CASE I. GIVEN THE HYPOTENUSE AND AN ANGLE. 1. In the right-angled spherical triangle ABC are given the hypotenuse BC 63° 30′, and the angle ABC 53° 42′; to fins the sides AB, AC, and the angle ACB. Construction. Draw the radius OF of the primitive BAD. Make OE the semi-tangent, and OF the tangent of 53° 42', then E is the pole of the hypotenuse, and F its centre, from which, with the secant of 53 42', describe the circle BCD. From B to I lay 63° 30′ on the primitive, draw a straight line from its extremity I to E, cutting BCD in C, and draw the radius OCA; then ACB is the triangle. The side AB is measured on the line of chords. OC measured on the line E B of semi-tangents, and subtracted from 90°, or AC reckoned on the line of semi-tangents from 90° backward, gives the arc AC. Extend the straight line IE to H, and HD, measured on the line of chords, gives the angle ACB. Calculation. The five parts of this triangle are BC, the angles at B and C, and the complements of AB and AC, which are AG and OC. Of these, BC and B are given; and of the things required, BA and C are adjacent to given things, and are therefore found by Equa. 1; AC being separated from given things, is found by Equa. 2. and By Equa. 1. R: cos. BC:: tan. B: cot. C; and R: cos. B:: tar. BC: cot. AG = tan. A B. And by Equa. 2. R: sin. B:: sin. BC: cos. CO sin. CA. And all the three are acute. For CA is of the same affection with B. And AB and C are acute, because BC and B are of the same affection. BC 63° 30' B 53° 42′ cos. 9-6495274 tan. 10-3022637 C 58° 43′ 28′′ cot. 97834924 AB tan. 10 0745951 AB acute 49° 53' 48" 2. Given the hypotenuse BC 126° 24', sin. 9-9517912 sin. 9.9062964 CA sin. 9-8580876 and the angle B 57° 22′; to find the rest. Ans. The angle C 132° 49′ 18", the sides AB 36° 10′ 59′′, and AC 137° 19′ 32′′. 3. Given the hypotenuse BC 72° 28′, and the angle B 138° 23'; to find the rest. Ans. The angle C 104° 58′ 58", the sides AB 112° 54' 32", and AC 140° 42′ 24′′. CASE II. GIVEN A SIDE AND THE ADJACENT ANGLE. find 1. In the spherical triangle ABC, right-angled at A, are given the side AB 51° 28', and the angle ABC 66° 48′; the hypotenuse BC, the side AC, and the angle at C. Construction. Draw the diameter GF the primitive ABD. Make OE the i-tangent of 66° 48', and OF its tanit. From F, with the secant of 66° 48' a radius, describe the circle BCD; ke BA 51° 28', and draw ACO, then 3C is the triangle. AC, or its complement CO, is measured the line of semi-tangents. Draw a e from E through C to I, and the disce of B from the point I, where it cuts AG, gives BC; and the distance of D from its other extremity gives angle at C. Calculation. The hypotenuse BC, and the side CA, being adjacent to en things, are found by Equa. 1., and the angle C by Equa. 2. Thus; 1. R: cos. AG = sin. AB:: tan. B: cot. OC tan. CA, e B; and cos. B: R:: cot. AG tan. AB: tan. BC, acute, for AB: cos. C, like A is like B. Also, 2. R: sin. B:: sin. AG cos. C 61° 16′ 52" tan. 10.2612906 BC tan. 10-5034441 C cos. 9-7578465 2. Given the side AB 126° 26', and the angle B 142° 48′′; find the rest. Ans. Hyp. BC 59° 32′ 45′′, side AC 148° "'′ 17′′, and the angle C 111° 2′ 34′′. 3. Given the side AB 57° 44′, and the angle B 112° 26'; to nd the rest. Ans. Hyp. BC 103° 32′ 46′′, the side AC 16° 1′ 26′′, and the angle C 60° 25′ 54′′. CASE III. GIVEN A SIDE AND THE OPPOSITE ANGLE. 1. In the spherical triangle ABC, right-angled at A, are ven the side AC 38° 27', and the opposite angle ABC 57° 48′; › find the hypotenuse BC, the side AB, and the angle at C. Construction. On OA the radius of the rimitive make OC 51° 33' the compleent of AC. With the tangent of 57° 3' describe an arc from O, and with the cant of 57° 48' from C cut that arc in ', from which centre describe the circle CD, then either ABC or ADC is the iangle. AB is measured on the line of hords, and BC and C as in the last case. Calculation. AB being adjacent to given hings, is found by Equa. 1., and BC B nd C by Equa. 2. They are all ambiguous, or have two values. G 1. R: cos. AG = sin. AB :: tan. B: cot. CO= tan. AC, and t B: tan. AC:: R: sin. AB or AD. == 2. R sin. B: sin. BC: cos, CO sin. CA, and (inver.) sin. B R:: sin. CA: sin. CB or CD. R: sin. OC cos. AC:: sin. C: c B, and (inver.) cos. CA: R:: cos. B: sin. ACB or ACD. tan. 10.2003431 sin. 9.9274695 R+cos. 19.720 R+tan. 19.8998271 R+sin. 19.7936727 B=37° 48′ AC=38227 AB 30° 0' 4" or 149 59′ 56′′ C=42° 32′ 37′′ or 137° 7′ 23′′ cos. 98 Sin. C 9.8327 Sine of AB=9.6989840 Sin. BC 9.8662032 2. Given the side AC 136° 28', and the angle B 127° 48 to find the rest. Ans. The hyp. BC 60° 39′ 24′′; the sid AB 47° 28′ 20′′; and the angle C 57° 43′ 1′′, or their su plements. 