To measure the required parts. The sides AB and BC (376) will be 48° and 64° 35′, NOTE. The right-angled spherical triangle abc (Plate V. fig. 14.) into which the quadrantal triangle ABC is transformed, may be constructed and measured exactly in the same manner as CASE IV. of right-angled spherical triangles was constructed and measured. See Plate V. fig. 9. (478) CASE II. Given a quadrantal side, and the other two sides, to find the rest. The quadrantal side Ac= 90° 0'+] Required the Given The side AB The side BC angles A, B, and c. BY RULE II. (476) Make CD=AC, then CAD, CDA, and adb are each of them a right angle, and AD measures the angle c (474); hence, if we take 90° from BC we get 23° 18′=BD; and ADB is a right-angled triangle. A 1. To find the LB, by Case IV. right-angled spherics. rad: cot AB:: tan BD: cos / B=78° 20′. But B is obtuse (453), and therefore=101° 40'. 2. To find AD, by Case VI. right-angled spherics. COS BD rad:: cos AB: cos AD-62° 26'. B But AD is obtuse (453), and therefore=117° 34′ = ▲ c. 3. To find the DAB, by Case V. right-angled spherics. sin AB : rad::sin BD : sin ▲ dab=25° 55'. Here/DAB is acute (453), and=25° 55′. Here the quadrantal side AC must represent the right angle b; AB the Le; and BC the La; ac the supplement of the B, ab the Lc, and be the LA. † This example was formed from Example 2. Case VI. of right-angled spherics. The B was made equal to the supplement of the hypothenuse, the two sides AB and BC were made equal to the angles of the right-angled triangle; and hence the legs of the right-angled triangle become angles in the quadrantal triangle. In a similar manner the other examples were made. 1. To find the hypoth, ac, by Case XVI. right-angled spherics. rad: cota=cot BC:: cot≤c=cot AB: cos ac=suppt. B =78° 20'. Here ac or suppt. B is acute (457); therefore B is obtuse=101° 40'. 2. To find ab, by Case XV. right-angled spherics. sina sin BC: rad::cos Lc=cos AB: cos ab cosc= 62° 26'. Here abc is obtuse (457), and therefore=117° 34'. 3. To find be, by Case XV. right-angled spherics. sinc=sin AB : rad: : cosa cos BC: cos bc = cos 2A= 64° 5'. Here bc A is obtuse (457), and therefore=115° 55'. If the two preceding cases be thoroughly understood, there can be no difficulty in solving those which follow; for which reason they are given as practical exercises. (479) CASE III. Given a quadrantal side, and its two adjacent angles, to find the rest. The quadrantal side Ac=90° 0′ Given The angle The angle A=54 43 Required AB, BC, c=42 12 and the angle B. Answer. AB=48°, BC=64° 35′, and the angle B=115° 20′. (480) CASE IV. Given a quadrantal side, one of the other sides, and the angle comprehended between them, to find the rest. The quadrantal side Ac= 90° 0′ Given The side AB=115 9 =115 55 } The angle CAB Answer. BC=113° 18′, the angle в=101° 40′, (481) CASE V. Given a quadrantal side, its and a side opposite to that angle, to find the rest. The quadrantal side Ac=90° 0' c=42 12 Given Its adjacent angle Required BC, and the an gles B and C. c=117° 34′. adjacent angle, Required BC, and the an And the opposite side AB=48 0 gles A and B. Answer. BC=64° 35', the angle A=54° 43′, and B=115°20'. But the required parts are ambiguous, and therefore are either acute or obtuse. (482) CASE VI. Given the quadrantal side, one of the other sides, and an angle opposite to the quadrantal side, to find the rest. The quadrantal side AC 90° 0' Given The side AB The angle B 9 =115 gles A and c. Answer. BC=113° 18', the angle A= 115° 55, and c= 117° 34'. CHAP. VIII III. PRACTICAL RULES FOR SOLVING ALL THE CASES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES, WITH A PERPENDICULAR; AND THEIR APPLICATION BY LOGARITHMS. RULE I. (483) The rule Prop. xxv. (392) will solve ten cases; see also the rules in Scholium, art. (408). RULE II. (484) When the three sides are given, to find the angles. Any one of the three sides may be called the base. Then, The tangent of half the base, Is to the tangent of half the sum of the sides, As the tangent of half the difference between the sides, Is to the tangent of the distance of a perpendicular from the middle of the base (411). According as this distance is less or greater than half the base, the perpendicular falls within