are given, AC-216; CB=117, the angle A=22° 37', to find the other parts. 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; either of the triangles ABC, or AB'C will answer all the conditions of the question. To find the sides and angles by computation. To find the angle B. Remainder 112° 09′ 05′′ or 22° 36′ 55′′, being ACB' and ACB. To find AB or AB', reverse the terms of the former propor The ambiguity in this, and similar examples, arises in consequence of the first proportion being true for both the triangles ACB and ACB'. As long as the two triangles remain, 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. To determine the value of Cd in terms of the given parts, we have the proportion sin. 90°, or R: AC sin. A: Cd, AC sin. A or Cd=3 or log. Cd=log. AC+log. sin. A—10. " R So that, when the logarithm of the opposide side CB is equal to log. AC+log. sin. A-10, there is one solution; when less, the problem is impossible; when greater, but less than the logarithm of AC, two solutions; and when greater than the logarithm of AC, but one. Ex. 3.--Given two angles of a plane triangle 22° 37′ and 134* 46', and the contained side 351: required the remaining parts. Answer.-Angle 22° 37'; sides 351 and 648. Example 4.-Given two sides of a plane triangle 50 and 40 respectively, and the angle opposite the latter equal to 32°; to determine the triangle. Answer-If the angle opposite to the side 50 be acute, it is = 41° 28', the third angle = 106° 32', and the remaining 106°32′, side = 72.36. If the angle opposite to side 50 be obtuse, it is 138° 32', the other angle = 9° 28', and the side = 12.415. CASE II. 60. When two sides and the included angle are given. The solution is made by Art. 44. Take the given angle from 180°, the remainder is the sum of the other two angles, which being divided by 2, gives half their sum. Then find half their difference: their half difference being added to half their sum, gives the greater angle, and being subtracted from it, gives the less. And as the greater angle is opposite the greater side, and the less angle opposite the less side, the three angles and two sides of the triangle become known: the third side may then be known by Case I. 62. Example 1.-Let there be given in any plane triangle ABC, AC=450, BC=540, and the included angle C=80°, to find the other angles and the remaining side. Let a+bs, and a-b-d: then by adding 2as+d, or a=s+d; and Example 2.-Given two sides of a plane triangle, 1686 and 960, and their included angle 128° 4′: to find the other parts. Answer.-Angles 33° 34′ 39′′, 18° 21′ 21′′, side 2400.03. CASE III. 62. When the three sides of a plane triangle are given, to find the angles. Let fall a perpendicular from the angle opposite the greater side, dividing the given triangle into two right-angled triangles; then find the difference of the segments by Art. 45. Half the 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 greatest hypothenuse, we have the sides and right angle of two right-angled triangles, to find the acute angles. Example 1.-The sides of a plane triangle (Pl. I. Fig. 4), are AC-40, AB=34, and BC-25. Required the angles. As AC AB+BC :: AB-BC: AD-CD, : And 13.3625-CD. = 9.727947 'Hence 90°-51° 34′ 40′′ 38° 25′ 20′′A; 90°-32° 18′ 35′′ =57° 41′ 25′′ C; and 51° 34′ 40′′+32° 18′ 35′′ 83° 53′ 15′′ =ABC. Example 2.-When the sides are 4, 5, and 6, what are the angles? Answer.-41° 24′ 35′′, 55° 46′ 16", and 82° 49′ 9′′. SOLUTION OF RIGHT-ANGLED TRIANGLES. 63. The unknown parts of a right-angled triangle may be found by either of the three last cases: or, if two of the sides be given, by means of the property, that the square of the hypothenuse is equal to the sum of the squares of the other two sides; or the parts may be found by Art. 46. Example 1.-In the right-angled triangle ABC, there are given the hypothenuse AC=25, and the base AB=24. Required the other parts. As AC, 25 To sin. B, 90° 1.397940 As radius 10.000000 To AB, 24 So is AB, 24 To sin. C, 73° 44′23′′ 9.982271 10.000000 1.380211 9.464889 40.845100 1.380211 So is tang. A, 16° 90°-73° 44′ 23′′=16° 15′ 37′′ To BC, 7.00004 =A. Example 2.-In a right-angled triangle, there are given one leg equal to 384, and its opposite angle 53° 8', to find the other leg and the hypothenuse. Answer.-Leg 280, hypothenuse 480. Example 3.-In a right-angled triangle, are given the base 195, its adjacent angle 47° 55' to find the rest. ELEMENTS OF SURVEYING. CHAPTER I. DEFINITIONS AND INTRODUCTORY REMARKS. 64. SURVEYING, in its most extensive signification, comprises all the operations necessary for determining the area, or content of any portion of the earth's surface, the lengths and direction of the bounding lines, and their accurate delineation on paper. 65. The earth being spherical, its surface is curved, and every line traced accurately on the surface, is also curved. 66. If large portions of the earth's surface are to be measured, such as states and territories, its curvature must be taken into account; and very material errors will arise if it be neglected. This method of measurement and computation is called Geodesic Surveying. 67. The radius of the earth, however, being large, the curvature of its surface is small, and when the measurement is limited to inconsiderable portions, the error is not sensible in supposing the surface a plane. This method of measurement and computation, is called Plane Surveying, which will be alone treated of in these Elements. 68. If at any point of the surface of the earth, a plane be drawn perpendicular to the radius passing through this point, such plane is tangent to the surface, and is called a horizontal plane. All planes parallel to such a plane, are also named |