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 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 CÊ in F, and with the centre F and radius FC or FE describe the semicircle CGE; draw DG perpendicular to СЕ: then DG will be a mean propor. tional between A and B. C B PROBLEM XVI. G DE 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. 1 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 angle, is what it wants of 180°. 5. The chord of an arc, is a right line drawn frow one extremity of the arc to the other: thus the line BE is the chord of the arc BAE or BDE, Fig. 31. 6. The sine of an arc, is a right line drawn from one extremity of the arc, perpendicular to the diameter which passes through the other extremity: thus BF is the sine of the arc AB or BD, Fig. 31. 7. The cosine of an arc, is that part of the diameter which is intercepted between the sine and the centre: thus CF is the cosine of the arc AB, and is equal to BI, the sine of its complement HB, Fig. 31. 8. The versed sine of an arc, is that part of the diameter which is intercepted between the sine and the arc: thus AF is the versed sine of AB; and DF of DB, Fig. 31. 9. The tangent of an arc, is a right line touching the circle in one end of the arc, being perpendicular to the diameter which passes through that end, and is terminated by a right line drawn from the centre through the other end: thus AG is the tangent of the arc AB, Fig. 31. 10. The secant of an arc, is the right line drawn from the centre and terminating the tangent; thus CG is the secant of AB, 31. 11. The cotangent of an arc, is the tangent of the complement of that arc; thus HK is the cotangent of AB, Fig. 31, 12. The cosecant of an arc, is the secant of the com. plement of that arc; thus CK is the cosecant of AB, 13. The sine, cosine, &c. of an angle is the same as the sine, cosine, &c. of the arc that measures the angle. PROBLEM I. To construct the lines of chords, sines, tangents, and secants, to any radius. Fig. 32. Describe a semicircle with any convenient radius CB; from the centre C draw CD perpendicular to AB and produce it to F; draw BE parallel to CF and join AD. Divide the arc AD into nine equal parts as A, 10; 10, 20; &c. and with one foot of the dividers in A, transfer the distances A, 10; A, 20; &c. to the right line AD; then will AD be a line of chords constructed to every ten degrees. Divide BD into nine equal parts, and from the points of division 10, 20, 30, &c. draw lines parallel to CB, and meeting CD in 10, 20, 30, &c. and CD will be a line of sines. From the centre C, through the divisions of the arc BD, draw lines meeting BE, in 10, 20, 30, &c. and BE will be a line of tangents. With one foot of the dividers in C transfer the distances from C to 10, 20, &c. in the line BE, to the line CF which will then be a line of secants. By dividing the arcs AD and BD each into 90 equal parts, and proceeding as above, the lines of chords, sines, &c. may be constructed to every degree of the quadrant. |