All the limitations are either contained in the 16th and 17th Sph. Trig. propofitions, or else they are easily deducible from them. Thus, because C is of the same affection with AB; if AC and C be of the same affection, AC and AB are of the same affection; and therefore BC is less than yo', (Prop. XVII.); this is the limi. tation in case 5. Again, if BC be less than 90°, AC and AB, or B and C, are of the same affection; and therefore, if AC be less than 90°, or of the same affection with CB, AB is less than 90° , and for the same reason, C is less than 90°. But if AC, BC be of different affection, that is, AC greater than 90°, BA is greater than 90°: and the same may be proved, in the fame manner, when BC is greater than 90°. Thus, the reason of the limitation in the 2d, 3d, 10th, and 12th cases, is maDifeft. PROP. XXI. proportional to the fines of their opposite angles. Let ABC be a spherical triangle, as the fine of AC to the fine of BC, so is the fine of the angle at B to the fine of the angle at A. If one of the angles, as ACB be a right angle; the fine of AC is to the fine of B as a the fine of the hypotenuse AB to a 30. 5. T. the radius or fine of the right angle ACB; that is, as the fine of BC to the fine of A. But if none of them be a right angle, let CD be the arch of a great circle, perpendicular to AB: and because ADC is a right angle, the fine of AC is to the radius, as the fine of CD to the fine of A a For the same reason, the radius is to the fine of CB, as the B fine of B to the fine of CD; therefore, by perturbate equality b, the line of AC is to the line of b 23. $. CB, as the fine of B to the fine of A. IN oblique angled triangles, if a perpendicular be drawn from the vertex to the base, or the base pro. duced ; and the segments of the base, or the segments of the vertical angle, contained between the perpendicular and the sides, be taken, with which to compare the other parts of the triangle: most of the cases of oblique angled triangles may be resolved by the following general proportion. GENE. Spb. Trig. GENERAL PROPORTION III. The fines of the parts adjacent to the perpendicular, are reciprocally proportional to the tangents of the parts adjacent to them; and directly proportional to the cofines of the parts separated from them. But the cofines of the parts adjacent to the perpen. dicular are to be used instead of their fines, when they are compared with the sides of the triangle. Let ABC be an oblique angled triangle, and let CD be pera pendicular to AB, and fall either within or without the triangle. First, Let the segments of the base AD, DB be taken, with which to compare the other parts of the triangle ABC, then the angles at A, B are the parts adjacent to them: and because ADC is a right angled triangle, the fine of AD is to the radius, as the tangent of CD to the tangent of Aa. For the same reason, the radius is to the fine of BD, as the tangent of B to the tanb 23. 5. gent of CD; therefore, by perturbate equality , the fine of AD is to the fine of BD, as the tangent of B to the tangent of 2 19. S. T. C 2. G. P. Also, the sides AC, CB are separated from AD and DB: and because ADC is a right angle, the cofine of AD is to the radius, as the cofine of AC to the cofine of CD C: and for the same reason, the radius is to the cofine of DB, as the cofine of DC to d 22. 5. the cofine of CB; therefore, by equality d, as the cosine of AD to the cofine of DB, so is the cofine of AC to the cofine of CB; which is the second part of the general proportion. Secondly, Let the vertical angles ACD, BCD be taken, with which to compare the other parts of the triangle ABC; then, the fides AC, CB are adjacent to the vertical angles: and because ADC is a right angle, the cosiné of ACD is to the radius, as b 23. S. as the tangent of CD to the tangent of CA 4, and for the same sph. Trig. reason, the radius is to the cofine of BCD, as the tangent of BC to the tangent of CD; therefore, by perturbate equality , { 1. G. P. as the cofine of ACD to the cofine of BCD, so is the tangent of BC to the tangent of CA; which is the first part of the proportion. Also, the angles at A, B are separated from the vertical angles ACD, BCD: and because ADC is a right angle, the fine of ACD is to the radius, as the cofine of A to the cofine of CD Ć: and for the same reason, the radius is to the fine of BCD, as the cofine of CD to the cofine of B ; therefore, by equality d, as the fine of ACD to the fine of BCD, so is the cofine of A to the cofine of B; which is the second part of the general proportion. C 2. G. P: d 22. 