tre; 2, the radius; 3, a segment measured from the centre to a point in the first: then, any two lines drawn from the two points to the circumference shall have the same ratio as those two segments of the first line which meet the circumference, beginning at the said points. Let F be the outer point, D the centre, E the sectional point, and A, B, C, points in the circumfe- C rence: Draw FB and produce it to G; join BA, BC, BD, BE. Now, since FD:DB::DB:DE, by hyp., and that these proportionals are about the same angle D, B the triangles FBD, BED are equiangular (a). A E D C Therefore FB:BD::BE: ED (b), and F alternately FB: BE :: BD, or AD : ED (c). But since FD: AD :: AD: ED, by hyp. by division FA: AD::AE: ED, (d); and alternately FA: AE:: AD, or BD: ED (c): Therefore FA: AE:: FB: BE (e). But FA, AE are the segments of the first line, which meet the circumference; and FB, BE are any two lines drawn from the points to the circumference. Cor. The chord AB bisects the angle FBE: for, in the triangle FBE, the segments of the base FA, AE have the same ratio as the sides FB, BE. Recite p. 3, 6. And for the same reason the chord BC bisects the exterior angle EBG, (p. A, 6). For since by hyp. FD: DC:: DC: DE by composition and by division FC:CD::CE: DE (p. 18, 5); FA: AD or CD:: AE: DE (p. 17, 5: Therefore, ex æquo FA:AE:: FC: CE (p. 22, 5). But it proves that FA: AE:: FB: BE; Therefore FB: BE:: FC: CE (p. 11, 5). G Th. If a chord, drawn through the extreme point of the diameter of a circle, meet a straight line in the circle or out of it, cutting the diameter or its exterior production, at right angles; the rectangle of the diameter and its segment, between the extreme point and perpendicular, is equal to the rectangle of the chord and its segment, between the same point and the cutting line. Given the chord AB and its segment AF; also the diameter AC, and its segment AD: then ABXAF=ACXAD. Join BC. Because ABC is an angle in a semicircle (a); it is equal to the right angle ADF; and the angles DAF, BAC are either identical or vertical, and therefore equal (b): therefore the triangles ABC, ADF are equiangular (c); and BA: AC:: DA: AF. But the rectangle contained by the means is equal to that contained by the extremes (d); therefore DAXAC=BAXAF. Wherefore, if a chord, &c. Recite (a) p. 31, 3; (d) p. 15, 6. Q. E. D. (b) p. 15, 1; (c) p. 4, 6; H Th. The perpendiculars drawn from the three angles of a triangle to the opposite sides, intersect each other in the same point. Given the triangle ABC; perpendiculars BD, CE intersecting each other in F, and AF joined; which last produced to G will fall perpendicular to BC. Join ED. Since AF subtends a right angle on either side of it, it is the diameter of a circle passing through the points A, E, F, D (a). Now the vertical angles EFB, DFC are E A D F C G Therefore equal (b); the right angles BEF, CDF are equal (c), and the triangles EFB, B DFC are equiangular (d); therefore FB: FE :: FC: FD, and alternately FB: FC:: FE: FD (e). also, the triangles EFD, BFC are equiangular: for their vertical angles are contained by proportional sides, as proved. Wherefore the angles FCB, EDF are equal: but EDF equals EAF as angles in the same segment (f); therefore EAF equals FCB: the vertical angles EFA, CFG are also equal (6); and the third angles AEF, CGF are equal (g). But AEF is a right angle; Wherefore, CGF is a right angle; and AG, passing through the point F, is perpendicular to BC. Recite (a) p. 31, 3; (b) p. 15, 1; (d) p. 4 and 6,6; (e) p. 16, 5; Q. E. D (c) ax. 10, 1; Cor. The triangles ADE, ABC are similar: for the triangles ABD, ACE, having right angles at D and E, and the angle A common, are equiangular, and have BA to AD as CA to AE; and alternately BA to CA as AD to AE. Therefore the two triangles BAC, DAE have the angle A common; and the sides about it proportionals; therefore they are equiangular (6, 6), and similar: and the rectangles BAXAF CAXAD are equal. END OF THE SIX BOOKS. ELEMENTS OF PLANE TRIGONOMETRY. ३ PLANE TRIGONOMETRY teaches the measurement of distance and elevation by means of angles. The principles of it are all contained in the preceding six Books. Lemma 1. An angle at the centre of a circle is to four right angles, as the arc on which it stands is to the whole circumference. Given ABC, acentral angle, and AC the arc which measures it (a); the angle ABC is to four right angles, as the arc AC is to the whole circumference ACF. Produce AB to meet the circle in E; and at right angles to AE draw DBF. E D HC K B GA Then, because ABC, ABD are central angles; and AC, AD their arcs;-ABC:ABD::AC: AD (b): moreover ABC: 4ABD::AC: 4AD (c). But ABD is a right angle, and AD is the arc which measures it: therefore the angle ABC is to 4 right angles as the arc AC is to the whole circumference. F Recite (a) def. 1, trig.; (b) p. 33 of b. 6; (c) def. 3, and p. 4 of b. 5. Cor. As the arcs AC and GH measure the same angle; AC is to the circumference ACE, as GH is to the circumference GHK: therefore equal angles at the centres of different circles stand on arcs which have the same ratio to their circumferences. Definitions. 