the opposite sides, CG is the difference between these sides, and DC is the difference between AF and FC, the segments of the base made by the perpendicular BF, let fall from the opposite angle, agreeably to the statement of the proposition. Had the perpendicular been let fall from either of the acute angles at A or C, it would have gone without the triangle, and required the opposite side to be produced, in order to meet it; when the proportion would have been this; as the base is to the sum of the opposite sides, so is the difference of these sides to the sum of the segments (of the base when produced) made by the perpendicular. WHEN the student is acquainted with the properties of triangles exhibited in the preceding propositions, he will be able to understand the process of solving the following cases of plane trigonometry, which are classed into those belonging to right-angled, and those belonging to oblique-angled triangles; and, first, of right-angled triangles. CASE I. Fig. 9, Plate 3. Given the two legs, or the base and the perpendicular. Required the angles and the hypothenuse. = In the plane triangle ABC, right-angled at A, are given the base AC 440 integers of any dimension, as feet, miles, &c. and the perpendicular AB 308 of the same integers, and the angles at B and C, as also the hypothenuse BC, are required. By the preceding 4th Prop. of Trigonometry, it was hown, that if the base of a right-angled triangle be made radius, the perpendicular will become the tangent of the opposite angle at the base. In this case, therefore, we state the proportion. As the base AC = 440, To the pependicular AB = 308, So So is the radius of the tables, which is always equal to the sine of 90 degrees, (pages 382 and 384, Fig. 1, Plate 3,) or 10,00000, To the tangent of the angle BAC. Or, thus, Log. 440 Log. 308 :: Sine 90,00; Tang. Having found in the tables the logarithms of the first and second terms, we write them down together with the standing logarithmic sine of radius, which is 10,00000; then, agreeably to what was said of the nature of logarithms, add the second and third terms together, and from their sum subtract the logarithm of the first term. The remainder being a logarithmic tangent, we search in the tables of tangents for the number equal or nearest to it, and in the column of degrees and minutes, we find 35 degrees, without any minutes, to be the quantity required for the angle BCA. As the three angles of any triangle, are together equal to two right angles (Geom. Prop. 7', and as the angle at A was made a right angle, the remaining two at B and C must together be equal to one right angle, or 90 degrees, (p. 383); subtracting then the angle at C 35 degrees, from 90 degrees, the remainder, 55 degrees, must be the quantity of the angle at B, as was required. Again, to find the length of the hypothenuse BC, we proceed in the following way; supposing the base AC still to be radius, the hypothenuse will become the secant of the angle at C, whence we have this proportion; as radius to the secant of the angle at C, so is the base AC to the hypothenuse BC; but as the secants are seldom inserted in logarithmic tables, such problems, where they can be introduced, being susceptible of a solution by means of other lines, we proceed with the following proportion, in which the hypothenuse itself being made radius, the perpendicular AB becomes the sine of the angle at C, (Trig, Prop. 5). As the sine of the angle at C= 35°,00′ = 9,75859 To radius So is the side AB To the hypothenuse BC 90°,00′ = 10,00000 = 308 = 2,48855 =337 = 2,72996 Here, by adding the logarithms of the second and third terms together, and subtracting from the sum that of the first term, the remainder 2,72996 is found in the tables to be the nearest number to that corresponding to the natural number 537, which is the measure of the hypothenuse, as required. It was formerly shown, (Geom. Prop, 18), that in all right-angled triangles the square constructed on the hypothenuse is equal to the sum of the two squares constructed on the opposite sides containing the right angle. From this property we may, in the present case of trigonometry, discover the measure of the hypothenuse BC, without having recourse to the logarithmic tables, by means of the following arithmetical operation.. Square Extract the square root) 298464(537 = hypoth. BC, Given the hypothenuse and one leg. Required the other leg and the angles. In the triangle ABC, right-angled at A, are given the bypothenuse BC= 537, and the perpendicular AB = 308; required the base AC, and the two angles ABC and BCA. Making the hypothenuse BC radius (Trig. Prop. 5), and C the centre, the perpendicular AB becomes the sine of the angle BCA; hence we have the proportion. Now, to find the base AC, we may employ the hypothenuse as radius, and make the centre at B, when the base will become the sine of the angle at B. As radius = 90°,00′ = 10,00000 To the sine of ABC 55°,00' = 9,91336 So the hypoth. BC = 537 = 2,72996 To the base AC = 440 = 2,61339 Or, by employing the perpendicular BA as radius, and making B the centre, we have the proportion, As radius = 90°,00' 10,00000 " To the tangent of ABC = 55°,00′ = 10,15477 So the perpend. BA To the base AC = 308 = 2,48855 = 440 = 2,64332 Or, the base may be found arithmetically, as was shown in the preceding case, by subtracting the square of the given perpendicular AB 308, from the square of the hypothenusc BC= 537, and extracting the square root of the remainder, which will be 440, as before. CASE III. Fig. 9, Plate 3. Given one leg and the angles. Required the other leg and the hypothenuse. = In the triangle ABC, right-angled at A, are given the base AC 440, and the angle ACB 35°,00', consequently the angle ABC = 55°,00 (Geom. Prop. 7); required the perpendicular AB and the hypothenuse BC. Making the given base AC radius, the perpendicular AB will become the tangent of the angle at C: hence we have this proportion; As |