in the polar triangle the hypothenuse a'=180°-115° 20' 64° 40', and one of the sides, c'=180°-42° 10'=137° 50', from which, by Napier's rule, we find B'=115° 23′ 20′′. C'-132° 2′ 13′′. b' =125° 15′ 36′′. Hence, taking the supplements of these arcs, we find the parts of the required triangle are AC 64° 36′ 40′′. AB=47° 57′ 47′′. B =54° 44′ 24′′. the Ex. 8. In the spherical triangle ABC, the side AC=90°, angle C 69° 13′ 46", and the angle A=72° 12' 4". Required (215.) In any spherical triangle, the sines of the sides are proportional to the sines of the opposite angles. In the case of right-angled spherical triangles, this proposition has already been demonstrated. Let, then, ABC be an oblique-angled triangle; we are to prove that sin. BC sin. AC :: sin. A : sin. B. Through the point C draw an arc A D of a great circle CD perpendicular to B (216.) Cor. 1. In any spherical triangle, the cosines of the sides are proportional to the cosines of the segments of the base, made by a perpendicular from the opposite angle. Cor. 2. The cosines of the angles at the base are propor. tional to the sines of the segments of the vertical angle. For, by Theorem I., Cor. 3, Also, cos. CD: R:: cos. A: sin. ACD. cos. CD: R:: cos. B: sin. BCD. cos. A cos. B:: sin. ACD: sin. BCD. Cor. 3. The sines of the segments of the base are reciprocally proportional to the tangents of the angles at the base. For, by Theorem II., Also, Hence sin. AD: R:: tan. CD: tan. A. sin. BD: R:: tan. CD : tan. B. sin. AD sin. BD :: tan. B : tan. A. : Cor. 4. The cotangents of the two sides are proportional to the cosines of the segments of the vertical angle. For, by Theorem II., Cor. 2, cos. ACD: cot. AC :: tan. CD: R. Also, cos. BCD cot. BC : cot. BC tan. CD: R. Hence cos. ACD: cos. BCD :: cot. AC: cot. BC. THEOREM IV. (217.) If from an angle of a spherical triangle a perpendicular be drawn to the base, then the tangent of half the sum of the segments of the base is to the tangent of half the sum of the sides, as the tangent of half the difference of the sides is to the tangent of half the difference of the segments of the base. Let ABC be any spherical triangle, and let CD be drawn from C perpendicular to the base AB; then tan. (BD+AD): tan. (BC+AC):: tan. (BC-AC): tan. (BD-AD). Let BC=a, AC=b, BD=m, and A AD=n. Then, by Theorem III., Cor. 1, 1 2 b C m n D cos. a: cos. b:: cos. m : cos, n. a Whence, Geom., Prop. 7, Cor., B. II., cos. b+cus. a : cos. b—cos.a :: cos. n+cos. m : cos.n—cos. m But by Trig., Art. 76, : cos. b+cos, a cos. b-cos. a:: cot. (a+b): tan. (a−b). Also, by the same Art., cos. n+cos. m : cos. n—cos. m :: cot. 1(m+n): tan. 1(m—n) Therefore cot. 1(a+b): cot. 1(m+n) :: tan. 1(a−b) : tan. 1(m—n). But, since tangents are reciprocally as their cotangents, Art. 28, we have cot. 1(a+b): cot. 1(m+n) :: tan. 1(m+n) : tan. 1(a+b) Hence tan. 1(m+n): tan. 1(a+b) :: tan. 1(a−b): tan. 1(m—n). (218.) In the solution of oblique-angled spherical triangles, six cases may occur, viz.: 1. Given two sides and an angle opposite one of them. 2. Given two angles and a side opposite one of them. 3. Given two sides and the included angle. 4. Given two angles and the included side. 5. Given the three sides. 6. Given the three angles. CASE I. (219.) Given two sides and an angle opposite one of them, to find the remaining parts. C In the triangle ABC, let there be given the two sides AC and BC, and the angle A opposite one of them. The angle B may be found by Theorem III. sin. BC sin. AC:: sin. A: sin. B. : From the angle C let fall the per pendicular CD upon the side AB. D B The triangle ABC is divided into two right-angled triangles, in each of which there is given the hypothenuse and the angle at the base. The remaining parts may then be found by Napier's rule. Ex. 1. In the oblique-angled spherical triangle ABC, the side AC=70° 10′ 30′′, BC=80° 5' 4", and the angle A=33° 15' 7". Required the other parts. sin BC sin. AC :: sin. A : sin. B=31° 34′ 38′′. : sin. A: sin. ACB :: sin. BC: sin. AB=145° 5′ 0′′. When we have given two sides and an opposite angle, there are, in general, two solutions, each of which will satisfy the conditions of the problem. If the side AC, the angle A, and the side opposite this angle are given, then, with the latter for radius, describe an arc cutting the arc AB in the points B and B'. The arcs CB, CB' will be equal, and each of the tri- A angles ACB, ACB' will satisfy the conditions of the problem. D B There is the same ambiguity in the numerical computation. The angle B is found by means of its sine. But this may be the sine either of ABC, or of its supplement AB'C (Art. 27). In the preceding example, the first proportion leaves it ambiguous whether the angle B is 31° 34' 38', or its supplement 148° 25′ 22′′. In order to avoid false solutions, we should remember that the greater side of a spherical triangle must lie opposite the greater angle, and conversely (Geom., Prop. 17, B. IX.). Thus, since in the preceding example the side AC is less than BC, the angle B must be less than A, and, therefore, can not be obtuse. If the quantity sought is determined by means of its cosine, tangent, or cotangent, the algebraic sign of the result will show whether this quantity is less or greater than 90°; for the cosines, tangents, and cotangents are positive in the first quadrant, and negative in the second. Hence the algebraic sign of each term of a proportion should be preserved whenever one of them is negative. Ex. 2. In the spherical triangle ABC, the side a=124° 53′, b=31° 19′, and the angle A=16° 26'. Required the remain ing parts. CASE II. B= 10° 19' 34" Ans. C=171° 48′ 22′′ c=155° 35' 22". (220.) Given two angles and a side opposite one of them, to find the remaining parts. C In the triangle ABC let there be given two angles, as A and B, and the side AC opposite to one of them. The side BC may be found by Theorem III. sin. B: sin. A :: sin. AC: sin. BC. From the unknown angle C draw A D B CD perpendicular to AB; then will the triangle ABC be divided into two right-angled triangles, in each of which there is given the hypothenuse and the angle at the base. Whence we may proceed by Napier's rule, as in Case I. Ex. 1. In the oblique-angled spherical triangle ABC, there are given the angle A=52° 20′, B=63° 40′, and the side b=83° 25'. Required the remaining parts. sin. B sin. A :: sin. AC : sin. BC=61° 19′ 53′′. : Then, in the triangle ACD, cot. AC: R:: cos. A: tan. AD=79° 18′ 17′′. Also, in the triangle BCD, Hence cot. BC: R:: cos. B: tan. BD=39° 3' 8". AB=118° 21' 25". To find the angle ACB. sin. BC sin. AB :: sin. A : sin. ACB=127° 26′ 47′′. When we have given two angles and an opposite side, there are, in general, two solutions, each of which will satisfy the conditions of the problem. If the angle A, the side AC, and |
