Pi. Trig, PRO P. VII. sides, is to the rectangle contained by half the fum, and half the difference of the base and the difference of the fides, fo is the square of the radius to the square of the fine of half the angle opposite to the base. Let ABC be a triangle, of which the side AC is greater than AB; and make AD equal to AC; therefore BD is the difference of the sides AC, AB: The rectangle BA, AC is to the rectangle contained by half the sum, and half the difference of CB, BD, as the square of the radius to the square of the fine of half the angle BAC. Join CD, and draw a AE, BF perpendicular to CD, and from the centre B, at the distance BC, describe a circle, meeting AD in G, H, and CD in K: and because AD is equal to AC, the angle ACD B 5.1. is equal to ADC, and AED, AEC A are right angles, and the fide AE is common to the triangles AED, AEC; € 26. 1. therefore DE is equal to EC, and Б the angle DAE to EAC: and because BF drawn through the centre cuts CK d 3. 3. at right angles, it bisects it d; therefore CK is double of CF: and CD K is double of CE; therefore the remainder DK is double of the remain der EF: and because the straight lines CK, GH in the circle, € 35. 3. cut one another, the rectangle CD, DK is equal e to the rect angle GD, DH; and the rectangles contained by their halves are equal; therefore the rectangle DE, EF is equal to the rectangle contained by half of DG, the sum of CB, BD, and half of DH their difference: but because BF is parallel to AE, DA is 12. 6. to AB, as DE to EF *; therefore the rectangles DA, AB, and 81.Def.6. DE, EF, are fimilar figures 8 upon AB, EF: and the squares of AD, DE are similar; therefore the rectangle DA, AB, or CA, h 22. 6. AB, is the rectangle DE, EF, as the square of AD to the square of DE ; that is, as the square of the radius to the square of the k s. P. T. fine of DAE \, half of BAC. Therefore the rectangle CA, AB is to the rectangle contained by half the sum, and half the difference of CB, BD as the square of the radius to the square of the fine of half the angle BAC. PROP. Pl. Trig. PROP. VIII. sides, is to the rectangle contained by half the f 3• 3, Let ABC be any triangle, the rectangle BA, AC is to the rectangle contained by half the sum of the three fides and its excess above the base BC, as the square of the radius to the {quare of the cofine of half the angle BAC. In BA produced, take AD equal to AC; and join DC, and D e & Cor. 36. C h 16.6. lar k to the rectangle DE, EF; and the squares ki. De!.6. of DA, DE are fimilar ; therefore the rectangle DA, AB, or CA, AB, is to the rectangle DE, EF, as the square of DA to the square of DE'; that is, as the square of the radius to the 122. ú. {quare of the cofine of ADC, half of the angle BAC m. Where- m l. P. fore, as the rectangle contained by the fides BA, AC is to the rectangle contained by half the sum of the three fides, and its excess above the base BC, so is the square of the radius to the {quare of the cofine of half the angle BAG. COR. 3. B Pl. Trig. Cor. Because the rectangle contained by half the sum of the three fides, and its excess above the base, is to the rectangle BA, AC, as the square of the coline of half BAC to the square 6. P. T. of the radius ?: and that the rectangle BA, AC is to the rect angle contained by half the sum, and half the difference of the bale and difference of the sides, as the square of the radius to 0 7. P. T. the square of the fine of half BAC °; therefore, by equality P, P 22. 6. the rectangle contained by half the sun of the three sides, and its excess above the base, is to the rectangle contained by half the sum, and half the difference of the base and difference of the fides, as the square of the cofine of half BAC to the square of its fine ; that is, as the square of the radius to the square 9 2. Cor. of the tangent 9 of half the angle BAC opposite to the 19. Def. base. SOLUTION of the Cases of OBLIQUE-ANGLED TRIANGLES. GENERAL PROPOSITION. N any IN triangle, of the three fides, and any two of the angles, any three being given, the other two and the remaining angle may be found. But if the three angles be given, the ratios only of the sides are given, beir.g the same with those of the angles opposite to them; and in this case, the sides cannot be found. If two angles of a triangle be given, the third is also given, being the supplement of their sum : and, on the contrary, if one of the angles be given, the sum of the other two is also given. The The four cases of this proposition may be resolved by the help. Pl. Trig. of some of the preceding propofitions ; as in the following Table. I :}P.3. Caies. Given. sought. solution. angles and a side. AC Sin. C:sin. B ::BA: AC.} two sides, and an The If AC be greater than AB, C acute or obtuse, by Cor. to 2 3 4 AB, AC and A, The The sum of BA, AC: diff. of BA, two fides and angles AC :: tan. of į sum of B, C : E; and R: tan. diff. of E and diff. of B, C. P. 6. LEM. 4. Sin. B : fin. A::AC: BC. three AC: F; and BC, F together are double of the greater seg ment BD. P.s. obtuse; if not, it is acute: then AC: CD::R: cos. C. AB, AC, the rect. AB, AC: BC, D :: R2: fin 2. 1 BAC. three sides. The rect. BA, BAC. k 1. 2. Pl. Trig. To construct a Table of Sines, Tangents, &c. Let ABC be a circle, of which the diameter is AC, and the centre D: and let AB be any arch of it; and bisect AB in a 30. 3. Fa; and join AB, BC, BD, DF: and let DF meet AB in E; 0 120 l. and draw 6 BG purpendicular to AC: and because the arch AB is double of AF, the angle ADB is C 33.6. double of ADFC: but ADB is double B d 20. 3. of ACB, therefore the angle ADF is equal to ACB; and DE is parallel to e 31.3. BC, or at right angles to AB e; theref 4. 6. fore AC is to CB, as AD to DEf; and E C D G A g 14. 5. double of DE 8: and because the tri. angles ADE, BCG are equiangular, AD is to DE, as BC to CG'; therefore the rectangle AD, CG is equal to the rectangle h 16.6. BC, DE “, that is, to double of the square of DE k: and DE is the cofine of AF!, and CG the sum of the radius CD, and DG 1 1. Cor. the cofine of AB. Therefore, the rectangle contained by the Def, 10. radius, and the sum of the radius and the cofine of Pl.T. arch, any is double of the square of the cofine of half that arch. If AB be equal to the radius AD, it is the side of a hexagon m Cor. 15. inscribed in a circle ; and the arch AB is the fixth part of the circumference; that is, it is an arch of 60°; and because then n 47. 1. the triangle ABD is equilateral, AG is equal to GD"; that is, it is the half of the radius AD. If, therefore, the radius be represented by unity, and the decimal notation be used, the sum of the radius and the cofine DG is represented by 1.5, and the rectangle contained by the radius and this sum is also represented by 1.5, because the radius is 1 ; therefore half of this rectangle is .75, which is therefore the square of DE the cofine of AF; that is, of 30°. Wherefore the cofine of 30° is the square-root of .75; that is, .8660254039344+. In the same manner, if the radius 1 be added to this number, half the sum .9330127019672+ is the square of the cofine of 15°; and therefore its square-root .96592582632789 + is the cosine of 15°. In like manner, the square-root of half the sum of i and the cofine of 15° is .99144485936634 + the cofine of go 30'. In the same manner are found the cosine of 3° 45', of 1° 57' 30', and so on : till at length, after twelve bisections of the arch of 60°, the cofine of 52" 44" 03"" 4.5"' is found to be -9999999673176965713. And if the square of the cosine be subtracted from i, which is the square of the radius; the remainder is the square of the fine of the same arch”. Thus, the fine 4. |