CHAPTER IV. ON THE SOLUTION OF TRIANGLES. 69. A TRIANGLE consists of six parts, namely, three sides and three angles; when three of these parts are given, (except they be the three angles), it will be shewn that the other three can, generally, be determined. The number of degrees in the angles of a triangle will be designated by the letters A, B, C, which are placed at the angular points of the triangle, and the lengths of the sides respectively opposite to the angles A, B, C, by the letters a, b, c. 70. The sines of the angles of a triangle are proportional to the sides respectively opposite. Let ABC be the triangle. Draw CD perpendicular to AB, or AB produced either way. 44 Then sin CAB = sin CAD = = DC CA (With reference to the third figure, see Art. 26.) In like manner, if a perpendicular were let fall from B upon the side opposite to it, it might be proved that, the magnitudes of the lines a, b, c, being represented by the number of units of length they respectively contain; for otherwise sin and a would not be quantities of the same kind, and consequently no ratio could exist between them. 71. Since (Euclid 1. 32), the interior angles of a triangle are together equal to two right angles, we have And if three of the parts of the triangle be given, the remaining three parts may be determined by these three equations. It is necessary however that one, at least, of the given parts be a side, or we have merely the ratios given which a, b, c bear to each other, and their magnitudes cannot be determined, because there are but two equations, for determining the three unknown quantities a, b, c. This also appears from the consideration that an indefinite number of equiangular triangles, of all possible degrees of magnitude, can be formed by drawing lines parallel to the sides of a given triangle. H 72. There is however one case, commonly called "the ambiguous case," in which the equations of the last article are not sufficient to determine the triangle when three of the parts are given. If two sides be given, and an angle opposite to one of them (a, b, A), the triangle can be determined only in the case where the side opposite to the given angle is the greater of the two sides which are given; i. e. when a is greater than b. The equations of the last article become If we can determine B from (ii), C and c are known from (i) and (iii), and the triangle is determined. Now the sine of an angle is equal to the sine of its supplement, and therefore there are two angles which satisfy (ii), the one greater and the other less than 90°. (1). Let a be greater than b; ... A> B. Eucl. 1. 18. Now B cannot be greater than 90°, for in that case A+B would be greater than 180°, which is impossible, (Eucl. 1. 17.); ... B<90°, and the lesser angle which satisfies (ii) is to be taken for the value of B. In this case there is nothing to guide us in taking the value of B from (ii), since the only limits, B+ A < 180°, and B>A, may be fulfilled whether B be greater or less than 90o. Thus if CB = CB', it is evident that both the triangles ABC, AB'C have a, b, A of the same values; also, in this case, B D A is less than the exterior angle CBB', or the exterior angle CB'D; i. e. A is less than CB'A, or CBA; .. a<b. Which agrees with what has just been asserted. 73. In the solution of triangles there are various artifices used, which will be pointed out in each case. We shall now prove some formulæ of great use in the solution of plane rectilineal figures. 74. To find the cosine of an angle of a triangle in terms of the sides. Let ABC be a triangle, and from C draw CD perpendicular to AB, or AB produced either way. Then, figs. 1, 2, CB2 = AC2 + AB2-2 AB. AD, (Eucl. 11. 13.) fig. 3, CB2 = AC2 + AB2 + 2 AB .AD, (Eucl. 11. 12.) |