Now the angle DAE in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to AE, is also perpendicular to AD : therefore AE is the tangent of CDA the half sum; and DF, the tangent of DAB, the half difference of the angles to the same radius AD, by the definition of a tangent. But the tangents A E, DF being parallel, it will be as BE : BD:: AE : D F ; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles to the tangent of half their difference. It is to be observed, that in the third term of the proportion the cotangent of half the given angle may be used instead of the tangent of the half sum of the unknown angles. c Example. In the plane triangle ABC (fig. 400.), Let AB = 345 ft. AC = 174-07 ft. A. A = 37° 20'. A Fig. ***. b Now, the side AB being 345 From 1809 OO The side AC 174-07 Take Z A 37 20 Their sum is 519-07 Sum of C and B 142 40 Their difference 170.93 Half sum of do. 7I 20 As the sum of their sides AB, AC = 519-07 - Log 2-715226 1052. Theohe M III. When the three sides of a triangle are giren. Let fall a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles, the proportion will be— As the base or sum of the segments Is to the sum of the other two sides, So is the difference of those sides To the difference of the segments of the base. Then take half the difference of these segments, and add it to the half sum, or the half base, for the greater segment; and for the lesser segment subtract it. Thus, in each of the two right-angled triangles there will be known two sides and the angle opposite to one of them, whence, by the first theorem, the other angles will be found. For the rectangle under the sum and difference of the two sides is equal to the rectangle under the sum and difference of the two segments. Therefore, forming the sides of these rectangles into a proportion, their sums and differences will be found proportional. So that the three angles are as follow, viz. 4. A 27° 4'; a B 37°20'; 1 C 115° 36. For AB being the given leg in the right-angled triangle ABC, from the centre A with any assumed radius AD describe an arc DE, and draw DF perpendicular to AB, or parallel to BC. Now, from the definitions, DF is the tangent and AF the secant of the arc DE, or of the angle A, f which is measured by that arc to the radius AD. Then, because of the E. parallels BC, DF, we have AD : AB::DF : BC, and :: AF : AC, which is the same as the theorem expresses in words. Note. Radius is equal to the sine of 90°, or the tangent of 45°, and is A D B expressed by 1 in a table of natural sines, or by 10 in logarithmic sines. Fig. 402. Example 1. In the right-angled triangle ABC, Note. There is another mode for right-angled triangles, which is as follows: — ABC being such a triangle, make a leg AB radius; or, in other words, from the centre A and distance AB describe an arc BF. It is evident that the other leg BC will represent the tangent and the hypothenuse AC the secant of the arc BF or of the angle A. In like manner, if BC be taken for radius, the other leg AB represents the tangent, and the hypothenuse AC the secant of the arc BG or angle C. If the hypothenuse be made radius, then each leg will represent the sine of its opposite angle; namely, the leg AB the sine of the arc AE or angle C, and the leg BC the sine of the arc CD or angle A. Then the general rule for all such cases is, that the sides of the triangle bear to each other the same proportion as the parts which they represent. This method is called making every side radius. 1054. If two sides of a right-angled triangle are given to find the third side, that may be found by the property of the squares of the sides (Geom. Prop. 32. ; viz. That the square of the hypothemuse or longest side is equal to both the squares of the two other sides together). Thus, if the longest side be sought, it is equal to the square root of the sum of the two shorter sides; and to find one of the shorter sides, subtract one square from the other, and extract the square root of the remainder. loss. The application of the foregoing theorems in the cases of measuring heights and distances will be obvious. It is, however, to be observed, that where we have to find the length of inaccessible lines, we must employ a line or base which can be measured, and, by means of angles, which will be furnished by the use of instruments, calculate the lengths of the other lines. Fig. 405. Sect. V. conic sections. 1056. The conic sections, in geometry, are those lines formed by the intersections of a plane with the surface of a cone, and which assume different forms and acquire different properties, according to the several directions of such plane in respect of the axis of the cone. Their species are five in number. 1057. DEFINItions.—1. A plane passing through the vertex of a cone meeting the plane 10. I 1. 12. 