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 45° 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 45° 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 contiguous to the angle. Thus: the given ang. 47° 30'. Nat. Sine .725374 Nat. Cos.=.688355 250 34417750 1376710 Base=172.088750 Perpend. 181.343500 CASE 11. The base and angles given; to find the perpendicular and hypothe nuse. PL. 5. fig. 5. 1 In the triangle ABC there is the angle A 42° 20', and of course the angle C47° 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 prop. 16. sect. 4.) in blank lines, as before. From a scale of equal parts lay 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. 3. Making BC the radius. T.C: AB:: Sec. C: AC That is, as the tangent of C-47° 40′ 10.040484 is to AB, So is the Secant of G =47° 40′ 2.278754 10.171699 12.450453 Or, having found one of the required sides, the other may be obtained, by one, or the other of the cors. to theo. 14. sect 4. 3d. By Gunter's Scale. 1. When AC is made the radius. Extend from 47° 40', to 900 on the line of sines; that distance will reach from 190 to 257, on the line of numbers, for AC, 2. When AB is made the radius, the first stating is thus performed: Extend from 45° on the tangents (for the tangent of 45° is equal to the radius, or to the sine of 90° as before) to 42° 20′; that extent will reach from 190, on the line of numbers, to 173, for BC. 3. When BC is made the radius, the second stating is thus performed: Extend from 47° 40′ on the line of tangents, to 45°, or radius; that extent will reach from 190 to 173, on the line of numbers, for BC; for the tangent of 47° 40', is more than the radius, therefore the fourth number must be less than the second, as before. The two first statings of this case, answer the question without a secant. Nat. S of C, side AB × R. Thus,739239) 190.000000 (257.02 &c.=AC. 147,8478 4215220 3696195 5190250 5174673 1557700 1478478 and, .673443 Nat. S. of A. 190-side AB. 60609870 |