mn, and lay it from r to s, and through Es draw the at line EsF; then the angle FED will be the C. he angle be given in degrees. Draw a straight line , and using a scale of chords from the centre E, with is the chord of 60°, describe an arc rs; then take e chord of the number of degrees required from the and from the centre r, with the chord of the given as a radius, cut the arc in s, and through the points Is draw the EF, and the FED will contain the number of degrees. If the angle be greater than 90°, be divided into two parts, and one part laid off, and he other part, from its extremity; it is generally convenient to lay off the chord of 60°, and then the of the excess above 60°. PROBLEM VII. draw a line parallel to a given line AB. E I. When the parallel line is to be at a given disC, from the given line. m any two points m and n line AB, with a radius to C, describe the arcs r Draw DE to touch these without cutting them, and be the parallel required. A m C B E II. When the parallel line is to pass through a point C. D C E m any point m in the line with the radius mC, describe c Cn. From the centre C he radius Cm, describe the Take the arc Cn in the sses, and apply it from m to o. Through C and o DE, and it will be the parallel required. 0. A B B.-This problem is more easily effected with a paruler. PROBLEM VIII. divide a given line AB into roposed number of equal parts. rough A and B draw Am and arallel to each other. In each A e lines Am, Bn, beginning at B, set off as many equal parts line AB is to be divided into. 5 m B Join A, 5; 1, 4; 2, 3; 3, 2; &c; and AB will be divid ed as required. Another Method. Through the point B draw the indefinite line CD, making any angle with AB. Take any point D in that line, and through it draw DE || to AB, and set off from D as many equal parts DK, KH, HG, &c., as the line AB is to be divided into. Through the points E, A, H B draw the line EA, and produce it to meet CD in C; then lines drawn from C, through the points F, G, H, &c., will divide the line AB into the required number of parts. PROBLEM IX. To find the centre of a given circle, or of one already described. D Let ABCD be the given circle; take any three points, A, B, C, and from the centres A and B, with a radius half the distance AB, describe arcs intersecting in r and s, and from the centres B and C, with a radius half the distance BC, describe arcs intersecting in m and n. Then through the points sr and mn, draw straight lines intersecting in o, then o will be the centre required. PROBLEM X. B To describe the circumference of a circle through any three given points A, B, C, provided they are not in the same straight line. Use the three points the same as in the last problem ; then a circle described from the centre o, with a radius equal to the distance oA, will pass through the other two, and will be the circle required. COR. Since the three points are not in the same straight line, they may be the three angular points of a triangle. Hence this problem also shows how a circle may be described about a given triangle. PROBLEM XI. To draw a tangent to a given circle,, that shall pass through a given point A. CASE I. When the point A is in the circumference of the circle. From the point A, to the centre of the circle, draw the radius oA. Then through the point A, draw BC to oA, and it will be the tangent required. CASE II. When the point A is without the circle. draw the straight line oA, and bisectA it in m. circle ABO, cutting the given circle PROBLEM XII. m B To two given straight lines A, B, to find a third proportional. From any point C draw ❘s, making any LFCG, and make CEA, CĎ and CG each B. Join ED, and c draw GF || to ED, then CF will be the third proportional required. For (CEA): (CG=B)=(CD=B): CF. G E D A B N.B.-If EG had been made B, then DF, by the same construction, would have been the third proportional. PROBLEM XIII. To three given straight lines, A, B, C, to find a fourth proportional. From any point D, draw two straight lines, making any. /FDH, and make DE=A, EF = B, DG=C. Join EG, and through F draw FH || EG; then GH will be the fourth proportional re quired. PROBLEM XIV. G Between two given straight lines, A, B, to find a mean proportional. Draw any straight line CD, in which take CE= A, and ED-B. Bisect CD in F, and from the centre F, with the radius A B D EF FD or FC, describe a semicircle. From E draw EG + to DC, meeting the semicircle in G, then EG is the mean proportional required. N.B.-Another method may readily be deduced from (Prop. 68) Geometry. PROBLEM XV. To describe an equilateral triangle on a given line AB. From the centres A and B, with the radius AB, describe arcs intersecting in C. Draw AC and BC, and ACB will be an equilateral triangle on the re- A quired line. B NOTE. An isosceles triangle may be described on AB, by taking as radius the length of one of the equal sides. For the method of describing a triangle, having its sides equal to three given lines, see Geometry, (Prop. 4.) |