1058. The primary axis of an ellipsis is called the major axis, as AB (fig. 407.); and a straight line DE drawn through its centre perpendicular to it, and terminated at each extremity by the curve, is called the minor axis. 1059. A straight line VQ drawn through the centre and terminated at each extremity by the curve is called a diameter. Hence the two axes are also diameters. . If a cone be cut by a plane parallel to the . If the plane of a conic section be cut by . If the line of the axis be cut in two points by the conic surface, . If a straight line be drawn in a conic section perpendicular to . The abscissa of an ordinate is that portion of the line of axis of the base or of the base produced is directing plane, the section is called a another plane at right angles passing . The point or points in which the line of the axis is cut by the conic surface is or are called the verter or vertices of the conic section. Thus the or by the surfaces of the two opposite cones, the portion of the line of the axis so as to meet the curve, such straight line Fig. 406. contained between the vertex and an ordinate to that line of , If the primary axis be bisected, the bisecting point is called the centre of the conic section. of THE Ellipsis. 1060. The extremities of a diameter which terminate in the curve are called the vertices of that diameter. Thus the points V and Q are the vertices of the diameter VQ. 1061. A straight line drawn from any point of a diameter parallel to a tangent at either extremity of the diameter to meet the curves is called an ordinate to the two abscissas. Thus PM, being parallel to a tangent at V, is an ordinate to the two abscissas VP, PQ. 1062. If a diameter be drawn through the centre parallel to a tangent at the extremity of another diameter, these two diameters are called conjugate diameters. Thus VQ and RS are conjugate diameters. 1063. A third proportional to any diameter and its conjugate is called the parameter or latus rectum. 1064. The points in the axis where the ordinate is equal to the semi-parameter are called the foci. 1065. ThroREM. I. In the ellipsis the squares of the ordinates of an aris are to each other as the rectangles of their abscissas. Let AVB (fig. 408.) be a plane passing through the axis of the cone, and AEB another section of the cone perpendicular to the plane of the former ; AB the axis of the elliptic section, and PM, HI ordinates perpendicular to it; then it will be PM2 : HIQ:: AP × P B : AH x HB. For through the ordinates PM, HI draw the circular sections KML, MIN parallel to the base of the cone, having KL, MN for their diameters, to which PM, HI are ordinates as well as to the axis of the ellipse. Now, in the similar triangles APL, AHN, AP : PL::AH : HN, And in BPK, BHM, BP: PK::BH : HM. Taking the rectangles of the corresponding terms, AP x BP : PL x PK::AH x BH : HN × HM. By the property of the circle, PL x PK= PM2 and HN × HM = HIQ. Therefore, AP × B P : PM2::AH x IIB : HIQ, or PM2 : HIQ::AP x BP : AH x HB. Coroll. 1. If C be the centre of the figure, AP × PB = CA*–CP2, and AH x HB = CAQ — CH2. Therefore PM2 : HIQ:: CAQ– CP2 : CA2–CH2. For AP=CA – CP, and PB = CA + CP; consequently AP × PB = (CA–CP) (CA + CP)= CA*–CP2; and in the same manner it is evident that AH x HB = (CA + CH)(CA–CH)= CA2–CH2. Coroll. 2. If the point P coincide with the middle point C of the semi-major axis, PM will become equal to CE, and CP will vanish ; we shall therefore have Coroll. 1. If a circle be described on each axis as a diameter, one being inscribed within the ellipse, and the other circumscribed about it, then an ordinate in the circle will be to the corresponding ordinate in the ellipsis as the axis belonging to this ordinate is to the axis belonging to the other; that is, CA: CE:: PG : PM, In the same manner it may be shown that CE : CA::pg : pm, or, alternately, CA : CE::pM : pg ; therefore, by equality, PG : PM ::pM : pg, or PG : Cp::CP : pg: therefore Cy G is a continued straight line. Coroll. 2. Hence, also, as the ellipsis and circle are made up of the same number of corresponding ordinates, which are all in the same proportion as the two axes, it follows that the area of the whole circle and of the ellipsis, as also of any like parts of them, are in the same ratio, or as the square of the diameter to the rectangle of the two axes; that is, the area of the two circles and of the ellipsis are as the square of each axis and the rectangle of the two; and therefore the ellipsis is a mean proportional between the two circles. Coroll. 3. Draw MQ parallel to GC, meeting ED in Q; then will QM = CG = CA; and let R be the point where QM cuts AB; then, because RMGC is a parallelogram, RM is equal to CG = CE ; and therefore, since QM is equal to CA, half the major axis and RM = CE, half the minor axis QR is the difference of the two semi-axes, and hence we have a method of describing the ellipsis. This is the principle of the trammel, so well known among workmen. If we conceive it to move in the line DE, and the point R in the line AB, while the point M is carried from A, towards E, B, D, until it return to A, the point M will in its progress describe the curve of an ellipsis. 1068. Theoh EM IV. The square of the distance of the foci from the centre of an ellipsis is equal to the difference of the square of the semi-ares. Let AB (fig. 4.12.) be the major axis, C the centre, F the focus, and FG the semi-parameter; then will CE*=CA*–CF*. For draw CE perpendicular to AB, and join FE. By Cor. 2. Th. II., CAQ : CE2:: CA2– CF4 : FG2, and the parameter FG is a third proportional to CA, CE; therefore CA2 : CE*:: CE* : FG%, and as in the two analogies the first, second, and fourth terms are identical, the third terms are equal ; consequently |