1. The measure of an angle is the arc of the circle on which it stands: thus, the arc AC is the measure of the angle ABC. 2. Degrees, minutes, seconds, &c., are the terms of circle measure. Example. Every circle great or small is divided into 360 degrees, each degree into 60 minutes, each minute into 60 seconds. Hence an angle and its arc contain the same number of degrees, &c. 3. The complement of an angle is its defect from a right angle. 4. The supplement of an angle is its defect from a semicircle. 5. The sine of an angle is a straight line drawn in the bosom (sinus) of the arc, from one end of it, perpending upon the diameter which meets the other end, as CD. The greatest sine is that of a quadrant or 90 degrees; which is equal to the radius. 6. The tangent is a straight line tou touching the arc, parallel to the sine, and intercepted by the two sides which contain the angle,as AE. The tangent of 45 degrees is equal to the radius. 7. The secant is one of the sides which contain the angle, produced to meet the tangent, through the point in which the sine meets the arc,-as BE. 8. The cosine, cotangent, and cosecant, are the sine, tangent and secant, of the complement of the arc or angle. 9. The sine, tangent and secant of an arc, or angle, are also the sine, tangent and secant of its supplement. 10. The subsine, or versed sine, is a segment of the diameter intercepted between the sine and tangent, -as AD. 11. The subtense*, or hypothenuse, is the side opposite to a right angle. These lines form several equiangular triangles, whose sides are therefore proportionals, viz: 1. Cosine is to sine as radius is to tangent : Ex. BD: DC:: BA: ΑΕ. 2. Tangent is to radius as radius is to cotangent: Ex. AE: AB:: BH: ΗΚ. 3. Cosine is to radius as radius is to secant: Ex. BD: BC:: BA: BE. 4. Sine is to radius as radius is to cosecant: Ex. CD: CB:: BH: BK. 5. Sine is to cosine as radius is to cotangent: Ex. BL: LC:: BH: HK. 12. The sine, subsine,* tangent and secant of an arc, taken as the measure of an angle, is to the sine, subsine, tangent and secant of any other arc measuring the same angle, as the radius of the first arc is to the radius of the other. Given the arcs AC and MN, as measures of the angle ABC; and let CD be the sine, DA the subsine, AE the tangent, and BE the secant, of the arc AC. Also, let NO be the sine, OM the subsine, MP the tangent, and BP the secant, of the arc MN. The first lines are to the second, each to each, as the radius BC is to the radius BN. E P A M N B * The terms subsine and subtense are more convenient than versed sine and hypothenuse. Cor. If numerical tables of sines, subsines, tangents and secants, of certain angles to a given radius, be constructed, they will show the ratios of the sines, &c., of the same angles to any radius whatever. Such are called Tris Trigonometrical Tables; in which the radius is represented as some term of the decimal series, 1, 10, 100, 1000, &c. Propositions. 1. Th. In a right-angled triangle, the subtense of the right angle is to either side, as radius is to the sine of the angle opposite to that side; and either of the sides is to the other, as radius is to the tangent of the angle opposite that other. Given the triangle ABC, right angled at A; the subtense BC, and the sides CA, AB. From the centre C, with any radius CD, describe the arc DE; draw DF at right angles to CE, and EG parallel to it, touching the arc in E, and meeting CB in G. Then DF is the sine, EG the tangent, and CG the secant of the angle C, or of the arc DE. G FEA B The three triangles DFC, GEC, BAC, are equiangular: because the angles at F, E, A are right angles; and the angle C is common to all. Therefore, 1. CB: BA:: CD: DF;-but CD is the radius, and DF is the sine of the angle C (def. 5, trig.); Therefore CB: BA:: R: sine of the angle C. 2. CA: AB::CE: EG;-but CE is the radius, and EG is the tangent of the angle C (def. 6, trig.); Therefore CA: AB:: R: tangent of the angle C. Wherefore in a right angled triangle, &c. Q. E. D. Cor. 1. Radius is to the secant as the side adjacent to the angle C is to the subtense. For CD, or CE: CG:: CA: CB. Cor. 2. Let Radius = 1; then the preceding analogies are as follows: Sine of the angle C=BA divided by CB. Cor. 3. If, in the triangle ABC, a perpendicular be drawn to the base BC; then (making AD radius), AD: DC:: R: tan. CAD, and AD: DB:: R: tan. BAD. Cor. 4. Since, in a right angled triangle, the A D B two acute angles are equal to the right angle, one of them is the complement of the other (def. 3 trig.), therefore any |