If from the sum of the second and third logs. that of the first be taken, the number will be the log. of the fourth; the number answering to which will be the thing required; but when the first log. is radius, or 10.000000, reject the first figure of the sum of the other two logs. (which is the same thing as to subtract 10.000000 ;) and that will be the log. of the thing required. 2. Making AB the radius. Secant A: AC:: R: AB. Secant A: AC :: T.A: BC. That is, As the secant of A=46° 30′ 10.162188 So is the tangent of A =46° 30′ 10.022750 12.420690 That is, as the secant of C-43° 30′ 10.139438 Or, having found one side, the other may be obtained by cor. 2. theo. 14. sect. 4. 3d. By Gunter's scale. The first and third terms in the foregoing proportions, being of a like nature, and those of the second and fourth being also like to each other; and the proportions being direct ones, it follows; that if the third term be greater or less than the first, the fourth term will be also greater or less than the second; therefore the extent in your compasses, from the first to the third term, will reach from the second to the fourth. Thus, to extend the first of the foregoing proportions; 1. Extend from 90° to 46° 30', on the line of sines; that distance will reach from 250 on the line of numbers, to 181, for BC. 2. Extend from 90° to 43° 30′, on the line of sines; that distance will reach from 250 on the line of numbers, to 172, for AB. If the first extent be from a greater to a less number; when you apply one point of the compasses to the second term, the other must be turned to a less; and the contrary. By def. 20. sect. 4. The sine of 90° is equal to the radius; and the tangent of 45o is also equal to the radius; because if one angle of a right angled triangle be 45°, the other will be also 45°; and thence (by the lemma preceding theo. 7. sect. 4.) the tangent of 45° is equal to the radius: for this reason the line of numbers of 10.000000, the sine of 90°, and tangent of 45o being all equal, terminate at the same end of the scale. The two first statings of this case, answers the question without a secant: the like will be also made evident in all the following cases. 4th. Solution by Natural Sines. From the foregoing analogies, or statements, it is obvious that if the hypothenuse be multiplied by the natural sine of either of the acute angles, the product will be the length of the side opposite to that angle; and multiplied by the natural cosine of the same angle, the product will be the length of the other side, or that which is conti guous to the angle. Thus : the given ang. 47° 30′. Nat. Cos..688355 250 34417750 1376710 Base=172.088750 Perpend. 181.343500 CASE II. The base and angles given; to find the perpendicular and hy. , pothenuse. PL. 5. fig. 5. In the triangle ABC there is the angle A 42o 20, and of course the angle C 47° 40′ (by cor. 2. theo. 5,) and the side AB 190, given; to find BC and AC. 1st. By Construction. Make the angle CAB (by prob. 16. sect. 4.) in blank lines, as before. From a scale of equal parts Jay 190 from A to B: on the point B, erect a perpendicular BC (by prob. 5. sect. 4.) the point where this cuts the other blank line of the angle, will be C: so is the triangle ABC constructed; let AC and BC be measured from the same scale of equal parts that AB was taken from, and the answers are found. |