SCHOLIUM. (C) Plane sailing in navigation is no B As thing more than the practice of right N angled trigonometry ; calling the hypothenuse the distance sailed, the perpendicular the difference of latitude, the W E base the departure, and the angle opposite to the base the course. In the annexed figure, wesn represents the horizon of the place c, from IS whenee the ship sails; CA the point of А B A the compass or rhumb she sails on, and a the place sailed to. Then we represents the parallel of latitude sailed from, and as the parallel of latitude arrived at. Hence Bc becomes the difference of latitude, AB the departure, ca the distance sailed, the angle ace the course, and the angle BAC the complement of the course. And this will always be the case whether the ship sails between the north and west, viz. between 'n and w; the north and east, viz. between N and E; the south and west, viz. between s and w; or south and east; viz. between s'and E. (D) The practice of Mercator's sailing is only the application of right-angled trigonometry, and similar triangles (Y. 34). What is called the difference of latitude in plane sailing, is here called the proper difference of latitude, to distinguish it from the line cd, which is called А. the meridional difference of latitude. The part E BD, by which cb is enlarged, is found by help of a table of meridional parts; being the sum, or difference of the meridional parts answering to the latitude sailed from and bound to; according as they are on different sides, or on the same side of the equator. ED is the difference of longitude. C B D 1 II. PRACTICAL RULES FOR SOLVING ALL THE CASES OF OBLIQUE TRIANGLES, WITH THEIR APPLICATION BY LOGARITHMS. (E) 1. When two of the three * given parts are a side and its opposite angle. Any one side of a triangle, * In every plane triangle there must be three parts given, to find the rest; and of these three given parts, one, at least, must be a side, because the same angles are common to an infinite number of triangles. As any other side, Is to the sine of its opposite angle. Is to its opposite side; (D. 35.) (F) An angle found by this rule is sometimes ambiguous, for trigonometry gives us only the sine of an angle, and not the angle itself, and the sine of every angle is also the sine of its supplement. The tables give only the acute value of an angle; the obtuse value is the supplement thereof. When the given side, opposite to the given angle, is greater than the other given side; then the angle opposite to that other given side is always acute. But when the given side opposite to the given angle, is less than the other given side, then the angle opposite that other given side may be either acute or obtuse, and consequently it is ambiguous. (G) II. When two sides and the angle contained between them are given.* The sum of the two given sides, other two angles. This half difference added to the complement of half the contained angle, gives the angle opposite to the greater side; and being subtracted, leaves the angle opposite to the less side, OR, angle two angles. This half difference, added to the half supplement, give the angle opposite to the greater side; and being subtracted, leaves the angle opposite to the less side. (E. 36.) The remaining side of the triangle may be found by Rule I. • If the two given sides be equal to each other, the triangle is isosceles, and each of the remaining angles will be equal to half the supplement of the given abgle. If the given angle be 90°, the required parts may be found by Case VI. of right angled triangles. (H) III. When the three sides of a triangle are given, to find the angles. Double the base, or longest side, the base. This distance added to balf the base gives the greater segment, and being subtracted therefrom, leaves the less segment. (C. 35.) The triangle being thus divided into two right angled triangles, each of which contains two given sides, the remaining angles may be found by Rule I. OR, From half the sum of the three sides of any plane triangle, subtract the side opposite to the angle sought, and note the half sum and the remainder. Then, to the logarithm of the half sum, add the logarithm of the remainder, and increase the index of the sum by 20; from the sum, thus increased, subtract the sum of the logarithms of the two sides containing the required angle, the remainder, divided by two, will give the logarithm cosine of half the angle required. (H. 37.) The remaining angles may