Solution. Construct the proportion sin A: BC= sin B: AC=1 BC sin B sin C: A B, or sin 2 sin. A (A+B): A B, (because sin C= sin (A+B).) ACABA C-AB cot 1 A: tang (B-C) And consequently since we know B-C and B+ C, it is easy to obtain the angles B and C; and we shall then have sin B: AC sin A: B C, which will give the third side BC. This third case may also be solved in the following manner: If B A had been the side given, we should have had sin (A + B) BA sin B: AC sin A: BC. CASE 2. Given two sides, and an angle opposite to one of the given sides, to find the third side and the other two angles, supposing that we know of what species they are. Solution. Suppose that the sides A B, AC (fig. 2.) are given, and the angle B; we shall first obtain Draw the perpendicular BF (fig. the angle C by this proportion AC 4.) upon the side A C, and let A B sin B AB: sin C, which willa, AC=b, sin_A=s, cos A=c, give the third angle. And we may we shall have R: as: BF= then obtain the side B C by the as proportion sin B: AC sin A: BC. But to find the side B C at once, draw the perpendicular AD (fig. 3.) on the side BC; call AB, a; AC, b; sin B, s; cos B, c; and we shall have R AB (a) = sin B (s): AD) =cos B (c) ; BD=0 = as R ac Ꭱ tang (B R =c:AF= ac R* a Therefore FC=b-co But in the right-angled triangle B FC, ac we have (FC (b) :BF R: tang C= R aRs bR-ac Similarly tang B = b Rs ¡as R a R-bc If the angle A is obtuse, C becomes negative, and we have tana Rs bR+ ac2 gent C= b Rs a R+bc and tang B The right-angled triangle B FC a2 s2 ac R a2 + From this expression we may deduce the following proportion: d-a+b d+a-b 2 X 2 =R2 ab: ab: (q-a) (q — b) = R2 : sino We may also calculate it by (a2 + b2 — d2), which gives 2 ab 2 ab: a2 + b2 — d2 = R : cos B. b B That is, to find an angle B in a triangle, of which we know the three sides, we must first find the sum of the squares of the two sides containing the angle required, and subtract from this sum the square of the third side; this will leave a 536 ab (g~~a) (q—b) 2 =tang2 cos B 2 q (q— d) 2 ab (g—a) (q—b) (g) (q-d) B 2 if radius=1; or tang? = R2 (q- a) (g—b), if radius= R. q (q-d) 2 s cos A sin A V2s sin A cos A = = 1/2 s cot A R V2s tang A R • he hypothenuse =√√x + y2 = os A sin A Every spherical angle is less n 180°. Any side of a spherical trigle is less than 180°. The sum of the sides of any erical triangle is always less in 360°. 3. If from the three angles of y spherical triangle, taken as ales, we describe three arcs, hose meeting forms a new sphecal triangle, the angles and sides this second triangle will be reprocally the supplements of the des and angles opposite to them the first triangle. 7. The sum of the angles of a pherical triangle may vary from 80° to 540° exclusively. the sines of the angles are proportional to the sines of the sides. 14. In any right-angled spherical triangle, radius is to the cosine of one of the sides of the right-angle, as the cosine of the other side is to the cosine of the hypothenuse. 15. If an oblique-angled spherical triangle be divided into two rightangled triangles by an arc perpendicular to its base, we shall always have the cosines of the segments of the base proportional to the cosines of the two adjacent sides. 16. In any oblique-angled spherical triangle, on the base of which we have let fall a perpendicular arc, and which lies within the triangle, the tangent of the half base is to the tangent of half the sum of the other two sides, as the tangent of the half difference of these sides is to the tangent of half the difference of the segments of the base. 17. In any right-angled spherical triangle, radius is to the cosine of one of the sides of the right-angle, as the sine of the oblique angle opposite to the other side is to the cosine of the other oblique angle. 18. If we let fall an arc perpendicularly on the base of an ob lique-angled triangle, the sines of the angles at the summit will be proportional to the cosines of the angles of the base. 19. In any right-angled spherical 8. Two spherical triangles are triangle, radius is to the cosine of equal, when their angles are re-one of the oblique angles, as the spectively equal. tangent of the hypothenuse is to the tangent of the side opposite to the other angle. 9. In every spherical triangle, equal sides are opposite to equal angles. 