sufficient to solve all the cases of Plane Trigonometry Of the six parts which make up a plane triangle, three must be given, and at least one of these a side, before the others can be determined. If the three angles only are given, it is plain, that an indefinite number of similar triangles may be constructed, the angles of which shall be respectively equal to the angles that are given, and therefore, the sides could not be determined. Assuming, with this restriction, any three parts of a triangle as given, one of the four following cases will al ways be presented. I. When two angles and a side are given. II. When two sides and an opposite angle are given. CASE I. When two angles and a side are given. 26. Add the given angles together, and subtract their sum from 180 degrees. The remaining parts of the triangle can then be found by Theorem I. EXAMPLES. 1. In a plane triangle, ABC, there are given the angle A = 58° 07′, the angle B=22° 37', and the side AB=408 yards. Required the other parts. B GEOMETRICALLY. 27. Draw an indefinite straight line, AB, and from the scale of equal parts lay off AB equal to 408. Then, at A, lay off an angle equal to 58° 07', and at B an angle equal to 22° 37', and draw the lines AC and BC: then will ABC be the triangle required. C The angle may be measured with the protractor (see page 270), and when so measured, will be found equal to 99° 16'. The sides AC and BC may be measured by referring them to the scale of equal parts (see page 268). We shall find AC=158.9 and BC=351 yards. ceeds 90°, we use the supplement 80° 44'. BC 351.024 (after rejecting 10) 2.545337. REMARK. The logarithm of the fourth term of a proportion is obtained by adding the logarithm of the second terin to that of the third, and subtracting from their sum the logarithm of the first term. But to subtract the first term is the same as to add its arithmetical complement and reject 10 from the sum (Int. Art. 13): hence, the arithmetical complement of the logarithm of the first term added to the logarithms of the second and third terms, miñus ten, will give the logarithm of the fourth term. 2. In a triangle ABC, there are given A= 38° 25', B= 57° 42', and AB=400: required the remaining parts. Ans. C=83° 53', BC=249.974, AC=340.04. CASE II. When two sides and an opposite angle are given. 28. In a plane triangle, ABC, there are given AC=216, CB=117, the angle A 22° 37', to find the other parts. = GEOMETRICALLY. B 29. Draw an indefinite right line ABB': from any point, as A, draw AC, making BAC-22° 37', and make AC=216. With C as a centre, and a radius equal to 117, the other given side, describe the arc B'B; draw B'C and BC: then will either of the triangles ABC or AB'C, answer all the conditions of the question. : sin B 45° 13′ 55′′, or ABC 134° 46′ 05′′ 9.851236. The ambiguity in this, and similar examples, arises in consequence of the first proportion being true for either of the angles ABC, or ABC, which are supplements of each other, and therefore, have the same sine (Art. 13). As long as the two triangles exist, the ambiguity will continue. But if the side CB, opposite the given angle, is greater than AC, the arc BB' will cut the line ABB', on the same side of the point A, in but one point, and then there will be only one triangle answering the conditions. If the side CB is equal to the perpendicular Cl, the arc BB' will be tangent to ABB, and in this case also there will be but one triangle. When CB is less than the perpendicular Cd, the arc BB' will not intersect the base ABB and in that case, no triangle can be formed, or it will be impossible to fulfil the conditions of the problem. 2. Given two sides of a triangle 50 and 40 respectively, and the angle opposite the latter equal to 32°: required the remaining parts of the triangle. Ans. If the angle opposite the side 50 is acute, it is equal to 41° 28' 59"; the third angle is then equal to 106° 31′ 01′′, and the third side to 72.368. If the angle opposite the side 50 is obtuse, it is equal to 188° 31′1′′, the third angle to 9° 28′ 59′′, and the remaining side to 12.436. CASE III. When the two sides and their included angle are given. 30. Let ABC be a triangle; AB, BC, the given sides, and B the given angle. Since B is known, we can find the sum of the two other angles for B A+ C=180° - B, and, }(A + C') = (180° — B). A We next find half the difference of the angles A and by Theorem II., viz., BC+BA : BC-BA :: tan }(A+0) : tan }(A — C'), in which we consider BC greater than BA, and therefore A is greater than C; since the greater angle must be op posite the greater side. Having found half the difference of A and C, by add ing it to the half sum, (4+ C), we obtain the greater angle, and by subtracting it from half the sum, we obtain the less. That is, Having found the angles A and C, the third side AU may be found by the proportion, EXAMPLES. 1. In the triangle ABC, let BC=540, AB=450, and the included angle B=80°: required the remaining parts GEOMETRICALLY. and from any At B make the 31. Draw an indefinite right line BC, point, as B, lay off a distance BC=540. angle CBA 80°: draw BA, and make the distance BA=450; draw AC; then will ABC be the required triangle. TRIGONOMETRICALLY. BC+BA=540+ 450 = 990; and BC-BA=540-450 = 90. A+ C=180° - B= 180° — 80° = 100°, and therefore, }(A + C) = {(100°) = 50°. - Hence, 50° +6° 11' 56° 11' A; and 50° - 6° 11′ = = 2. Given two sides of a plane triangle, 1686 and 960, and their included angle 128° 04': required the other parts. Ans. Angles, 33° 34′ 39′′; 18° 21′ 21′′; side 2400. CASE IV. 32. Having given the three sides of a plane triangle, to find the angles. |