either of them be found by Rule I. OR, From half the sum of the three sides of any plane triangle şubtract each of the sides containing the angle sought, and note the two remainders. Then, add the logarithms of these two remainders together, and increase the index of their sum by 20; also, from balf the sum of the three sides, subtract the side opposite to the angle required, and add the logarithm of the half sum to the logarithm of the remainder; the difference between these logarithmical sums, divided by 2, will give the logarithmical tangentt of half the angle sought. (I. 38.) Either of the remaining angles may be found by Rule I. * If the triangle be equilateral, each of its angles will be 60°; if it be isosceles, the perpendicular will bisect the base, and the angles may be found by Case IV. of right angled triangles. + If an angle be near 90°, or if it be a very small angle, it ought to be determined by a rule wbich produces a tangent or a co-tangent. For the sines of arcs near 90° vary by almost imperceptible increments; and the co-sines of very small arcs, or those near 180°, vary by decrements which are not easily assigned.' To illustrate these observations by an example; suppose that in working any problem, by a table calcu. lated to seconds, and carried to 7 places of decimals, the result of the operation should produce 9.9999998 a sine or co-sine. By the tables this is the sine of any arc be (I) CASE I. Given the angles and one side of a triangle, to find the rest. The angle a= 32°.15' The angle b=1140.24 Required the sides ac and BC BY CONSTRUCTION. Make AB equal to 98 by a scale of equal parts, draw Bc making the angle B equal to 1149.24'. (R. 27); also make the angle A equal to 329.15', draw ac meeting Bc in c. Then ac measured by a scale of equal parts is 162, and BC 95. BY CALCULATION. See Rule 1. To the angle A = 32°.15' add the angle B = 114°.24', the sum is 1460.39', which subtract from 180° leaves the angle c=33.21. To find AC. To find BC sine L C=339.21' 9.74017 9.74017 : AB=98 1.99123 : AB=98 1.99123 : ; sine 2 31149.24 9.95937 :: sine LA=320.15 9.72723 : side AC=162.34 2.21043 : side BC=95.12 1.97829 (K) BY GUNTER'S SCALE. 1. Extend the compasses from 33o.21' to 65o.36' the supplement of B, on the sines, that extent will reach from 98 to 162 on the line of numbers. 2. Extend the compasses from 330.21' to 32°.15' on the line of sines, that extent will reach from 98 to 95 on the line of numbers. Given S4A=270.59" PRACTICAL EXAMPLES. AB=142.02 Ans. BC=70 Lc=1070.49' { tween 89o.56'.19", and 899.57.8"; or the co-sine of any arc between 2.52" and 3.41"; here is therefore a choice of 49" or arcs, and no reason can be given why one should be preferred to another: but the logarithmical differences between the tangents and co-tangents are never so small as to leave any doubt respecting the accuracy of the result, the smallest difference for 1". being 42, at which time the arc is 450. 2. In the plane triangle ABC, LA=790.23 AC=103.4 Ans. 3 AB=91.87 LC=46°.15! (L) CASE II. Given two sides and an angle opposite to one of them, to find the rest. The angle c=339.21' Given The side AB=98 Required the angles A and B and the side ac. BY CONSTRUCTION. Make bc=9512 by a scale of equal parts, draw ca making the ä angle c=33°.27 (R. 27.) With the side AB in your compasses, taken from the same scale of equal parts, and B as a centre, describe the arc aa cutting ac in the point A. Then AC measured by the scale of equal B parts will be 162, and the angles A and B measured by a scale of chords (S. 27.) will be 32o.15' and 114.24, Here agreeably to the observation (F. 53.) the angle a is acute and not ambiguous; but had AB been less than BC, the arc aa would evidently have cut ac in two points on the same side of Be. K. BY CALCULATION. See Rule I, AB=98 1.99129 sine LC=330,21. 9.74017 : sine (c=330.21' 9.74017 : AB=98 1.99123 : : BC=95.12 1.97827 :: sine ZB=1149.24' or : sine LA=320.15' 9.95997 9.72721 sine 650.36' hence ZB=1149.24 side AC=162.34 2.21043 E or} (M) BY GUNTER'S SCALE. 1. Extend the compasses from 98 to 95 on the line of numbers, that extent will reach from 33°21' to 320.15' on the line of sines. 2. Add the angles A = 32°.15' and c= 33°21' together, and the sum will be 650.36'; then extend the compasses from 33° 21' to 650.36' on the line of sines, that extent will reach from 98 to 162:3 on the line of numbers. The angle c=330.21 Given The side BC=959.12 Required the angles A and B and the side AC. |