Or, by the arithmetical complement; Arith. comp. of log. of AB 7,84164 Sine of ACB=9,92111 BAC 42°, 30'-9,82968 Having thus found the angle at A, by adding it to the given angle at C, and subtracting the sum from two rightangles 180 degrees, (Geom. Prop. 7,) we obtain the angle at B=81°,00', which was required. Then, to find the unknown side AC, we say; As the sine of the angle BCA=56°,30′ = 9,92111 To the sine of the angle ABC=81°,00 9,99462 Or, by the arithmetical complement; Arith. comp. of sine of BCA 56°,300,07889 Sine of ABC= 81°,00 Log. of AB 9,99462 144 2,15836 AC 170,6 = 2,23187 CASE III. Fig. 10, Plate 3. Two sides, and the angle contained between them, being given, to find the other side and angles. In the plane triangle ABC, are given the sides AB = 144, and BC= 116,6, as also the angle ABC contained between these two sides 81°,00': required the other angles at A and C, and the side AC. = In Prop. 6 of Trigonometry it was demonstrated, that in any plane triangle the sum of the two sides opposite to the base is to the difference between these sides, as the tangent of half the sum of the two angles at the base is to the tangent of half the difference between these angles: hence we have this proportion; As AB+ BC: AB-BC :: Tang. BCA + BAC 2 The sum of AB≈ 144, and BC116,6, is 260,6, and their difference is 27,4; then subtracting the given angle at B=81o,00 from two right angles, or 180°, we have 99°, the half of which is 49°.30′: hence we have this proportion; In Prop. 3 of Trigonometry it was shown, that in any two unequal quantities, if half their sum be added to half their difference, the sum will be equal to the greater quantity; and that it from half their sum be subtracted half their dif ference, the remainder will be equal to the less quantity: consequently, knowing now half the sum and half the difference of the two angles at the base, the angles themselves may be found as follows: Half And, agreeably to Case 1, the remaining side, or base, AC, will be found by stating the following proportion: Sine of ACB 56°, 30′ = 9,92111 ABC 81°, 00' = 9,99462 Or, by the arithmetical complement : Arith. comp. of sine of 56,30' 0,07889 = CASE IV. Fig. 10, Plate 3. The three sides being given, to find the angles. In the oblique-angled plane triangle ABC, are given the three sides AB 144, BC 116,6, and AC=170,6; required the angles. Having constructed the triangle agreeably to the given dimensions, from the angle at B, opposite to the base, let tall fall the perpendicular BD, (Geom. Prop. 24,) cutting the base into two segments, AD and DC. By Prop. 7 of Trigonometry it appears, that a line drawn perpendicular to the base of a triangle from the opposite angle, will cut the base in such a proportion, that the whole base will be to the sum of the two opposite sides, as the difference between these two sides is to the difference between the segments of the base. Hence, in the present case, we have this proportion; as the base AC 170,6, to the sum of the opposite sides AC 144+ BC116,6=260,6, so is the difference of these sides AB-BC= 27,4, to the difference between the segments of the base, AD-DC. Arith. comp. of log. of AC =170,67,76815 Log. of AB+BC = 260,62,41597 AD-DC 41,8671,62187 Having thus found the difference between the segments of the base, formed by the perpendicular BD, the segments themselves are found in this way. To half the sum of these segments, that is, to half the whole base AC, add half the difference now found, and the sum will be the greater segment, which is always opposite to the greater side; and from half the whole line subtract half the difference, when the remainder will be the less segment opposite to the least side. Less segment DC = 64,3665 Having thus found the side AD, and the other side AB being given in the triangle ABD, right-angled at D, because BD stands perpendicularly on AD, we have only to apply the the 2d case of right-angled Trigonometry, to find the angle at A. BA = 144 = 2,15836 Log. Sine of BAD = 47°,30′ = 9,86776 And subtracting this angle from 90 degrees, we have 42,50' for the angle at A. By a similar process, making BC radius, we discover the angle at C to be 56°, 30'; and adding these two angles together, the difference between their sum and 180 degrees is 81°,00', equal to the great angle ABC. PRACTICAL |