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. (X) When the three angles are given, to find the sides. The co-tangent of half the sum of the angles at the base, Is to the tangent of half their difference; As the tangent of half the verticle 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. (W. 180.) 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 rightangled triangles, the remaining parts must be found by the preceding rules. (Y) 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′ The side BC=63°.50' Required the B. DETERMINATION OF THE SPECIES. A perpendicular in this case d D 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 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. SOLUTION. Sine BC sine Ac:: sine ▲ A: sine в=59°.16′. Here в is ambiguous=59°.16'. or 120°.44'. BY FORMULA† 11. page 200. Log sine B(log sine a + log sine b)-log sine a=59°.16'. This example is ambiguous, vide Table I. page 207. Given EXAMPLE II. The side AC 57°.30' The side BC 115°.20' Required the B. Answer. The B=48°.30′, not ambiguous. (Z) CASE II. Given two sides and an angle opposite to one of them, to find the angle contained between these sides. The side AC-80°.19' Given 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 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 ZACB, 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 ▲ A× cot ≤ ACD. Hence, cot Arad::cos ac cot ACD=78°.4' acute. 2. In the triangle CDB, rad x cos BCD=tang DC × cot BC. In the triangle ADC, rad x cos ACD=tang 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 + less than 180°, the LACB is ambiguous; viz. LACB= ACD ACD + BCD131°.32′, or ▲ ACB ACD- / BCD=24°.36′. In using the Formula, the three angles of the triangle are represented by ^, », c, and their opposite sides by a, b, c. The perpendicular is not regarded. BY FORMULA III. page 201. Log tan (log cos b+log tang A)-10=11°.56'. acute. Log sine +c=(log tang b+log sine p)-log tang a=36°.32′. Then (+c)--36°.32'-11°.56' 24°.36'= c. This example is ambiguous, though not shewn by the formula, vide Table I. page 207. OR, Log cot (log cos b+log tang A)-10=78°.4' acute. Log cos (9+c)=(log cot a + log cos 4)-log cot b=53°.28'. (4+c)+9=53°.28′+78°.4'—131°.32′ = ▲c. Then Given The side BC115°.20' Required the c. Answer. ACD=125°.52′ obtuse, BCD 64°.12′ acute, and ACB=61°.40', not ambiguous. (A) CASE III. Given two sides and an angle opposite to one of them to find the other side. Given The side AC-80°.19′ The side BC 63°.50 Required the side AB. DETERMINATION OF THE SPECIES. 1. The segment AD is acute or obtuse, according as the LA is of the same, or of dif ferent species with AC. DC is of the same species with the A. DB is acute or obtuse, according as BC is of the same, or of different species, with DC. 2. If DB be less than AD, and their sum less than 180°, then AB is ambiguous; but if their sum be not less than 180°, their difference is the true value of AB, not ambiguous. If DB be not less than AD, and at the same time their sum be less than 180°, this sum is the true value of AB, not ambiguous. SOLUTION. 1. In the triangle ADC, find the side ad. ** * Hence, cos AC cos BC::COS AD COS DB 46°.6′ acute. Because DB is less than AD, and AD+DB less than 180°, AB is ambiguous; viz. AB AD+DB=120°.46′, or AB=AD-DB= 28°.34. BY FORMULA 111. page 201. Log tang (log cos A+log tang b)-10=74°.40′ acute. Log cos (c-)=(log cos a+log cos 9)-log cos b=46°.6'. Then, (c)+9=46°.6′+74°.40' 120°.46′-AB. This example is ambiguous, though not shewn by the formula, vide Table I. page 207. Given EXAMPLE II. The side AC 57°.30' The side BC=115°.20Required the side ab. Answer. AD=136°.53′ obtuse, DB 54°.27′ acute, and AB AD-DB=82°.26′ not ambiguous. (B) CASE IV. Given two angles, and a side opposite to one of them, to find the other opposite side. Given The LA 51°.30' The ZB=59°.16 Required the side AC. The side BC 63°.50′ DETERMINATION OF THE SPECIES. A perpendicular in this case is unnecessary. 1. If BC + (AC acute), BC + (AC obtuse) and ♣ + B, be each of the same species, with respect to 180°, AC is ambiguous; but, if only two of these sums be of the same kind, that value of AC must universally be taken which agrees with the sum of the angles; in all such cases ac is not ambiguous. SOLUTION. Sine A sine BC::sine B sine AC-80°.19', Because BC+(AC acute), BC+(AC obtuse) and A+B, are each less than 180°, AC is ambiguous, being either 80°.19', or 99°.41'. BY FORMULA II. page 201. Log sine b=(log sine B+ log sine a)-log sine a=80°.19′. This example is ambiguous, vide Table II. page 207. Answer. Ac=57°.30′ acute, and not ambiguous. (C) CASE V. Given two angles, and a side opposite to one of them, to find the side adjacent to these angles. Given The ZA=51°.30' The B=59°16′ Required the side ab. DETERMINATION OF THE SPECIES. 1. DC is of the same species as the A. The species of AD cannot be determined. DB is acute, or obtuse, accord A A a ing as BC and DC are of the same, or of different species. 2. When the angles A and B are of the same species; if DB+ (AD acute), and DB+(AD obtuse), be each less than 180°, AB is ambiguous; but if only one of these sums be less than 180°, that sum is the true value of AB, not ambiguous. When the angles A and B are of different species; if both AD acute, and AD obtuse, be greater than DB, then AB is ambiguous; but if only one of them be greater, that value diminished by DB leaves AB not ambiguous. 1. In the triangle BDC, Thus, rad x cos / B SOLUTION. find the segment DB. * Cot BC X tang DB. Cot BC rad::cos B: tang DB=46°.7′ acute. * 2. In the triangle ADC, rad x sine AD Cot LAX tang DC. In the triangle BDC, rad × sine DB=cot B x tang DC. cot B cot/A:: sine DB sine AD=74°.40′, or 105°.20', Because A and B are of the same species, and DB +(AD acute), DB + (AD obtuse), are each less than 180°, AB is ambiguous, viz. ABAD+DB=120°.47', or 151°.27'. Log tang BY FORMULA III. page 202. (log cos Blog tang a)-10=46°.7' acute. Log sine (c-4)=(log sine + log tang B)-log tang A=74°.39′. Then (c-4)+4=74°.39′+46°.7′=120°.46′AB. This example is ambiguous, though not shewn by the formula, vide Table II. page 207. The Given The EXAMPLE II. LA=126°.37' ZB 48°.30 Required the side ab. The side BC115°.20' Answer. DB 54°.27' acute, AD 43°.6', or 136°.54', and AB=82°.26', not ambiguous. |