2 COS a a COS COS 2 2 arith. co. log tang A (61° 25') 189.736269 log cot ACD (67° 38' 59') 9.614006 log R 10. log cos AC 9.350275 Hence the side AC=77° 3' 18". We have before found it 77° 5', in the solution of the first case, and this difference arises from the quantities neglected in the valuation of the logarithms. 112. This case may also be solved a single proportion. For retaining the notation used in the last paragraph of the preceding case, let a', b', c', A, B, C, represent the sides and angles of the supplemental triangle ; and call a'+b+c=s. Then according to . ' ' the last problem A' sin S. sin (S' - a') 2 sin 6 sin c But by the definition of the supplemental triangle A' 180-a -4) =cos (906– ) =sin Similarly sin b'=sin (180"—B)=sin B; and sin c'=sin C. 1800-A +180°-B + 1800-C = sin 2 sin (2700-S)=-cos S (S-) =cos (S-A). Substitute these values in the above expression for cos. have cos S. cos (S-A), if radius=1, or sin B. sin C if radius=R sin B sin C The expression under the radical sign is always positive ; for sinA+B+C=S has been shewn (70) to be greater than 90°, and 2 less than 270o, and therefore cos S is negative, and consequently -cos S, positive. B+C-A Again, S-A= but since in every spherical triangle 2 the sum of any two sides is greater than the third side, we have Again, sin S=sina' +b+c' sin { +} and sin (S-d')=sin{270..-..-(180--A)} = = sin {900.–S–A)} ( = A and we sin e sin 2 b' +c'> a'; therefore 180°-B +180°_C> 180o--A, and consc B+C-A quently by transposition and reduction < 90°; whence 2 cos (S—A) is positive. Example. Given the three angles of an oblique spherical triangle respectively 61° 25', 82° 36', and 82° 4'; required one of the sides; for example that opposite the angle 82° 36'. By the preceding formula we have side sought sin R?.--cos S. cos (S-82° 36') sin 610 25' x sin 820 4' log cos (113. 2) = 9.592473 log cos (30° 26') 9.935618 212960 5 arith. co. log sin (61° 25') 10.056445 1190 2-S arith. co. log sin (82° 4') = 10.004177 820 36 2|19.588713 30. 26-S-820 36 side required 9.794356 2 Whence the side is 77° 3' nearly, as before. log R? = 20. log sin EXAMPLES OF THE CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. 1. In the oblique-angled spherical A triangle ABC, given BC=114° 30', AC = 56° 40', and the angle BAC= 1250 20ʻ; required the other parts of the B С triangle Answer, 2 B=48° 30', _ C=629 54, side AB=830 11'. 2. In the oblique-angled spherical triangle ABC, given AB=45°30', BC=720 40', and the angle A=135°; required the rest ? Answer, L C=31° 53' 35", L B=240 31 25", AC=34° 4' 46". 3. In the oblique-angled spherical triangle ABC, given BC=1140 30', CA=56° 40', and the angle C=62° 54'; required the rest ? Answer, L B=730 31', LA=1250 20', side AB-830 12'. 4. In the oblique-angled spherical triangle ABC, given AB=41° 45, BC=730 20', and the angle B=30° 30'; required the rest ? Answer, L A=130° 8' 45", 4 C=31° 33' 43', AC=40, 12' 59". 3 5. In the oblique-angled spherical triangle ABC, given the angle B=52° 20', the angle C=63° 40', and the side AB=83° 25'; required the rest? Answer, AC=61° 19' 53", BC=118° 21' 25', LA=127° 26° 47'. 6. In the oblique-angled spherical triangle ABC, given the angle C=48° 20, the angle B=125° 20', and the side AC=114° 30°; required the rest? Answer, AB=56° 40', BC=83° 11', _ A_62° 54'. 7. In the oblique-angled spherical triangle ABC, given the angles B=53° 30°, C= 60° 15', and the side BC=115° 40'; required the rest? Answer, AB=75° 56' 31", AC=67° 2' 39", A-126° 18' 55". 8. In the oblique-angled spherical triangle ABC, suppose the angle A=125° 20', the angle C=48° 30', and the side AC=83° 12', required the rest ? Answer, BC=114° 30', AB= 65° 30', L B=62° 54'. 9. In the oblique-angled spherical triangle ABC, given AB=810 10, AC=6020, BC=112° 25'; required the angles ? Answer, LA–122. 11' 6", L B=520 42' 11", LC=64° 46' 36". 10. In the oblique-angled spherical triangle ABC, given AB=56° 40', AC=83° 13', and BC=114° 30'; required the angle A opposite the side BC? Answer, LA=1250 18' 56'. 