(D) CASE VI. Given two angles, and a side opposite to one of them, to find the third angle. Given The ZA 51°.30' The B=59°.16′Required the angle c. The side BC=63°.50' DETERMINATION OF THE SPECIES. 1. DC is of the same species as the A. The species of the ACD cannot be determined. The BCD is acute, or obtuse, according as the side BC C A A BA is of the same, or of different species with DC. D 2. When the angles A and B are of the same species; if BCD +(ACD acute), and BCD+(ACD obtuse), be each less than 180°, ACB is ambiguous; but if only one of these sums be less than 180°, that sum is the true value of ACB, not ambiguous. When the angles A and B are of different species; if both ACD acute, and ACD obtuse, be greater than BCD, then ACB is ambiguous; but if only one of them be greater, that value diminished by BCD leaves ACB, not ambiguous. SOLUTION. 1. In the triangle BDC, find the BCD. * Thus, rad x cos BC=cot / BX cot BCD. * Cot Brad::cos BC cot BCD=53°.26' acute. 2. In the triangle ADC, rad × cos ▲ A=COS DC × sine Acd. In the triangle BDC, rad × cos / B⇒COS DC × sine BCD. Hence, cos 4B COS ZA:: sine BCD sine ACD=78°.4', or101°.56'. Because A and B are of the same species, and BCD + (ACD acute), BCD+(ACD obtuse), are each less than 180°; ACB is ambiguous, 1310.30', or 155°.22'. BY FORMULA III. page 202. Log cot (log cos a+ log tang B)—10—53°.26′ acute. Log sine (c-p)(log cos A+ log sine 4)-log cos B=78°.3'. Then (c-)+4=78°.3′+53°.26′131°.29′ This example is ambiguous, though not shewn by the formula, vide Table II. page 207. Answer. The BCD64°.11' acute, the ACD = 54°.8', or 115°.52′ ambiguous, and the ACB=61°.41′, not ambiguous. (E) CASE VII. Given two sides, and an angle contained between them, to find an opposite angle. Given The side AC 80°.19' The side AB=120°.47'Required the B. DETERMINATION OF THE SPECIES. 1. AD is acute, or obtuse, according as AC is of the same, or of different species with the LA. A D 2. The B is of the same, or of different species with the A, according as AD is less, or greater than AB. SOLUTION. 1. In the triangle ADC, find AD. * Thus, rad x cos Acot AC × tang AD. CoL AC rad::cos. A tang AD74.40′ acute. * 2. In the triangle BDC, rad x sine DB=tang DC X cot / B. In the triangle ADC, rad x sine AD tang DC X cot A. Hence, sine AD sine DB:: cota: cot B59°.16′ acute. Because AD is less than AB, the B is of the same species with the A, and consequently it is acute. Log tang BY FORMULA VII. page 203. (log cos a+log tang b)-10=74°.40′ acute. Log cot B=(log cot a+log sine c~)-sine 59°.16'. EXAMPLE II. Given The side AC 57°.30' The side AB 82°.27'Required the B. Answer. AD=136°.54′ obtuse, DB54°.27′, and the B= 48°.30' acute. (F) CASE VIII. Given two sides, and the angle contained between them, to find the third side. The side AC Given 80°.19′ The side AB=120°.47'Required the side BC. DETERMINATION OF THE SPECIES. 1. AD is acute or obtuse, according as AC is of the same, or of different species with the LA. d لا D 2. CD is of the same species as the A; therefore, according as the A and DB are of the same, or of different species, BC is acute or obtuse. SOLUTION. 