In the last problem the given angle is acute, and the side opposite to it less than the other given side, therefore the angle C may be acute or obtuse; but the side and angle answering to the acute value of C has been already found. Now it remains to find the side and angle of the triangle answering to the obtuse value of C, which is thus found : The acute value of C, found in the foregoing calculation, is 60° 31', consequently its obtuse value is 180°-60° 31'=119° 29'; then 119° 29′+46° 30′, taken from 180°, gives 14° 1′= to the remaining angle ABC (pl. 14, fig. 5). To find the side AC, answering to the obtuse value of the angle C. In the triangle ABC there is the angle A 46° 30′, AB 230, and the angle B 37° 30′ given, to find AC and BC. 1st. By Construction. Draw a blank line, upon which set AB 230 from a scale of equal parts; at the point A of the line AB make an angle of 46° 30', by a blank line; and at the point B of the line AB make an angle of 37° 30', by another blank line; the intersection of those lines gives the point C: then the triangle ABC is constructed. Measure AC and BC from the same scale of equal parts that AB was taken, and you have the answer required. 2d. By Calculation. By cor. 1, theo. 5, sect 4, 180° and BC. the sum of the angles A A 46° 30' B 37 30 180°-84° 00′=96° 00′=C. By def. 27, sect. 4. The sine of 96° = the sine of 84°, which is the supplement thereof; therefore, instead of the sine of 96°, look in the tables for the sine of 84°. Extend from 84° (which is the supplement of 96°) to 46° 30′ on the sines; that distance will reach from 230 to 168, on the line of numbers, for BC. Extend from 84° to 37° 30′, on the sines; that extent will reach from 230 to 141, on the line of numbers, for AC. CASE III. Two sides and a contained angle given, to find the other angles and side. PL. 5. fig. 16. In the triangle ABC, there is AB 240, the angle A 36° 40′, and AC 180 given, to find the angles C and B and the side BC. 1st. By Construction. Draw a blank line, on which, from a scale of equal parts, lay AB 240; at the point A of the line AB make an angle of 36° 40', by a blank line; on which from A lay AC 180, from the same scale of equal parts; measure the angles C and B and the side BC, as before, and you have the answers required. 2d. By Calculation. By cor. 1, theo. 5, sect. 4, 180° the angle A 36° 40′= 143° 20′, the sum of the angles C and B: therefore, half of 143° 20′ will be half the sum of the two required angles C and B. By theo. 2 of this sect. As the sum of the two sides AB and AC=420 is to their difference, So is the tangent of half the sum of = 60 } =71° 40' the two unknown angles C and B By theo. 4. To half the sum of the angles C and B=71° 40′ The sum is the greatest angle, or ang. C=95 00 Subtract, and you have the least angle, or B=48 20 The angles C and B being found, BC is had as before, by theo. 1 of this sect. Thus, S.B: AC:: S.A: BC. 48° 20' 180:: 36° 40′ : 143.9. 3d. By Gunter's Scale. Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 71° 40′; that distance laid from 45° on the same line will reach to 23° 30′, the half-difference of the required angles. Whence the angles are obtained, as before. The second proportion may be easily extended, from what has been already said. CASE IV. PL. 5. fig. 17. The three sides given, to find the angles. In the triangle ABC, there is given AB 64, AC 47, BC 34; the angles A, B, and C are required. 1st. By Construction. The construction of this triangle must be manifest, from prob. 1, sect. 4. 2d. By Calculation. From the point C let fall the perpendicular CD on the base AB, and it will divide the triangle into two right-angled ones, ADC and CBD, as well as the base AB into the two segments AD and DB. AC 47 BC 34 Sum 81 Difference 13 By theo. 3 of this sect. As the base or the longest side AB is to the sum of the other sides AC and BC, So is the difference of those sides 64 81 13 to the difference of the segments of the base AD, DB, 16.46 Subtract, and their difference will be the least seg-23.77 ment, DB, In the right-angled triangle ADC, there is AC 47 and AD 40.23 given, to find the angle A. This is resolved by case 4 of right-angled plane trigonometry, thus: AD: R:: AC: sec. A 40.23 90° : : 47 : 31° 08′ : Or it may be had by finding the angle ACD, the complement of the angle A, without a secant, thus: 90° AC: R: AD: S.ACD 44 90° 40.23 : 58° 52' : 58° 52′ 31° 08′, the angle A. Then by theo. 1 of this sect. BC: S.A:: AC: S.B 34: 31° 08′ :: 47: 45° 37' By cor. 1, theo. 5, sect. 4, 180°. -the sum of A and B=C. The first proportion is extended on the line of numbers; and it is no matter whether you extend from the first to the third or to the second term, since they are all of the same kind : if you extend to the second, that distance applied to the third will give the fourth; but if you extend from the first to the third, that extent will reach from the second to the fourth.* The methods of extending the other proportions have been already fully treated of. RULE 2. Either of the angles, as A, may be found by adding together the arithmetical complements of the logarithms of the two sides AB, AC, containing the required angle, the log. of the halfsum of the three sides, and the log. of the difference between the half-sum and the side opposite the required angle; then half the sum of these four logarithms will be the logarithmic co-sine of half the required angle.† It is required to find the angle A, in the last problem, by this rule, the sides remaining the same. * The reader is referred to Hutton's Mathematics, vol. ii. New-York edition, for the method of investigating Plane Trigonometry analytically. + The demonstration of this rule is evident from theo. 5, and the nature of logarithms; but in working the proportion by logarithms, we omit |