11. In the oblique-angled spherical triangle ABC, given the angle A= 125° 20', the angle B=48° 30', and the angle C=68° 54'; required the sides? Answer, AB=830 124", BC=114° 29' 56", AC=560 39' 29". 12. In the oblique-angled spherical triangle ABC, given the angles A=129° 30', B--54° 35', and C=63° 5'; required the sides? Answer, AB --82° 19, BC-120° 57' 4", AC=64° 55' 36". ', = " NAPER'S RULES OF THE CIRCULAR PARTS. The proportions upon which the solution of the various cases of right-angled spherical triangles depend, are simple, and perfectly adapted to logarithmic computation; but they are not easily remembered. All these cases may be solved by Naper's Rules of the Circular Parts, which supply an artificial memory to the computist; and in the whole compass of the mathematical science it will not be easy to find rules equally ingenious and conducive to facility and brevity of computation. The nature and application of these rules will readily be understood from the following explanation: In a right-angled spherical triangle, the right angle is not considered; the complements of the other two angles, the complement of the hypothenuse, and the two sides, making in all five quantities, are called by Naper circular parts. Any one of the circular parts may be called a middle part (M), then the two circular parts imme. diately on the right and left of M are called adjacent parts ; and the other two remaining circular parts, each separated from M, the middle part, by an adjacent part, are called opposite parts: The circular parts are reckoned in order round the triangle, so that to every one considered as a middle part, there are two adjacent and two opposite parts. In every question proposed for solution, three of the circular parts are concerned, of which two are given, and one required; and of these three the middle part must be such, that the other two may be equi-distant from it, that is, either both adjacent, or both opposite parts. This arrangement being made, the value of the part required may be found by the following Rule. The product of the radius and the sinc of the middle part, is equal to the product of the tangents of the adjacent, or to the product of the cosines of the opposite parts. Note. It is obvious that the cosine of a complement is a sine, and the tangent of a complement a cotangent, and ricc versa. Here the circular parts concerned are the legs and hypothenuse, and it is evident that if AB be made the middle part, BC and AC will be opposite parts; consequently rad * cos AB=cos BC x cos AC whence AC=54° 59' 49". 2. To find the angle A. Here BC is the middle part, and AB and the angle A are opposile parls ; therefore rad * sin BC=sin AB x sin A whence A is found=45° 41' 24". 3. To find the angle B. Here the angle B is the middle part, and the leg BC, and the hypothenuse AB are adjacent parts; therefore rad x cos B=tang BC X cot AB whence B is found-65° 45' 57". By the above rule of Naper's we are enabled to solve all the cases of right-angled spherical triangles; and also those cases of oblique-angled spherical triangles in which we have directed a per. pendicular to be drawn from an angle to the opposite side, provided that two of the given parts remain in one of the two triangles thus formed, Trigonometry is a branch of mathematical science which is indispensible on the calculations of remote and inaccessible objects; and hence its use in geography, in ascertaining the various distances and position of places on the earth ; in navigation, in directing the course, the latitude and longitude of a ship. Spherical Trigonometry is particularly applied to the sublime science of astronomy, in discovering the positions, magnitudes, and distances of the heavenly bodies. It may also be applied to the useful arts, as in architecture the theorems of Naper will be found useful in ascertaining the angles which two adjacent planes of a roof at a hip make with each other, the inclination of the planes being given to the horizon. In short, a catalogue of its applications would be too formidable to be inserted in this place. a |