third fide. AC=60 Required ang. A ang. B and AB. Draw AC equal 60, and BC equal 50 equal parts, meeting in C at an angle of 45°; then join AB, and it is done : For if you take AB in your compasses, it will measure 43.1 on the same scale of equal parts; also angle A will measure 55° 7', and angle B 79° 53', from the line of chords. Plate 3. fig. 49 LOGA |