1. In the triangle ADC, find AD. Thus, rad x cos ▲ A=cot ACX tang AD. Cot AC rad::cos A: tang AD=74°.40′ acute. * 2. In the triangle BDC, rad × cos BC=cos DC × COS DB. Log tang BY FORMULA I. page 203. (log cos a+log tang c)-10=133°.45′ obtuse. Log cos a=(log cos c+log cos bp)-cos 463°.50. Log tang OR, (log cos a+ log tang b)-10-74°.40' acute. Log cos a=(log cos b+log cos c~)-cos 63°.50'. LA=126°.37' Answer. AD 136°.54' obtuse, AD-AB-DB=54°.27' acute, and BC 115°.20' obtuse. (G) CASE IX. Given two angles, and the side adjacent to both of them, to find a side opposite to one of the given angles. The LA 51°.30' Given The 1. The ACB=1310.30 Required the side BC. The side ac 80°.19′. DETERMINATION OF THE SPECIES. ACD is acute or ob tuse, according as the A is of the same, or of different species with the side Ac. c D 2. The A is of the same species as CD; therefore, according as the ▲▲ and the BCD are of the same, or of different species, BC is acute or obtuse. SOLUTION. 1. In the triangle ADC, find the ACD. * Thus, rad x cos AC=cot A× cot ▲ ACD. Cot Arad::cos AC cot ACD=78°.4′ acute. 2. In the triangle BDC, rad x cos BCD tang DC × cot BC. In the triangle ADC, rad × cos ACD=tang DC X cot AC. Hence, cos / ACD COS/BCD::cot AC: cotBC=63°.50'acute. Because the A is of the same species as the BCD, the side BC is acute. Log cot BY FORMULA VII. page 204. (log tang A+ log cos b)-10=78.4' acute. Log cot a=(log cot b+log cos c~)-log cos 463°.50′. 64°.11' acute, and the side BC 115°.20′ obtuse. (H) CASE X. Given two angles, and the sides adjacent to both of them, to find the other angle. The A = 51°.30' Given The ACB = 131°.30 Required the angle B. 1. The DETERMINATION OF THE SPECIES. ACD is acute or ob tuse, according as the A is of the same, or of different species with the side AC. D B 2. The angles A and B are of the same, or of different species, according as the ACD is less or greater than the ACB. SOLUTION. 1. In the triangle ADC, find the ▲ ACD. * Thus, rad x cos AC-cot A× cot ▲ ACD. Cot Arad::cos AC: cot ACD=78°.4' acute. * 2. In the triangle BDC, rad x cos BCOS DC X sine BCD. In the triangle ADC, rad × cos A⇒COS DG X Sine ACD. Hence, sine / ACD: sine BCD::COS A: Cos / B59°.16'acute. Because the ACD is less than the Z ACB, the B is of the same species as the▲ A, viz. acute. Log cot BY FORMULA II. page 205. (log tang a+log cos b)-10=78°.4' acute. Log cos B=(log cos A+ log sine c~)-log sine =59'.16". EXAMPLE II. The LA 126°.37' Given The 4 ACB The side AC 61°.41' Required the B. Answer. The ACD 125°.52′ obtuse, ACD LACB= (I) CASE XI. Given the three sides, to find the angles. The side AC 80°.19′′ Given The side BC= 63°.50 Required the angles A, B, and c. The sum of the sides AC and BC being less than 180°, the perpendicular falls nearest to the less side BC; therefore if it falls without the triangle, the rule determines ED in fig. 2. tang AB tang 60°.23'.30" = tang (AC+BC)=tang 72°.4.30" tang (AC-BC)=tang 8°.14'.30′′= Here ED is less than half the base, therefore the perpendicular falls within the triangle, and consequently the angles A and B are acute. 4 ED B 10.24544 The segment AD≈ AB+ED=74°.40′, and DB=46°.7. SECONDLY. In the triangle ADC, the segment AD and the hypothenuse AC are given to find the A=51°.31', and the ▲ ACD. In the triangle BÓC, the segment BD and the hypothenuse BC are given to find the ▲ B=59°.17', and the BCD. Lastly, ACD+ 2 BCD = 2 ACB=131°.30′. OR THUS, Let BC be considered as the base. Then, the sum of the sides AB and Ac being greater than 180°, the perpendicular falls nearest to the greater side AB; R |