PROBLEM VI. · At a given point B, in a given right line LG, to make an angle equal to a given angle A, Fig. 23. With the centre A and any distance AE, describe the arc DE, and with the same distance and centre B describe the arc FG; make HG equal to DE, and through B and H draw the line BH; then will the angle HBG be equal to the angle A. PROBLEM VII. To bisect any right lined angle BAC, Fig. 24. In the lines AB and AC, from the point A, set off equal distances AD and AE; with the centres D and E and any distance more than half DE, describe two arcs cutting each other in F; from A through F draw the line AG, and it will bisect the angle BAC. PROBLEM VIII. To describe a triangle that shall have its sides respectively equal to three right lines D, E and F, of which any two must be together greater than the third, Fig. 25. Make AB equal to D; with the centre A and distance equal to E, describe an arc, and with the centre B and distance equal to F describe another arc, cutting the former in C; draw AC and BC, and ABC is the triangle required. PROBLEM IX. Upon a given line AB to describe a square, Fig. 26. At the end B of the line AB, by problem 3, erect the perpendicular BC, and make it equal to AB; with A and cutting each other in D; draw AD and CD, then will ABCD be the square required. PROBLEM X. To describe a circle that shall pass through the angular points A, B and C, of a triangle ABC, Fig. 27. By problem 1, bisect any two of the sides, as AC, BC, by the perpendiculars DE and FG; the point H where they intersect each other will be the centre of the circle; with this centre, and the distance from it to either of the points A, B, or C, describe the circle. PROBLEM XI. To divide a given right line AB into any number of equal parts, Fig. 28. Draw the indefinite right line AP, making an angle with AB, also draw BQ parallel to AP, in each of which, take as many equal parts AM, MN, &c. Bo, on, &c. as the line AB is to be divided into; then draw Mm, Nn, &c. intersecting AB in E, F, &c. and it is done. PROBLEM XII. To make a plain diagonal scale, Fig. 29. Draw eleven lines parallel to, and equidistant from each other; cut them at right angles by the equidistant lines BC; EF; 1, 9; 2, 7; &c. then will BC, &c. be divided into ten equal parts; divide the lines EB, and FC, each into ten equal parts, and from the points of division on the line EB, draw diagonals to the points of division on the line FC: thus join E and the first division on FC, the first division on EB and the 'second on FC, &c. Note.-Diagonal scales serve to take off dimensions or numbers of three figures. If the first large divisions be parts; and the divisions in the altitude, along BC, will be 100th parts. If HE be tens, EB will be units, and BC will be 10th parts. If HE be hundreds, BE will be tens, and BC units. And so on, each set of divisions being tenth parts of the former ones. For example, suppose it were required to take off 242 from the scale. Extend the dividers from E to 2 towards H; and with one leg fixed in the point 2, extend the other till it reaches 4 in the line EB; move one leg of the dividers along the line 2, 7, and the other along the line 4, till they come to the line marked 2, in the line BC, and that will give the extent required. PROBLEM XIII To find a third proportional to two given right lines A and B. Draw two right lines, CD, CE B containing any angle; make CF A equal A, and CG, CH each equal B; join FG and draw HL parallel to it: then will CL be the third proportional required. PROBLEM XIV. D H E G L To find a fourth proportional to three given right lines, A, B and C. PROBLEM XV. To find a mean proportional between two given right lines A and B. Draw any right line CE and in it take CD equal A, and DE equal B; bisect CE in F, and with the centre F and radius FC or FE describe the semicircle CGE; draw DG perpendicular to CE: then DG will be a mean propor. tional between A and B. C A B PROBLEM XVI. DF To divide a given right line AB into two parts that shall have the same ratio to each other as two given lines C PLANE TRIGONOMETRY. DEFINITIONS. 1. PLANE TRIGONOMETRY is the art by which, when any three parts of a plane triangle, except the three angles, are given, the others are determined. 2. The periphery 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, &c. 3. The measure of an angle is the arc of a circle, contained between the two lines that form the angle, the angular point being the centre; thus the angle ABC, Fig. 30, is measured by the arc DE, and contains the same number of degrees that the arc does. The measure of a right angle is therefore 90 degrees; for DH, Fig. 31, which measures the right angle DCH is one fourth part of the circumference, or 90 degrees.
Note. The degrees, minutes, seconds, &c. contained in any arc, or angle, are written in this manner, 50° 18′ 35"; which signifies that the given arc or angle contains 50 degrees, 18 minutes and 35 seconds. 4. The complement of an arc, or of an angle, is what it wants of 90°; and the supplement of an arc, or of an |