67. Solution of an Isosceles Triangle. An isosceles triangle is reducible to two right-angled B D triangles, and can be solved if an angle and a side, or if two unequal sides be given in magnitude. If ABC be an isosceles triangle where the sides AB, AC are equal to one another, and consequently the angles EABC, ACB are equal to one another. If AD be drawn perpendicular to BC, it will bisect this side BC at right angles. 1. When an angle is given, the other angles are also at once known, since the three angles of a triangle are two right angles. Then BC = 2BD = 2AB sin 4 = 2AB sin (90°— B), 2 and BC is known from AB, or AB from BC. 2. When two unequal sides are given 68. Area of an Isosceles Triangle. The area of the isosceles triangle is ¦ AD . BC or if a perpendicular BE be drawn upon AC, the area = AC. BE, =AC. AB sin A, = AB2 sin A. 69. Examples for Practice. 1. If the lengths of the sides of an isosceles triangle are 183, 183, and 197 yards, its angles are 65° 6' and 57° 27'. 2. The equal sides of an isosceles triangle being each 200 feet, and the equal angles each 30° 15', the remaining e of the triangle is 345'534 feet in length. CHAPTER IV. OBLIQUE-ANGLED TRIANGLES. 70. We proceed now with triangles wherein it is not a given fact that any angle is a right angle. Such angles are for distinction sometimes called oblique-angled triangles. For their treatment the following two propositions are required. I. Let ABC be a triangle. Since two at least of its angles are acute, let CAB be one of them, and let CBA be either acute as in Figure 1, or obtuse as in Figure 2, or a right angle as in Figure 3. In Figures 1 and 2 draw CD perpendicular to AB, meeting it, produced if necessary, in D. Hence, in any triangle the sines of any two angles have the same ratio which the sides opposite to them have, whether the angles be both acute, or one be obtuse, or one be a right angle. This is expressed by the equations 71. II. If ABC be a triangle wherein A and B are acute angles (Fig. 1, 70), and if from the point C a straight line be drawn perpendicular to AB, meeting it in D, and if AC be greater than BC, to prove that AB = AC-BC AC+CB AD-DB Since BC2—BD2 = AC2—AD2, each being equal to CD2 (Euclid, i. 47), :. AD2-BD2 AC2-BC2, :. (AD+DB)(AD−DB) = (AC+BC)(AC—BC), or AB. (AD—DB) = (AC+BC)(AC−BC). This proposition is expressed sometimes in words, calling the side AB the base of the triangle, that the base has to the sum of the other sides the same ratio which the difference of these sides has to the difference of the segments of the base. 72. If the angles A and B are not both acute, so that the perpendicular CD falls on AB produced (Fig. 2, 70), if AC be greater than BC it may be similarly proved that but this proposition will not at present be called into use. 73. Hence, if the sides of the triangle be given, the lengths of the two segments into which the perpendicular divides the base can be computed. Let the base be 24 feet and the sides 23 and 25 feet 74. Obs. The sides of a triangle ABC usually have their lengths designated by the symbols a, b, c, in the order of the angles to which they are opposite, viz. BC= a, AC = b, AB = C. Then the last propositions may be stated in the forms 75. Methods shall now be given for solving any obliqueangled triangle, by reducing it if need be to two right-angled triangles, and then making its solution result from the solution of these right-angled triangles by the methods already set forth. 76. A triangle has six parts, three angles and three sides. The three angles together make two right angles (Euclid, i. 32), and of the sides no one must be so great as the sum of the other two (Euclid, i. 20). If, in compliance with these conditions, any values be given at random to any three of the parts, it will be found that, with two exceptions, it is always possible to lay down on paper one, and only one, triangle which has these parts. Thus the assignment of three parts in general makes the triangle determinate. The three parts may be : I. Three angles. II. Two angles and one side, which side may be either : (1) the side between the two angles, or (2) a side opposite one of the two angles. III. An angle and two sides, which sides may be either : (1) the sides containing the angle between them, or (2) sides not containing the angle between them. IV. Three sides. Each of these four cases and their sub-cases are now to be considered. 77. I. Let three angles be given, together two right angles. This is one of the two cases where three given parts do not fix and determine the triangle, because an endless number of triangles can be formed all equiangular and similar, any one therefore having the three given parts. In all these triangles the sides have the same ratios among themselves, but there is nothing to fix the actual lengths of any of them. The shape of the triangle is known, but not its magnitude. 78. II. (1) Let two angles be given and the side between them. Let A and B be the angles given; then the angle C = 180° — (A+B) is at once known. Also, let c denote the given side; then a = c sin A sin C' .. log a log c+ log sin A-log sin C log b = = = log c+L sin A–L sin C (43). Thus the sides a and b can be found, and they complete the solution of the triangle. C |