CASE III. Given two sides of a triangle, with their included angle, to find the third side and the two remaining angles. Let ABC be a triangle, B the given angle, and c and a the given sides. Knowing the angle B, we shall likewise know the sum of the other two angles C+A=180°-B, and their half sum (C+A)=90-B. We shall next A b compute the half difference of these two angles by the tion (Theorem V.), B C propor c+a: c―a :: tang (C+A) or cot B: tang (C—A,) in which we consider c>a and consequently C>A. Having found the half difference, by adding it to the half sum (C+A), we shall have the greater angle C; and by subtracting it from the half-sum, we shall have the smaller angle A. For, C and A being any two quantities, we have always, C=} (C+A) +} (C—A) A} (C+A)—1 (C—A). Knowing the angles C and A to find the third side b, we have the proportion. sin A sin B::a:b : Ex. 1. In the triangle ABC, let a=450, c=540, and the included angle B 80°: required the remaining parts. c+a=990, ca=90, 180°-B=100°C+A. Hence, 50° +6° 11′=56° 11'=C; and 50°--6° 11′ =43° 49′ Ex. 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. Given the three sides of a triangle, to find the angles. We have from Theorem IV. the formula, sin † A—R√( (p—b) (p—c) bc in which p represents the half sum of the three sides. Hence, 2 log. sin A=2 log. R+log. (p—b)+log. (p—c)—log. c— log. b. Ex. 1. In a triangle ABC, let b=40, c==34, and a=25: required the angles. In a similar manner we find the angle B=83° 53′ 18′′ and the angle C=57° 41′ 24′′. Ex. 2. What are the angles of a plane triangle whose sides are, a=60, b=50, and C =40? Ans. 41° 24′ 34′′, 55° 46′ 16′′ and 82° 49' 10". APPLICATIONS. Suppose the height of a building AB were required, the foot of it being accessible. On the ground which we suppose to be horizontal or very nearly so, measure a base AD, neither very great nor very small in comparison with the altitude AB; then at D place the foot of the circle, or whatever be the instrument, with which we are to measure the angle BCE formed by the horizontal line CE parallel to AD, and by the visual ray direct it to the summit of the building. Suppose we find AD or CE-67.84 yards, and the angle BCE 41° 04': in order to find BE, we shall have to solve the right angled triangle BCE, in which the angle C and the adjacent side CE are known. As R ar.-comp. 0.000000 9.940183 1.831486 1.771669 Hence, EB-59.111 yards. To EB add the height of the instrument, which we will suppose to be 1.12 yards, we shall then have the required height AB=60.231 yards. If, in the same triangle BCE it were required to find the hypothenuse, form the proportion log. 0.122660 10.000000 1.831486 1.954146 Note. If only the summit B of the building or place whose height is required were visible, we should determine the distance CE by the method shown in the following example; this distance and the given angle BCE are sufficient for solving the right angled triangle BCE, whose side, increased by the height of the instrument, will be the height required. As sine ABD 40° 05" ar.-comp. - log. 0.191920 9.769913 2.769710 2.731543 If for another inaccessible object C, we have found the angles CAD-35° 15', ADC=119° 32', we shall in like manner find the distance AC-1201.744 yards. 3. To find the distance between two inaccessible objects B and C, we determine AB and AC as in the last example; we shall, at the same time, have the included angle BAC=BAD— DAC. Suppose AB has been found equal to 538.818 yards, AC 1201.744 yards, and the angle BAC 68° 40′ 55′′; to get BC, we must resolve the triangle BAC, in which are known two sides and the included angle. = As AC+AB 1740.562 ar.-comp. log. 6.759311 Is to AC-AB 662.926 2.821465 Now, to find the distance BC make the proportion, As sine B 84° 47′ 51′′ ar.-comp. Is to sine A 68° 40′ 55′′ So is AC 1201.744 log. 0.001793 9.969218 3.079811 3.050822 4. Wanting to know the distance between two inaccessible objects which lie in a direct line from the bottom of a tower of 120 feet in height, the angles of depression are measured, and found to be, of the nearest, 57°; of the most remote, 25° 30′ required the distance between them. Ans. 173.656 feet. 5. In order to find the distance between two trees, A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distance of a third point C from each, was measured, viz. CA=588 feet and CB =672 feet, and also the contained angle ACB=55° 40': required the distance AB. Ans. 592.967 feet. 6. Being on a horizontal plane, and wanting to ascertain the height of a tower, standing on the top of an inaccessible hill, there were measured, the angle of elevation of the top of the hill 40°, and of the top of the tower 51°: then measuring in a direct line 180 feet farther from the hill, the angle of elevation of the top of the tower was 33° 45′ : required the height of the tower. Ans. 83.9983 feet. 7. Wanting to know the horizontal distance between two inaccessible objects A and B, and not finding any station from which both of them could be seen, two points C and D, were chosen, at a distance from each other equal to 200 yards, from the former of which A could be seen, and from the latter B, and at each of the points C and D a staff was set up. From C a distance CF was measured, not in the direction DC, equal to 200 yards, and from D, a distance DE equal to 200 yards, and the following angles were taken, viz. AFC=83° ACF= 54° 31', ACD=53° 30', BDC=156° 25′, BDE=54° 30′, and BED=88° 30': required the distance AB. Ans. 345.46 yards. 8. From a station P there can be seen three objects, A, B and C, whose distances from each other are known, viz. AB= 800, AC=600, and BC=400 yards. There are also measured the horizontal angles, APC=33° 45', BPC=22° 30'. It is required, from these data, to determine the three distances PA, PC and PB. Ans. PA=710.193, PC-1042.522, PB=934.291 yards. |