D B C But in the right-angled triangle BAD, RX BD we still have cos ABDand the angle ABD being AB supplemental to ABC, or B, we have Rx BD cos B-cos ABD: == =R2x AB hence by substituting the value of BD, we shall again have AB2+BC2-AC2 2AB × BC cos B=RXx Scholium. Let A, B, C, be the three angles of any triangle; a, b, c, the sides respectively opposite them: by the theorem, a2 + c2_b2 2ac we shall have cos B-Rx and cos C-R×· 2ab Recurring to the formula R2-R cos A-2 sin2 A (Art. XXIII.), or 2sin2A-R2-RcosA, and substituting for cosA, we shall have 2sin2 AR2-R3× b2+c2-a2 2bc R2 × 2bc--R2(b2+c2—a3) 2bc =R2 × (a+b—c) (a+c—b) =R2x A a2-(b-c)2 sin For the sake of brevity, put (a+b+c)=p, or a+b+c=2p; we have a+b-c=2p-2c, a+c-b=2p-2b; hence 2bc ·(a+b—c) (a+c- JA=R√ ((a+b—c) a2b2-c2+2bc 2bc +omb)). sin JA-R✓ ((Pb) (pc)). }A=R√ (p—b) Hence THEOREM V. C-A 2 In every rectilineal triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides, to the tangent of half their difference. : For, AB: BC: sin C: sin A (Theorem III.). Hence, AB+BC: AB-BC :: sin C+sin A: sin C-sin A. But sin sinC+sin A: sin C-sin A:: tang 2 tang (Art. XXIV.); hence, AB+BC AB-BC tang the property we had to demonstrate. With the aid of these five theorems we can solve all the cases of rectilineal trigonometry. C+A 2 : tang C-A LOGARITHMS. B C which is Scholium. The required part should always be found from the given parts; so that if an error is made in any part of the work, it may not affect the correctness of that which follows. SOLUTION OF RECTILINEAL TRIANGLES BY MEANS OF It has already been remarked, that in order to abridge the calculations which are necessary to find the unknown parts of a triangle, we use the logarithms of the parts instead of the parts themselves. Since the addition of logarithms answers to the multiplication of their corresponding numbers, and their subtraction to the division of their numbers; it follows, that the logarithm of the fourth term of a proportion will be equal to the sum of the logarithms of the second and third terms, diminished by the logarithm of the first term. Instead, however, of subtracting the logarithm of the first term from the sum of the logarithms of the second and third terms, it is more convenient to use the arithmetical complement of the first term. The arithmetical complement of a logarithm is the number which remains after subtracting the logarithm from 10. Thus 10—9.274687=0.725313: hence, 0.725313 is the arithmetical complement of 9.274687, It is now to be shown that, the difference between two logarithms is truly found, by adding to the first logarithm the arithmetical complement of the logarithm to be subtracted, and diminishing their sum by 10. Let a the first logarithm. b = the logarithm to be subtracted. c = 10-b-the arithmetical complement of b. Now, the difference between the two logarithms will be expressed by a-b. But from the equation c=10-b, we have C -10——b: hence if we substitute for b its value, we shall have which agrees with the enunciation. When we wish the arithmetical complement of a logarithm, we may write it directly from the tables, by subtracting the left hand figure from 9, then proceeding to the right, subtract each figure from 9, till we reach the last significant figure, which must be taken from 10: this will be the same as taking the logarithm from 10. Ex. From 3.274107 take 2.104729. Common method. 3.274107 2.104729 a-b=a+c-10, Diff. 1.169378 By ar.-comp. 3.274107 ar.-comp. 7.895271 sum 1.169378 after re jecting the 10. We therefore have, for all the proportions of trigonometry, the following RULE. Add together the arithmetical complement of the logarithm of the the first term, the logarithm of the second term, and the logarithm of the third term, and their sum after rejecting 10, will be the logarithm of the fourth term. And if any expression occurs in which the arithmetical complement is twice used, 20 must be rejected from the sum. SOLUTION OF RIGHT ANGLED TRIANGLES. b A с Let A be the right angle of the proposed right angled triangle, B and C the other two angles; let a be the hypothenuse, b the side opposite the angle B, c the side opposite the angle C. Here we must consider that the B two angles C and B 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 C=cot B. This being fixed, the unknown parts of a right angled triangle may be found by the first two theorems; or if two of the sides are 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. As hyp. a To sin B As hyp. a To side b Ex. 1. In the right angled triangle BCA, there are given the hypothenuse a=250, and the side b=240; required the other parts. or, When logarithms are used, it is most convenient to write the proportion thus, ar.-comp. log. EXAMPLES. : R sin B: a b (Theorem I.). 250 240 250 240 7.602060 2.380211 - 10.000000 73° 44′ 23′′ (after rejecting 10) 9.982271 But the angle C=90°-B=90°-73° 44′ 23′′=16° 15′ 37′′. or, C might be found by the proportion, ar.-comp. log. 16° 15' 37" To find the side c, we say, log. As R To tang. C 16° 15′ 37′′ To side c 70.0003 ar. comp. a C Or the side c might be found from the equation a2= b2+c2. For, hence, c2=a2-b2=(a+b)× (a—b): 2 log. c=log. (a+b)+log. (a—b), or log. Log.c 70 Ex. 2. In the right angled triangle BCA, there are given, sideb=384 yards, and the angle B 53° 8': required the other parts. To side or, B 53° 8' As tang Is to R So is side b 384 To find the third side c. R: tang B::c: b (Theorem II.) log. c 287.965 2.690196 1.000000 2) 3.690196 1.845098 ar.-comp. required the other parts. 9.875010 10.000000 2.584331 2.459341 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, : ar. comp. log. As sin B 53° 8' Is to R So is side b 384 To hyp. a 479.979 Ex. 3. In the right angled triangle BAC, there are given, side c=195, angle B=47° 55', 0.096892 10.000000 2.584331 2.681223 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. |