5. IN PROP. XXII. from the vertex, perpendicular to the base; the tangent of half the sum of the segments of the bale, is to the tangent of half the sum of the sides, as the tangent of half the difference of the sides to the tangent of half the difference of the segments of the base. Let ABC be a spherical triangle, and let the arch CD be pera pendicular to AB; the tangent of half the sum of AD, DB is to the tangent of half the sum of the sides AC, CB, as the taugent of half the difference of AC, CB, to the tangent of half the difference of AD, DB. Let AL be the common section of the planes AC, AB, and o the centre; make CE, CF each equal to CB; therefore AE is the sum of AC, CB, and AF is their difference : and if AC be equal to CL, A is the pole of CD; therefore AD is equal to DL; and the tangent of balf the difference of AC, CB is the cotangent of half their sum, and the tangent of half the difference of AD, DB is the cotangent of half their sum. But if AC be not equal to CL, join EF, and let it meet AL in M; and join BM meeting the arch AB again in G; and join CO, OD, and let them meet EF, BG in H, K; and join BH ; and through A draw a AN parallel to OC, meeting the circle ACL a 31. 1. again in P; and join EA, EP, PL: and because EC is equal to CF, OC bisects EF at right angles in H 6: and because CB is bCor.30.3. equal to CE, CH is the verfcd fine of CB, and therefore OHB is a right angle : and OHE is a right angle ; therefore OH is c 4. Def. LI perpendicular P.T, e 18. II. Sph. Trig. perpendicular to the plane MBH"; the plane MBH is therefore perpendicular to the plane COD pafling through OH: and the d 4. 11. plane ADB is also perpendicular to the plane COD, because ADC is a right angle ;' therefore their common section BG is f 19.11. perpendicular ? to the plane COD; the angle OKB is therefore 8 3. 3. a right angle, and BK is equal to KG 8, and the arch BD to the h 30. 3. arch DGb; therefore AB, AG are the sum and difference of AD, DB : and because AN is parallel to OC, ANE is a right K 29. 1. angle \; therefore, if EN be made the radius, AN is the tan gent of AEN, half of the angle at the centre standing upon m 20,3 ÅE ", and PN is the tangent of PEN, or the cotangent It Def. EPN, that is, of half the arch AE: but becaufe the angle o 31. 3. APL in the semicircle is equal o to the right angle ANM; AM is P 2.6. to ML, as AN to NPP; that is, as the tangent of half AF to the cotangent of half AE. In the same manner, if a straight 1 1. P. T. 1 P. T. line be drawn through A, parallel to OD, it may be proved, that AM is to ML, as the tangent of half AG to the cotangent of half AB; therefore, as the tangent of half AF to the cotangent of 911. 5. half AE, fo e is the tangent of half AG to the cotangent of 1 16. 5. half AB; and, alternately, the tangent of half AG is to the tangent of half AF, as the cotangent of half AB to the cotangent of half AE ; that is, as the tangent of half AE to the tangent of half AB, because the tangents of two arches are re5.Cor.1o. ciprocally proportional to their cotangents. Wherefore, &c. Def. P. T. Cor. Hence, the tangent of half the base is to the tangent of half the sum of the fides, as the tangent of half the difference of the fides to the tangent of half the difference of the segments of the base, when the perpendicular falls within, or to the tangent of half the sum of the fegments when it falls without the Triangle. LEMMA I L E M M A. Spb. Trig. Na plane triangle, the rectangle contained by half of the base and the distance of the perpendicular from the middle of the base, is equal to the rectangle contained by the fines of half the sum and half the difference of the arches of any circle, which are subtended by the fides of the triangle. 210. 1 Let ABC be a triangle, and bife&t the base BC in D, and draw AE perpendicular to BC: and let FGH be any circle in which are placed FG equal to AB, and GH to AC: the rectangle BD, DE is equal to the rectangle contained by the fines of half the sum and half the difference of the arches FG, GH. Join FH, and bisect * it in K; and draw GM, KN perpendicular to KH ", and GOP parallel to it; therefore the angle b 11.12.1. FGP is equal to GFHd; and the arch FP equal to GH 4: and C 31. 1. because KV bisects FH at right angles, it passes through the d 29.1. centre ', and therefore it bisects & the straight line GP and the e 27. 3. arches FGH, PNG". Wherefore FK is the line * of FN halff Cor. 1.3. g 3. 3. the sum of the arches h 30. 3. FG, GH, and KM P G k 3. Def. or GO is the line k of half PG the dif. ference of the arches FG, GH: but four times the rectangle B E o км H FK, KM is equal to twice the rectangle FH, KM; that is, to the difference of the {quares of FG, GHỊ, or of BA, AC; and four times the ic, e. rectangle BD, DE is also equal to the difference of the squares of BA, AC '; therefore the rectangle BD, DE is equal to the rectangle FK, KM, contained by the fine of half the sum of the arches FG, GH and the fine of half their difference. P. T. |