3. Given the angle B 84° 21', and the side AC 78° 40';: find the rest. Aus. The hyp. BC 80° 9′ 34′′; the side A 29° 34′ 42′′; and the angle C 30° 3′ 54′′, or their suppe ments. CASE IV. WHEN THE HYPOTENUSE AND A SIDE ARE GIVEN 1. In the spherical triangle ABC, right-angled at A, ar given the hypotenuse BC 64° 42', and the side AC 47° 48 to find the side AB, and the angles at B and C. Construction. Lay AC 47° 48' on the primitive, and draw the radii OC, OA. On the former lay the secant of 64° 42′ from O to H, from which, with the tangent of 64° 42', cut OA in B, and describe the circle CBD, then ABC is the triangle. E Let F be the centre of CBD, then OF measured on the line of tangents gives the angle ACB. Lay the semi-tangent of it from O to E. Lay a ruler from B through E; the arc of the primitive between it and D is the measure of the angle at B, and OB measured on the line semi-tangents gives the complement of AB. Calculation. The angle at C being adjacent to the given things found by Equa. 1; the other two, being separated from them, are fe by Equa. 2. 1. Tan. CB cot. AG tan. AC:: R: cos. C acute, since A CB are alike. 2. Sin. CB: R:: cos. AG = sin. AC: sin. B, like AC. S AG=cos. AC: R:: cos. BC: sin. OB=cos. BA acute, because Al CB are alike. . Given the hypotenuse BC 121° 12', and the side AC 56° ; to find the rest. Ans. The angles C 155 0' 34", and 76° 25′ 31′′; and the side AB 158° 48′ 57′′. 3. Given the hypotenuse BC 72° 28', and the side AC 123° ; to find the rest. Ans. The angles C 118° 47′ 19′′, and 118° 44′ 1′′, and the side AB 123° 18′ 46′′. CASE V. GIVEN THE SIDES ABOUT THE RIGHT ANGLE. 1. Given the sides AB 47° 38', and AC 67° 30', about the ht angle BAC of the spherical triangle ABC; to find the potenuse BC, and the angles at B and C. Construction. Make AB 47° 38′ on : primitive, and draw the radius OA, which make OC = 22° 30', the comement of AC taken from the line of ni-tangents, and having drawn the diaeter BD, describe the circle BCD; en ABC is the triangle. Let F be the centre of BCD, then OF easured on the line of tangents gives e angle at B. Make OE its semi-tannt, then E is the pole of BCD, and C and C are measured as in the 2d Case. F 12 E B G Calculation. The angles at B and C being adjacent to given things, e found by Equa. 1., and the hypotenuse BC by Equa. 2. C. 1. Cos. AG = sin. AB: R:: cot. OC tan. AC: tan. B, like Cos. OC Isin. AC: R:: cot. AG =tan, AB : tan. C, like B. 2. R sin. OC = cos. AC:: sin. AG =cos. AB: cos. BC = cute, for AB, AC are like. C 67° 30' R + tan. 20-3827757 sin. 9-9656153 sin. 9-8685548 R + tan. 20 0399770 372° 59' 2" tan. 10.5142209 cos. 9-5828397 cos. 9-8285778 C tan. 10:0743617 BC cos. 9.4114175 2. Given the sides about the right angle AB 108° 44′, and AC 67° 42'; to find the rest. Ans. The angles C 107° 25' 13", and B 68° 46′ 25′′, and the hyp. BC 97°. 3. Given the sides about the right angle AC 127° 48', and AB 71° 25'; to find the rest. Ans. The angles B 126° 19′ 29", and C 75° 7' 21", and the side BC 101° 15′ 49′′. CASE VI. WHEN THE TWO OBLIQUE ANGLES ARE GIVEN. 1. In the spherical triangle ABC, right-angled at A, are given the angles at B 39° 48', and at C 67° 12'; to find the hypotenuse BC, and the sides AB and AC. The hypotenuse and the sides are measured as before. Calculation. The hypotenuse being adjacent to the given angles is foun by Equa. 1., and the sides by qua. 2. 1. Tan. B: cot. C:: R: cos. BC acute, for B and C are alike. 2. Sin. B: R:: cos. C: sin. AG R: cos. B: sin. OC = cos. AC, like B. = cos. AB, like C; and sin. C sin. 9-964 C 67° 12' R + cot. 19-6236227 R + cos. 19-5882892 tan. 9.9207329 sin. 9-8062544 R+cos. 19.8852! B 39° 48' BC 59° 41′ 59′′ cos. 9-7028898 AB cos. 9-7820348 AC cos. 9-920855 AB 52° 44' 35" AC=3% 2. Given the angles B 112° 38', and C 63° 40′; to find the sides. Ans. The hyp. BC 101° 54′ 34′′, the sides AC 115 25′ 44", and AB 61° 16′ 30′′. 3. Given the angles C 102° 28′, and B 118° 30′; to find the sides. Ans. The hyp. BC 83° 6′ 20′′, the sides AB 104 13' 11", and AC 119° 15′ 14′′. SOLUTION OF OBLIQUE-ANGLED SPHERICAL TRIANGLES. WHEN the three sides or the three angles are not the given parts, the solution may always be obtained by drawing a per pendicular from the extremity of a given side and opposite given angle, and then computing by Napier's rules of the cir cular parts. The following Table contains the proportions for the sol tion of the 12 cases of oblique-angled spherical triang ABC represents any spherical triangle in which the perpendi cular AD either falls within the triangle or meets the base B produced beyond C. NOTE. The cases referred to are those of the preceding Table. 2 |