draw AH perpendicular to AB, meeting FL produced in H, and, therefore, (III. 16. cor.) touching the circle in A; and, lastly, draw DK perpendicular to DE meeting FL produced in K. Then the arc AF is the measure of the angle ACF; FG is the sine of this arc and angle; FI, or its equal CG, their cosine; AG their versed sine, and DI their coversed sine; AH their tangent, and CH their secant; DK their cotangent, and CK their cosecant. From these definitions we derive immediately the following important corollaries. Cor. 1. The sine of an arc AF is half the chord of its double. For if FG be produced to meet the circumference in N, FN is bisected (III. 3.) in G, and (III. 30.) the arc FAN in A. Cor. 2. The sine of the quadrantal arc AD, or of the right angle ACD, is the radius CD. Cor. 3. If AF be half of AD, and consequently ACF half a right angle, the tangent AH is equal to the radius. For A being a right angle, H must be half a right angle, and (I. 6.) AH equal to AC. = Cor. 4. Put the arc AF A, and the radius = 1. Then (I. 47.) FG2 + CG2 = CF2; that is, sin A+ cos3 A = 1. In like manner, we find from the right-angled triangles CAH, CDK, that CH CA+ AH, and CK = CD+DK; that is, sec2A=1+tan A, and cosec2A = 1 + cot A. Cor. 5. In the similar triangles CGF, CAH, CG: CF, or CA:: CA: CH; that is, the cosine of an arc is to the radius, as the radius to its secant. Hence also (VI. 17.) CG.CH = CA2; that is, cos A sec A 1. It would be found in like manner from the triangles CIF, CDK, that sinA cosecA = 1, CI being equal to the sine FG. = Cor. 6. In the same triangles CGF, CAH, the cosine CG is to the sine GF, as the radius CA to the tangent AH; whence (VI. 16.) cos A tan A = sin A. The triangles CIF, CDK give in like manner sin A cotA = cos A. Cor. 7. The radius is a mean proportional between the tangent of an arc and its cotangent. For the triangles CAH, CDK are similar; and therefore HA: AC:: CD, or CA : DK. (VI. 17.) tan A cot A = 1, Hence Cor. 8. The sine of an arc or angle, and the sine of its supplement are equal. So likewise are their cosines, tangents, and cotangents, and their secants and cosecants. B A D G Let AF be an arc, FG, AH its sine and tangent, and CG, DK its cosine and cotangent. Make the angle BCM equal to ACF; draw the perpendicular MO; and produce MC both ways to meet HA, KD produced in P and Q. Then (1. def. 38. and I. 13.) the angles BCM, ACM, or ACF, ACM are supplements of each other; as are also the arcs BM, AM, or AF, AM, since (III. 26.) BM, AF are equal. Now the triangles CGF, COM are equiangular and have the sides CF, CM equal; therefore (I. 26.) MO is equal to FG, and CO to CG; and MO, FG are the sines of AM, AF, and CO, CG their cosines. Again, the triangles ACP, ACH are equiangular, and have AC common; therefore (I. 26.) AP is equal to AH, and CP to CH; and AP, AH are the tangents, and CP, CH the secants of AM, AF. In like manner it would be proved, by means of the triangles CDQ, CDK, that DQ, the cotangent of AM, is equal to DK, the cotangent of AF, and that their cosecants CQ, CK are equal. PROP. I. THEOR. P In a right-angled triangle, the hypotenuse is to either of the legs, as the radius to the sine of the angle opposite to that leg, or to the cosine of the adjacent angle: (2.) either of the legs is to the other, as the radius to the tangent of the angle opposite to the latter and (3.) either of the legs is to the hypotenuse, as the radius to the secant of the contained angle. Let ABC be a triangle, right-angled at C; then (1.) c:b:: R: sin B, or cosA; (2.) a:b:: R: tanB; and (3.) a: c : sec B. D G R: FEC From B as centre, with any radius, describe an arc cutting AB, BC in D, E; and through D and E draw (I. 12. and 11.) DF, EG perpendicular to BC. Then (APP. III. def. 5. and 6.) FD, EG, and BG are respectively the sine, tangent, and secant of the arc DE, or of the angle B: and, since C is a right angle, A and B (APP. III. def. 4.) are complements of each other; and therefore (APP. III. def. 7.) sinB cosA. Again, since the angle B is common to the triangles ABC, DBF, GBE, and the angles at C, F, E right angles, these triangles (I. 32. cor. 5.) are equiangular. Hence (VI. 4.) in the triangles ABC, DBF, BA: AC :: BD: DF; that is, c: b:: R: sinB, or cosA. Again, (VI. 4.) in the triangles ABC, GBE, pro BC: CA: BE: EG; that is, a: b:: R: tan B; and BC: BA: BE: BG; that is, a: c:: R: sec B. Cor. Hence (VI. 16.) R b c sinB = c cosA; that is, the duct of either leg and the radius is equal to the product of the hypotenuse and the sine of the angle opposite to that leg, or of the hypotenuse and the cosine of the adjacent angle. When R= 1, this becomes simply b c sinB = c cos A. = PROP. II. THEOR. THE sides of a plane triangle are proportional to the sines of the opposite angles. Let ABC be any triangle; a:b:: sinA: sinB; a:c:: sinA sinC; and b c :: sin B : sin C. : : Draw AD perpendicular to BC: then AD is a leg of each of the right-angled triangles ADB, ADC; and therefore, by the last corollary, R.AD = AB sin B, and R.AD AC sinC. Hence (I. ax. 1.) AB sin B AC sinC, or c sinB = b sinC: whence (VI. 16.) b: c :: sinB : sinC: and, by drawing perpendiculars from B and C to the opposite sides, it would be proved in a similar manner, that a: c :: sinA : sinC, and a: b:: sinA : sin B.