Secondly. If the perpendicular falls A without the triangle, we shall have AC2=AB2+ BC2+2BC× BD (Prop. 13. III.); hence BD= AC2-AB-BC 2BC But in the right-angled triangle BAD, we still have sin BAD or cos ABD= ABD being supplemental to ABC or B, we have cos B cos ABD RX BD (Art. 11.); hence by substituting the value of BD, we shall again have cos B-Rx AB2+BC2—AC2 2ABX BC * XLVI. Let A, B, C be the three angles of any triangle; a, b, c the sides respectively opposite them: by the last Article, we shall have cos B=RX a2+c2-b2 And the same principle when applied to each of the other two angles, will, in like b2+c2—a2 manner give cos A=R× These three formulas are of themselves sufficient for solving all the problems of rectilineal trigonometry; because, when three of the six quantities A, B, C, a, b, c, are given, we have by these formulas the equations necessary for determining the other three. The principles already explained, and whatever others may be added to them, can, therefore, only be consequences of these three principal formulas. R2 Accordingly the value of cos B gives sin2 B-R2-cos2 B-R2. 4a2c2 sin B b 4a2 c2—(a2+c2—b2)2 4a2 c2 (2 a2 b2+2 a2 c2+2 b2 c2-a1-b-c1): hence R 2 abc√(2 a2 b2+2 a2 c2+2 b2 c2———-a2———ba—c1). The second member being a function of a, b, c, in which these three letters all occur under the very same form, we may evidently change two of these letters at will, and thus have sin B sin A sin C b a C ; which is the principle of Art. 44. And from this, the principles of Art. 42 and 43 are easily deducible. XLVII. In any rectilineal triangle, the sum of two sides is to their difference, as the tangent of half the sum of the angles opposite those sides is to the tangent of half the difference of those same angles. From the proportion AB: AC:: sin C: sin B (see the figures in pp. 307, 308.); we derive AC+AB: AC-AB:: sin B+ sin C: sin B-sin C. But, according to the formulas of Art. 29, we hav B-C sin B+sin C: sin B-sin C :: tang hence B+C : tang 2 B+C B-C 2 ; AC+AB: AC-AB:: tang 2 : tang which is the property we had to demonstrate. With this small number of principles, we are enabled to solve all the cases of rectilineal trigonometry. SOLUTION OF RIGHT-ANGLED TRIANGLES. XLVIII. Let A be the right angle of the proposed rightangled triangle, B and C the other two angles; fet a be the hypotenuse, b the side opposite the angle B, c the side opposite the angle C. Here we must consider that the two angles A and A are complements of each other; and that consequently, according to the different cases, we are entitled to assume sin C =cos B, sin B=cos C, and likewise tang B=cot C, tang Ccot B. This being fixed, the different problems concerning rightangled triangles are all reducible to the four following cases: FIRST CASE. XLIX. Given the hypotenuse a, and a side b, to find the other side and the acute angles. For determining the angle B, we have (Art. 43.) the proportion ab: R sin B. Knowing the angle, we shall also know its complement 100°—B=C; we might also find C directly by the proportion a: b:: R: cos C. As to the third side c, it may be found in two ways. Having found the angle B, we can either (Art. 43) form the proportion R: cot Bbc; or the value of c may be obtained directly from the equation c2—a2-b2, which gives c=√(a—b2), and consequently log c=4 log (a+b)+§log (a—b). SECOND CASE. L. Given the two sides b and c of the right angle, to find the hypotenuse a, and the angles. We shall have the angle B (Art. 43.) from the proportion cb:: R: tang B. Next we shall have C-100°-B. We might also find C directly by the proportion b:c:: R: tang C. Knowing the angle B, we shall find the hypotenuse by the proportion sin B: R::b: a; or a may be obtained directly from the equation a=/(b2+c2); but as b+c2 cannot be decomposed into factors, this expression is incommodious in calculating with logarithms. THIRD CASE. LI. Given the hypotenuse a and an angle B, to find the other two sides b and c. Make the proportions R'sin B they will give the values of b and c. equal to the complement of B. FOURTH CASE. ab, R: cos B:: a: c;" As to the angle C, it is LII. Given a side b of the right angle with one of the acute angles, to find the hypotenuse and the other side. Knowing one of the acute angles, we shall likewise know the other; hence we may look upon the side b and the opposite angle B as given. To determine a and c, we shall then have the proportions sin B: Rb: a, R: cot B:: b: c. 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 other three, will all be reducible to the four following cases. FIRST CASE. LIII. Given the side a and two angles of the triangle, to find the two other sides b and c. Two of the angles being known will give us the third; then the two sides b and c will result from the proportions (Art. 44.) sin A: sin B::a: b, sin A: sin C:: a: c. SECOND CASE. LIV. Given the two sides a and b, with the angle A opposite to one of them, to find the third side c and the other two angles B and C. The angle B may be had by the proportion lue of sin B, we may either take B=M, or B=200°-M. This ambiguous solution will not occur, however, except we have at once A an acute angle and ba. If the angle A is obtuse, B cannot be so; hence we shall have but one solution; and if, A being acute, we have ba, there will equally be only one solution, because in that case we shall have MA, and by making B=200°-M, we should find A+B 200°; which it cannot be. Knowing the angles A and B, we shall also know the third angle C. Then we shall obtain the third side c by the proportion sin A: sin C:: a: c. We might also deduce c directly from the equation cos A R But this value will not admit of being computed by logarithms, except by help of an auxiliary angle M or B, which brings it back to the foregoing solution. THIRD CASE. LV. Given two sides a and b, with their included angle C, to find the other two angles A and B with the third c. Knowing the angle C, we shall likewise know the sum of the other two angles A+B=200°-C, and their half-sum (A+B) -100°-C. Next we shall compute the half-difference of these two angles by the proportion (Art. 42.) a+ba-b: tang (A+B) or cot C: tang (A—B), in which we consider ab, and consequently AB. Having found the half-difference, by adding it to the halfsum (A+B), we shall have the greater angle A; by subtracting it from the half-sum, we shall have the smaller angle B. For, A and B being any two quantities, we have always A = } (A+B)+1⁄2 (A + B), B = {} (A+B)—§ (A — B). Knowing the angles A and B, to find the third side c, we have the proportion sin A: sin C::a: c. P * LVI. In trigonometrical calculations, it often happens that two sides a and b are known by their logarithms; in that case, to avoid the trouble of seeking the numbers which correspond to them, we need only seek the angle by the proportion 6: a:: R: tang. The angle will be greater than 50°, since we supposed ab; subtract from 50° therefore, and form the proportion R tang (4-50°): cos C: tang (AB); from which, as formerly, the value of (A B) may be determined, and afterwards that of the two angles A and B. This solution is founded on the property, that, tang (50°—✨) R° tung ¢R° tang 50° R2+tang tang 50° = R2 R ; now tanga, and tang 50°—R; b hence tang (4—50° — R(ab); hence a+b; a-b:: R: tang a+b (-50°): cot C: tang (A—B). As for the third side c, it may be found directly by means of But this value is inconvenient for calculating with logarithms, unless the numbers which represent a, b, and cos C are very simple. We may observe that the value of c might also be put under these two forms: c= √ [ (a — b)2 = 4 a ¿sin3 ↓ C cos2 R2 C which is easily verified by means of the formulas sin2 CRR cos C, cos2 & CR2+R cos C. These values will especially be useful, if it is required to compute c with great precision, the angle C and the line ab being at the same |