CASE 4. Given the three sides, to find the angles. RULE 1. Consider the longest side of the triangle as the base, and on it let fall a perpendicular from the opposite angle. This perpendicular will divide the base into two parts, called segments, and the whole triangle into two right angled triangles. Then, So is the difference of those sides, To the difference of the segments of the base.* * DEMONSTRATION. Let ABC, Fig. 48, be a triangle, and CD be perpendicular upon AB. About Casa centre with the less side BC for a radius, describe a circle, meeting AC produced, in G and E, and AB in F. Then it is evident that AE is equal to the sum of the sides AC, BC, and that AG is equal to their difference; also because CD bisects FB (3.3), it is plain that AF is the difference of the segments of the base; but AB × AF = AE × AG (36.3. cor.); therefore AB: AE :: AG: AF (16.6); that is, the base, is to the sum of the sides, as the difference of the sides, is to the difference of the segments of the base. Cor. If AF be considered the base of the triangle AFC, then will CD be a perpendicular on the base produced; AE will be equal to the sum of the sides AC, FC, and AG will be equal to their difference; also AB will be equal to the sum of the segments AD, FD. But by the preceding demonstration and (16.5), AF: AE :: AG: AB; hence when the perpendicular falls without the triangle, the base, is to the sum of the sides, as the difference of the sides, is to the sum of the segments of the base. A rule might, therefore, be given, making either side of a triangle, the base; and such a rule would be rather more convenient, in some cases, than the one above: but then, on account of the perpendicular, sometimes falling within and sometimes without the triangle, it To half the base, add half the difference of the segments and the sum will be the greater segment; also from half the base, subtract half the difference of the segments, and the remainder will be the less segment. Then, in each of the two right angled triangles, there will be known two sides, and an angle opposite to one of them; consequently the other angles may be found by Case 2d. EXAMPLES. 1. In the triangle ABC, are given AB = 426, AC = 365, and BC= 230; required the angles. By Construction, Fig. 49. 230, Draw AB = 426; with AC = 365 in the dividers, and one foot in A, describe an arc, and with BC = and one foot in B describe another arc, cutting the former in C; join AC, BC, and ABC will be the triangle required. The angles measured by a scale of chords, will be A = 32° 39', B = 58° 56′ and C = 88° 25'. By Gunter's Scale. Extend the compasses from 426 to 595 on the line of numbers, that extent will reach on the same line from 135 to 188.6 the difference of the segments of the base. Whence the segments of the base are found as before. To extend the other proportions, proceed as directed in Case 2d. 2. In a triangle ABC, there are given AB = 64, AC = 47, and BC= 34; required the angles. Ans. Angle A = 31° 9′, B = 45° 38', and C = 103° 13'. 3. In a triangle ABC, are given AC = 88, AB = 108, and BC= 54, to find the angles. Ans. Angle A = 29° 49', B = 54° 7', and C 96o 4'. RULE 2 Add together the arithmetical complements of the lo garithms of the two sides containing the required angle, the logarithm of the half sum of the three sides, and the logarithm 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 cosine of half the required angle.* * DEMONSTRATION. Let ABC, be a triangle of which the side AB, is greater than AC: make AD = AC, join DC, bisect it in E, and join AE; draw EH parallel and equal to CB; join HB and produce it to meet AE produced in G. 1 Now in the triangles AED, AEC, all the sides of the one are equal to the sides of the other, each to each; therefore (8.1) the angle EAD EAC, and AED AEC; consequently AED is a = = D EXAMPLES. 1. In the triangle ABC, are given AB = 426, AC = 365, and BC= 230; required the angle A. Because EH is equal and parallel to BC, BH is also equal and parallel to EC (33.1); now in the triangles EDF and HBF, the angle EFD= BFH, the angle FED = FHB (29.1) and ED = EC — BH; therefore (26.1) EF = FH, and FD = FB. Again the angle HGE = DEA a right angle; if therefore with the centre F, and radius FH, a circle be described, it will pass through the point G FE = AD + AD + DB 1 1 AD + AB Now 2 AF 2 AD + 2 DF ACAB; therefore AF AC + AB; Also FK — } IK — † EH BC; hence by adding equals to equals, AF + FK = AB+BC, or AK = 1⁄2 (AC + AB + BC); again AI: IK (AC+ AB+ BC) - BC. = = AC + AK But (Dem. to rule, case 1st.) AD: AE :: sin. AED: sin. ADE:: rad. cos. EAD (cos. 1⁄2 BAC). Also, AB: AG :: sin. AGB: sin. ABG:: rad. cos. BAG (cos. BAC). Hence (C. 6.) AB × AD:: AG x AE:: rad.: (cos.BAC). |