10. In any right-angled spherical triangle, each of the oblique angles is of the same species as the side opposite to it. 20. If we let fall an arc perpen. dicularly on the base of an oblique. angled spherical triangle, the cosines of the angles at the summit will always be reciprocally proportional to the tangents of the adjacent sides. 11. In a spherical triangle, if the two sides of the right-angle are of the same species, the hypothenuse will be less than 90°; and if they are of different species, the hypothenuse will be greater than 90°. 12. In every right-angled sphe-gent of this last side. rical triangle, radius is to the sine of the hypothenuse, as the sine of one of the oblique angles is to the sine of the side opposite to it. 21. In any right-angled triangle, radius is to the sine of one of the sides of the right-angle, as the tangent of the oblique angle opposite to the other side is to the tan 13. In any spherical triangle, 22. In every right-angled spherical triangle, radius is to the cosine of the hypothenuse, as the tangent of one of the oblique angles is to the cotangent of the other angle. Solution of the different Cases of The angle B CD is found by subOblique-angled Spherical Tri-tracting ACD from AC B, when angles. the arc falls perpendicularly withHere there are as many varieties in the triangle; but when this arc as there are different combinations falls without, we must subtract the among the six parts of a triangle, angle ACB from the angle A CD, taken four by four. But there are in order to find B CD. fifteen of these combinations, and each admits of three different cases. Consequently, there are 45 problems to solve relating to oblique-angled triangles. But, as the resolution of one frequently includes that of several others, they may be reduced to the twelve following cases: Case 1. In an oblique-angled spherical triangle ABC, (fig. 6.) given the angles B and A, and the side opposite to the angle A, to find the side A C opposite to the other angle. Case 2. Given the two angles A, (fig. 7, 8.) and B, with the side B C, opposite to the angle A, to find the third angle. Let fall the perpendicular arc CD, and in the triangle B C D we shall have R: cos BC tang B: cot BCD and then Case 5. Given as before, the two R:cos A C=tang A: cot ACD cos BCD: cos ACD = tang AC: Case 6. Suppose any two sides given, A C and B C fig. 9, with one of the angles opposite to those sides, such as the angle A; to find the angle B, opposite to the other side. This problem may be resolved by the well known proportion sin BC sin AC sin A: sin B This problem falls within the first case; they differ only by an inversion of the terms in the proportion which resolves both. Case 7. The same things being given, required the third side A B. cos B: cos Asin B C D : sin A CD. R: cos A tang AC: tang AD Adding these two angles, or sub-cos AC: cos BC= cos AD: cos BD. tracting one from the other, according as the given angles A and B are of the same or of different species, we shall have the angle CR: cos A C=tang A: cot A CD required. Case 3. Given two angles A and B, fig. 7 or 8, with the opposite side BC, as in the preceding case, to find the side contained between these two angles. The perpendicular arc CD forms the two right-angled triangles ACD and DCB, the last of which gives R: cos B tang BC: tang BD Besides we have tang A tang B= sin B D : sin A D. Consequently, we shall have A B = ADD B, according as the angles given are of the same or of different species. Case 4. Given the two angles A and C, fig. 7 or 8, with the side on which these angles are formed, to find the third angle B First, R: cos AC And then, sin AC D: sin B C D= cos A: cos B. tang A: cot A CD Case 8. On the same suppositions, to find the angle C, contained be tween the given sides A C and BC. and, tang BC: tang AC=cos ACD: BCD. Whence we deducè A CB= ACD B CD, according as the sides AC and B C are of the same or of different species. The supplemental triangle is equally proper to resolve this problem by reducing it to the third case. Case 9. Given the two sides A C and AB, and the angle A, contained between them, to find the other side B C. R: cos A tang AC: tang A D Case 10. The same things being Case 11. If the three sides were given, how should we find one of the three angles, A for example? |