Or the side e might be found from the equation Ex. 2. In the right angled triangle BCA, there are given, sideb=384 yards, and the angle B-53° 8': required the other -parts. Note. When the logarithm whose arithmetical complement is to be used, exceeds 10, take the arithmetical complement with reference to 20 and reject 20 from the sum. To find the hypothenuse a. R sin B: a b (Theorem I.). Hence, As sin B 53° 8' ar. comp. Ex. 3. In the right angled triangle BAC, there are given, side c=195, angle B=47° 55', required the other parts. Ans. Angle C=42° 05′, a=290.953, b=215.937. SOLUTION OF RECTILINEAL TRIANGLES IN GENERAL. Let A, B, C be the three angles of a proposed rectilineal triangle; a, b, c, the sides which are respectively opposite them; the different problems which may occur in determining three of these quantities by means of the other three, will all be reducible to the four following cases. CASE I. Given a side and two angles of a triangle, to find the remaining parts. First, subtract the sum of the two angles from two right angles, the remainder will be the third angle. The remaining sides can then be found by Theorem III. I. In the triangle ABC, there are given the angle A=58° 07', the angle B=22° 37', and the side c=408 yards: required the remaining angle and the two other sides. This angle being greater than 90° its sine is found by taking that of its supplement 80° 44'. 2. In a triangle ABC, there are given the angle A=38° 25′ B=57° 42', and the side c=400: required the remaining parts. Ans. Angle C-83° 53′, side a=249.974, side b=340.04. CASE II. Given two sides of a triangle, and an angle opposite one of them, to find the third side and the two remaining angles. X* The ambiguity in this, and similar examples, arises in consequence of the first proportion being true for both the triangles ACB, ACB'. As long as the two triangles exist, the ambiguity will continue. But if the side CB, opposite the given angle, be greater than AC, the arc BB' will cut the line ABB', on the same side of the point A, but in one point, and then there will be but one triangle answering the conditions. If the side CB be 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 arc BB' will not intersect the base ABB', and in that case there will be no triangle, or the conditions are impossible. 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 be 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 be obtuse, it is equal to 138° 31′ 01′′, the third angle to 9° 28′ 59′′, and the remaining side to 12.436. 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 B C compute the half difference of these two angles by the proportion (Theorem V.), 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) 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, c-a-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 in which p represents the half sum of the three sides. Hence, (p R2 (pb) bc sin2A=R2 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. Angle A=38° 25′ 18′′. 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′′. Suppose the height of a building AB were required, the foot of it being accessible. |