Let ABC be the proposed triangle; AD the perpendicular, let fall from the vertex A on the opposite side BC: there may be two cases. First. If the perpendicular falls within B the triangle ABC, the right-angled triangles ABD, ACD, will give, R: sin B:: AB: AD. In these two propositions, the extremes are equal; hence, sin C: sin B::AB: AC. Secondly. If the perpendicular falls A without the triangle ABC, the rightangled triangles ABD, ACD, will still give the proportions, R: sin ABD :: AB: AD, from which we derive D B THEOREM IV. hence BD= ABD, we have sin C sin ABD::AB: AC. But the angle ABD is the supplement of ABC, or B; hence sin ABD=sin B; hence we still have sin C: sin B::AB: AC. Let ABC be a triangle: then will cos B-R First. If the perpendicular falls within X 2BC In every rectilineal triangle, the cosine of either of the angles is equal to radius multiplied by the sum of the squares of the sides adjacent to the angle, minus the square of the side opposite, divided by twice the rectangle of the adjacent sides. D AB2+BC2-AC2. 2AB × BC. D R: cos B::AB: BD; But in the right-angled triangle Secondly. If the perpendicular falls without the triangle, we shall have AC2-AB2+BC2+2BC x BD; hence AC-AB2-BC2 BD= 2BC or by substituting the value of BD, 2ABX BC D B с A But in the right-angled triangle BAD, RX BD we still have cos ABD- = ; and the angle ABD being AB supplemental to ABC, or B, we have Rx BD cos B-cos ABD: = AB hence by substituting the value of BD, we shall again have 2ABX BC A For the sake of brevity, put (a+b+c)=p, or a+b+c=2p; we have a+b-c=2p-2c, atc-b-2p-26; hence sin ¿A=R √ ((p—b) (p—c)). bc THEOREM V. 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 C+A sinC+sin A: sin C-sin A :: tang 2 (Art. XXIV.); hence, C+A 2 C-A tang the property we had to demonstrate. With the aid of these five theorems we can solve all the cases of rectilineal trigonometry. : tang B C-A LOGARITHMS. 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. X It is now to be shown that, the difference between two loga rithms is truly found, by adding to the first logarithm the arith metical complement of the logarithm to be subtracted, end dimin. ishing 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 compleinent 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 à 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 So is R To sin 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. As hyp. a So is R or, When logarithms are used, it is most convenient to write the proportion thus, R sin B: a b (Theorem I.). · 250 EXAMPLES. 250 7.602060 2.380211 10.000000 73° 44′ 23′′ (after rejecting 10) 9.982271 ar.-comp. log. 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" As R To side c 70.0003 To find the side c, we say, log. ar. comp. a |