or without the triangle. When the sum of the two sides is less than 180°, the perpendicular falls nearest to the less side, when greater than 180° it falls nearest to the greater side; consequently the greater segment is joined to the greater side in the former case, and to the less side in the latter. The sum of half the base, and the fourth term found by the above proportion, gives the greater segment, their difference gives the less. The triangle being thus divided into two right-angled triangles, the remaining parts must be found by the proper rules. RULE III. (485) When the three angles are given, to find the sides. The cotangent of half the sum of the angles at the base, Is to the tangent of half their difference, As the tangent of half the vertical angle, Is to the tangent of the excess of the greater of the two vertical angles (formed by a perpendicular), above half the aforesaid vertical angle (413). If the sum of the base angles be less than 180°, the perpendicular and the less segment are nearest the greater base angle, if greater than 180° they are nearest the less base angle. The sum and difference of this fourth term, and half the vertical angle, gives the greater and less vertical angle formed by the perpendicular. The triangle being thus divided into two right-angled triangles, the remaining parts must be found by the preceding rules. (486) CASE I. Given two sides, and an angle opposite to one of them, to find an angle opposite to the other. Given The side AC=80° 19' Required the B. The LA=51 30 DETERMINATION OF THE SPECIES. d A perpendicular in this case is unnecessary. If AC+BC, A+ (в acute), and A+ (B obtuse), be each of the same species with respect to 180°, B is ambiguous: - But if only two of A D these sums be of the same species, that value of в must universally be taken which agrees with the sum of the sides; in all such cases B is not ambiguous (319). SOLUTION. sin BC sin AC:: sin A: sin B-59° 16'. Here B is ambiguous 59° 16′ or 120° 44′, vide Table I. page 208. EXAMPLE II. The side AC 57° 30′ Given The side BC=115 20 The LA=126 37 Required the B. Answer. The B=48° 30', not ambiguous. (487) CASE II. Given two sides, and an angle opposite to one of them, to find the angle contained between these sides. Given The side AC=80° 19′ ́ { The side BC=63 50 Required the c. DETERMINATION OF THE SPECIES. 1. If AC and the A be of the same species, the ACD is acute. The perpendicular CD is of the same species as the LA. If BC and DC be of the same species, the ▲ BCD is acute. D d B :D In using the Formula, the three angles of the triangle are represented by A, Bc, and their opposite sides by a, b, c. The perpendicular is not regarded. 2. If the BCD be less than the ACD, and their sum less than 180°, then the ACB is ambiguous; but if their sum be not less than 180°, their difference is the true value of the LACB, not ambiguous. If the BCD be not less than the ACD, and at the same time their sum be less than 180°, this sum is the true value of the LACB, not ambiguous. SOLUTION. 1. In the triangle ADC, find the ACD. * Thus rad x cos AC=cot LAX cot LACD. Hence cot A: rad: : cos AC: cot ACD=78° 4′ acute. * 2. In the triangle CDB, rad x cos BCD=tan DC X cot BC. In the triangle ADC, rad × cos ACD=tan DC X cot ac. 'Hence cot AC cot BC: : cos ACD: cos BCD=53° 28′ acute. Because the BCD is less than the ACD, and ▲ BCD+ ACD less than 180°, the ACB is ambiguous; viz. ≤ ACB= ▲ ACD + 4 BCD=131° 32', or ACB=LACD-4 BCD=24° 36'. BY CASE II. page 201. Log tan log cos b+log tan A-10; hence =11° 56′ acute. Log sin (+c) = log tan b+ log sin log tan a= +9.77481, from which we have +c=36° 32′ or 143° 28'. Then c=(+c) —p=24° 36′ or 131° 32′. This example is ambiguous, vide Table I. page 208. EXAMPLE II. The side AC 57° 30′ Given The side BC=115 20 The LA=126 37 } Required the C. Answer. LACD=125° 52′ obtuse, BCD=64° 11' acute, and ▲ ACB=61° 41′ not ambiguous. (488) CASE III. Given two sides, and an angle opposite to one of them, to find the other side. The side AC=80° 19′ Given The side BC=63 50 The LA=51 30 DETERMINATION OF THE SPECIES. 1. The segment AD is acute or obtuse, according as the LA is of the same, or of different species with AC. DC is A of the same species with the C B LA. |