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 A22° 37', to find the other parts. = TRIGONOMETRICALLY. To find the angle B. 117 ar. comp. 216 22° 37' GEOMETRICALLY. 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. B d BC 7.931814 AC 2.334454 :: sin A 9.584968 : sin B' 45° 13′ 55′′, or ABC 134° 46′ 05′′ 9.851236. B 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 Cd, 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 are 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 138° 31' 01", 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). We next find half the difference of the angles A and C by Theorem IÌ., viz., A BC+BA BC-BA :: tan (A+C): 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 adding 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, }(A + C) + }(A − C')= A, and 3(A + C) — }(A — C') = C. Having found the angles A and C, the third side AC may be found by the proportion, sin A sin B :: BC : A C EXAMPLES. 1. In the triangle ABC, let BC=540, AB=450, and the included angle B=80°: required the remaining parts GEOMETRICALLY. 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+450990; and BC-BA=540-450=90. A+ C = 180° — B = 180° — 80° = 100°, and therefore, (A+C)=(100°) = 50°. sin C : sin B :: AB AC : 450 6° 11' . ar. comp. Hence, 50° 6° 11' 56° 11'=A; and 50° - 6° 11' = 43° 49' C. To find the third side AC. 43° 49' ar comp. 80° and from any At B make the 7.004365 1.954243 10.076187 9.034795. 0.159672 . 9.993351 2.653213 2.806236. 640.082. 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. Let fall a perpendicular from the angle opposite the greater side, dividing the given triangle into two rightangled triangles: then find the difference of the segments of the base by Theorem III. Half this difference being added to half the base, gives the greater segment; and, being subtracted from half the base, gives the less segment. Then, since the greater segment belongs to the right-angled triangle having the greater hypothenuse, we have two sides and the right angle of each of two right-angled triangles, to find the acute angles. 26.6375. 90° 51° 34' 40" D : CD — BD, = 13.275. In the triangle DAC, to find the angle DAC AC 34 ar. comp DC :: sin D : sin DAC 8.468521 1.425493 10.000000 9.894014. In the triangle BAD, to find the angle BAD, AB 25 ar. comp. BD :: sin D : sin BAD 13.3625 90° . 32° 18′ 35′′ - = Hence, 90° DAC 90° - 51° 34′ 40′′ and, 90° BAD 90° 32° 18′ 35′′ and, BAD+DAC=-51° 34′ 40′′ +32° 18′ 35′′ = 83° 53′ 15" A. == = EXAMPLES. 8.602060 1.125887 10.000000 9.727947. 38° 25′ 20′′ C, 57° 41′ 25′′ = B, 2. In a triangle, of which the sides are 4, 5, and 6, what are the angles? Ans. 41° 24′ 35"; 55° 46' 16"; and 82° 49' 09". SOLUTION OF RIGHT-ANGLED TRIANGLES. 34. The unknown parts of a right-angled triangle may be found by either of the four last cases; or, if two of the sides are given, by means of the property that the square of the hypothenuse is equivalent to the sum of the squares of the two other sides. Or the parts may be found by Theorems IV. and V. 1. In a right-angled triangle BAC, there are given the hypothenuse BC=250, and the base AC= 240 required the other parts. Ans. B 73° 44′ 23′′; C=16° 15′ 37′′; AB=70.0003. B 2. In a right-angled triangle BAC, there are given AC 384, and B= 53° 08′: required the remaining parts. Ans. AB 287.96; BC= 479.979; C=36° 52'. |