Note 4. The triangle being carefully constructed, and marked by suitable letters of reference, as A and B for example, at the extremities of the base, c at the vertex, then, according to the nature of the problem, write down the requisite proportion in four distinct lines, with the letters of reference to each term, and the given numbers to the three first: against these numbers place their respective logarithms; find the logarithm to the fourth term by the directions in note 2, and ascertain its angular or lineal value, by means of the tables. *** The last three notes are not restricted to the present case, but extend in their application to the usual practice of plane trigonometry. Example I. 3. In a plane triangle are given two angles equal to 58°7', and 22° 37', respectively, and the side between then 408 yards. Required the remaining angle and sides. Construction. On an indefinite right c line, set off, from a convenient diagonal scale, the distance AB = 408. From the /> point A draw a right line Ac, to make with AB an angle of 58° 7'; and from A B the point B another line, turned towards the former, and to make with BA an angle of 22° 37'. The intersection c, of these two lines determines the triangle; and the sides ac, Bc, measured upon the scale of equal parts from which AB was laid down, are found to be 159 and 351 respectively. Computation. Two of the angles being known, their sum 80° 44' taken from 180°, the sum of the angles in . a plane triangle, leaves 99° 16' for the third angle c. This, being an obtuse angle, its sine is to be found in the table by taking that of its supplement 80°44', which (chap. i. 19) is the same. Hence, , Ac and Bc, therefore, are 351-02 yards, and 158.98 yards, respectively. y 4. In the preceding operation, instead of adding together the logs, of the second and third terms, and subtracting that of the first from their sum, the work has been performed thus:—The right hand figure of the upper line was taken from 10, and each of the other figures from 9, and their several remainders added to the numbers below them in the respective columns. This is easily effected in practice by making those remainders emphatical in adding downwards. Thus, in the operation for Bc, we begin at the right hand, and adding downwards say, ten and 2 are 12, and 8 are 20; set down 0: carry 2, and four are 6, and 1 make 7: nothing added to 6 and 7, gives 13; set down 3: carry 1, and seven are 8, and 6 are 14, and 9 are 23; set down 3: carry 2, and five are 7, and 8 are 15: and so on, to the left hand column, which added in the same way amounts to 12; of which the 2 are put down, and the 10 rejected, to compensate for what has been borrowed in the process by the arithmetical complement. This method is very easy in practice, and is found less liable to produce error than that in which the arith. comp. is put down at once from the tables. Erample II. 5. In a plane triangle ABC are given Ac = 216; ch = 117; the angle A = 22° 37'; to find the rest. Construction. Draw an indefinite right line ABB', from an assumed point A in which C = 117 describe an arc BB', it will cut the line AB in two points, from each of which drawing lines Bc, Boc, there will be formed two triangles Abc, AB'c, each of which answer the conditions of the question. The required lines and angles being measured, give AB = 117 . . . . .AcB = 22.”.... Abe = 1343% AB' = 282 . . . . ACB = 1 12° ... ..AB"c = 45+ °. To AB" . . . . . . 281-79 . . . . . . 24499222 6. Remark. The ambiguity in this and similar examples, does not, as has been often affirmed, depend upon the circumstance that an angle and its supplement have the same sine, but solely upon this, that in con structing triangles from analogous data, when the side cB has its length between certain limits, that is, between the length of the side Ac, and that of the perpendicular cal from c on the third side AB, it must necessarily cut the indefinite right line ABB' in two points. fo limits there is no ambiguity; for when cis is proposed to be less than ca, the problem is impossible; and when cB exceeds ca, the angle A being all along supposed given, the line ABB' can only be cut in one point. Hence the practical maxim may be thus expressed:—when cB is proposed less than Ac sin A, the data are erroneous; when it is given greater than Ac there can be only one triangle; between those limits the problem is ambiguous. Er. 3. Given two angles of a plane triangle 22° 37' and 134°46', and the side between them 351. To find the remaining angle and sides. Ans. Angle 22°37.3 sides 351 and 648. Er. 4. Given two sides of a plane triangle 50 and 40 respectively, and the angle opposite to the latter equal to 32°; to determine the triangle. Ans. If the angle opposite to the side 50 be acute, then is it = 41° 28′, the third angle 106° 32', and the remaining side 72°36. If the angle opposite to side 50 be obtuse, it is - 138° 32', and the other angle and side 9°28' and 12.415 respectively. CASE II. 7. When two sides and the included angle are given, The solution is effected by means of chap. ii. prop. 15, Take the given angle from 180°, the remainder will be the sum of the other two angles. To the tangent of half their difference. Half the difference added to half the sum of those angles, gives the greater of them; and taken from half the sum, leaves the less *. All the angles becoming known by this process, the third side is found by the rule in case 1. Example I. 8. Let there be given in a plane triangle ABC, Ac = 450, Bc = 540, and the included angle c = 80°; to find the remaining angles and side. Leaving the construction to be effected by the pupil, I shall proceed to the computation. Bc + Ac = 990, Bc – Ac = 90, 180° – c = 100°. As BC + Ac ......990 .. 2:9956352 To Bø – Ac...... 90 ... 1954.2425 So is tan (A + B) 50°.. 10.076.1865 9. The two sides of a triangle are 40 and 32, and the included angle 90°, required the other angles and side. In this example the operation may be considerably shortened by working without, ...of with, the loga |