* B D C Schol. If any angle C be a right angle, the proof is included in that of the preceding proposition; as it was there shown that c : b: R sin B, or c: b:: sinC: sin B, because, by the second corollary to the definitions of this book, the sine of a right angle is equal to the radius. PROP. III. THEOR. THE sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides, is to the tangent of half their difference. Let ABC be a triangle, a, b any two of its sides, of which a is the greater, and A, B the angles opposite to them; a + ba~ -b:: tan (A + B) : tan (AB). From C as centre, with the greater side a as radius, describe the circle DBE, cutting AC produced in D and E, and BA produced in F; join BD, BE, CF; and draw EG parallel to AB, meeting DB produced in G. D See the Notes at the end of the volume. G Then because DC and CE are each equal to a, DA is equal to a+b, and AE to a―b. Also (I. 32.) the exterior angle DCB is equal to A + B: and DEB, which is at the circumference, is (III. 20.) half of DCB, which is at the centre: therefore DEB = = ↓ (A + B). Again, (I. 5.) the angle F is equal to B; `and, (I. 32.) in the triangle ACF, the exterior angle A = ACF + F = ACF+B; and, consequently, ACF = A-B; and (III. 20.) ABE, or (I. 29.) its equal BEG (AB). = Now, since (III. 31.) EBD, being in a semicircle, is a right angle, as also (I. 13.) EBG; if a circle were described from E as centre, with EB as radius, DBG (III. 16. cor.) would touch it, and (App. III. def. 6.) DB would be the tangent of DEB, and BG of BEG; and therefore DB, BG will evidently be proportional to the tangents of those angles to any other radius. Lastly, since BA (const.) is parallel to GE, we have (VI. 2.) DA: AE::DB: BG; that is, a + b: a - b :: tan (A + B): tan (AB). PROP. IV. THEOR. In a plane triangle, the cosine of half the difference of any two angles, is to the cosine of half their sum, as the sum of the opposite sides to the third side: and (2.) the sine of half the difference of any two angles is to the sine of half their sum, as the difference of the opposite sides to the third side. Let ABC (see the figure for the last proposition) be any plane triangle: then cos (AB): cos (A + B) :: a+b: c; and sin(AB) sin (A + B) :: a-b: c. But For it was shown in the preceding proposition, that BED = (A+B), and ABE = } (À — B); and since DBE is a right angle, DBA is the complement of ABE, and D of BED. in the triangle ABD, by the second proposition of this book, sin ABD sinD:: AD: AB; that is, (App. III. def. 7.) cos (AB): cos (A + B) :: a+b: c. Again, (APP. III. 2.) in the triangle ABE, sinABE: sinAEB:: AE: AB; that is, sin(A — B) : sin} (A + B) : : a —b: c. : PROP. V. THEOR. IN any plane triangle, the sum of the segments of the base made by a perpendicular from the vertex, is to the sum of the other sides, as the difference of those sides to the difference of the segments. Let ABC be a triangle, and AD a perpendicular from the vertex to the base: the sum of the segments BD, DC is to the sum of the sides AB, AC, as the difference of AB, AC to the difference of BD, DC. For (II. 5. cor. 4.) the rectangle under the sum and difference of AB, AC is equal to the rectangle under the sum and difference of BD, DC; and therefore (VI. 16.) the sum of BD, DC A B C D Α is to the sum of AB, AC, as the difference of AB, AC to the difference of BD, DC. Schol. If the perpendicular fall within the triangle, the segments make up the base, and their difference is less than the base; but if the perpendicular fall without the triangle, as it does when one of the angles at the base is obtuse, the base is the difference of the segments, and their sum is greater than the base. LEMMA. Let 2s be put to denote tudes a, b, c then s-a (a+c—b), and s—c = b), and s—c = the sum of three magni- (a - For, since 2s = a+b+c, by subtracting successively 2a, 26, 2c, and halving the remainders, we obtain those expressions. THE rectangle under two sides of a triangle is to the rectangle under the excesses of half the perimeter above those sides, as the square of the radius to the square of the sine of half the contained angle. Let ABC be a triangle, and let s = {(a+b+c): then bc: (s—b) (s—c) :: R: sin2 A. G Produce the less side AC through C, making AD equal to AB; join BD; and draw AE, CF perpendicular, and CG parallel, to BD: then (I. 47. cor. 5.) AE bisects BD and the angle A. Now (II. 5. cor. 4.) the rectangle under the sum and difference of BC, CD is equal to the rectangle under the sum and difference of BF, FD, that is, of BD and twice EF: therefore the rectangle under half the sum and half the difference of BC, CD is equal to the rectangle BE.EF. But (APP. III. 1.) AB: BE::R : singA, and E AC: CG or EF:: R: sinA; whence (V. Sup. 15.) AC.AB: BE.EF, or (BC+ CD). (BC-CD):: R: sin2 A; or bc: (a+c—b). § (a + b − c) : : R2 : sin2 + A, |