Right angled Triangles. (12.) From what has already been said respecting the Trigonometrical Lines, it will appear obvious that the tables, to which we have adverted, and in which the lengths of these lines, corresponding to every value of the angle to which they refer are registered, are nothing more than tables of right angled triangles, computed on the hypothesis that one of the three sides is the unit of length, and numerically expressed by 1. Thus, let ABC be a right angled triangle, of which a side AB, about the right angle, is one inch or one foot, or any length, represented numerically by 1; then, with A as centre, and radius AB, describe the arc Ba. It is plain, from the definitions, that BC is the tangent, and AC the secant of the angle A. Suppose this angle to contain, say, 36° 17'; then, in order to find the numerical values of the unknown sides BC, AC, we have only to refer to that page of G the tables in which 36° 17' occurs; and, in juxtaposition with this value, and under the head tangent, we shall find the length of BC: and under secant the length of AC; and thus all the parts of the triangle will become known. But if, instead of a side about the right angle, the side opposite to it be the unit, then, making this unit the radius, as before the unknown sides BC, AB will be the sine and cosine respectively of the angle A; so that, having found the page of the tables where the value of the angle A is, we shall see against this value and under the heads sine and cosine, the respective lengths of BC and AB. Now, from being possessed of these ready means (a table) of finding two sides of a right angled triangle, of known angle A, when the third side is 1, we may easily find the two sides of a similar right angled triangle, when the third side is any numerical value whatever, by an obvious application of the third proposition of the Sixth Book. If, for instance, the third side of the proposed triangle is 10 times 1 or 10, then we know, by that proposition, that each of the two remaining sides will be 10 times the corresponding side (computed in the tables) of the other triangle. And generally, if the given side of the proposed triangle be n times 1 or n, then each of the other sides will be n times that corresponding to it of the similar triangle computed in the tables. (13.) Hence, when an angle of a right angled triangle and a side are given, to determine another side we shall have to proceed thus: 1st. We must seek in the tables for the triangle similar to that proposed. 2d. We must take out the side corresponding to that whose length is sought. 3d. We must multiply this tabular side by the given side. The product will be the numerical value sought. The similar triangle in the table is found by means of the given angle, as expressed in degrees and minutes. The degrees will be found at the top of the page, and the minutes along the margin; and opposite to these, across the page, are registered all the particulars required.* Thus, if in the proposed triangle the given side is the hypothenuse, then the tabular triangle similar to it, and whose hypothenuse (the radius) is unity, will have for its perpendicular the sine of the given angle A (see fig. 2, above,) and for its base the cosine. Hence, multiplying the given hypothenuse by the sine of the given angle will give the perpendicular, and multiplying the hypothenuse by the cosine will give the base. If in the proposed triangle the base is given in conjunction with the angle A, then the tabular triangle similar * The use of a TrigonometricalTable will be easily learnt by actual examination, after consulting the preliminary explanation with which every such table is accompanied. See "Young's Trigonometrical Tables." to it, and whose base (the radius) is unity, will have for its perpendicular the tangent (see fig. 1, above,) and for its hypothenuse the secant of the given angle; so that we must multiply the given base by the tangent of the given angle to get the perpendicular, and by the secant to get the bypothenuse. It appears, therefore, that whether a side, as AB, or the hypothenuse AC be given, if with this given part as radius, and the vertex of the given angle as centre, an arc be described, the trigonometrical names of the remaining sides, in reference to this arc, will be those of the sides corresponding to them in the tabular triangle; so that, by sketching such an arc, or by conceiving it to be sketched about the given side, we shall immediately see what name, whether that of sine, cosine, tangent, &c. the tabular line has, and can take it out of the table accordingly. Suppose, for instance, the base AB is 64 feet, and the angle A 27o 19', and that the perpendicular BC is required. Then, describing the arc Ba with the given radius AB, we see at once that the sought side BC takes the name tangent. We seek, therefore, in the table for the tangent of 27° 19′, which we find to be 5165059, and multiplying this by the given radius, there results 33 0563776 for the length of BC, in feet. To distinguish the given line thus taken in our figure for a radius from the tabular radius, unit, it is as well to call it the geometrical radius, and the lines connected with the resulting arc, the geometrical sine, the geometrical cosine, &c.; for then the necessary directions for every example in right angled triangles will all be comprehended in one short precept, viz. (14.) The geometrical radius multiplied by any trigonometrical line produces the corresponding geometrical line, or line in the figure. Or more briefly, thus: Geom. rad. x Trig. line Geom. line. A few examples will fully illustrate what has now been said. 1. Given the angle A EXAMPLES. 53° 8', and the base AB = 288 to find the perpendicular and the hypothenuse. I. To find the perpendicular. Making the given side radius, the required side BC becomes the tangent of the angle A; therefore, by the precept (14), Α' = ABX tan A=288 × 1·33349384.045 BC. II. To find the hypothenuse. C B Preserving the same radius, AC will be the secant of the angle A to that radius; therefore, 2. Given the two perpendicular sides AB 472, BC= 765, to find the hypothenuse and angles. I. To find the angle A. Making AB radius, and applying the precept (14), we and referring to the tables, we find the angle whose tange is this number to be 58° 19' 32" ;* and, consequently, the angle C, which is the complement of this, is 31° 40′ 28′′. II. To find the hypothenuse. Preserving the same radius, AC will be the secant of the angle A, just determined; hence AB X sec A= AB cos A = 472 898-89. The hypothenuse might also have been found from the 47th of the First Book of the Elements, which gives AC=√AB2+ BC2. 3. Given the base AB = 288, and the perpendicular 384, to find the hypothenuse and angles BC A=53° 8', C=36° 52′, AC=480. 4. Given the hypothenuse AC = 645, and the base AB= 500, to find the other parts A 39° 10′, C=50° 50′, BC=407.37. = (15.) The preceding examples furnish so many immediate applications of the precept (14); and in each of which * In the ordinary Trigonometrical Tables the sines, cosines, &c. are computed only from minute to minute, so that when, as above, the arc corresponding to any trigonometrical line contains minutes and parts of a minute or seconds, these seconds must be determined by proportion. In the present example the nearest tangent, below 1.6207627 in the table, is 1-6201920; and the arc to which it belongs is 58o 19'; the arc 58° 20' which is 60" greater than the former, has for its tangent 1.6212469, exceeding the other tangent by 10549; hence, as this difference is to that between tan 580 19' and the proposed tangent, viz. to the difference 5707, so is 60" to 32", the in<rement of the arc corresponding to the increment of the tangent. |