Nat. Sof C.739239)127.954170(173.09=BC. 73 9239 5403027 2283540 6502300 CASE III. The angles and perpendicular given, to find the base and hypothenuse. Pl. 5. fig. 6. In the triangle ABC, there is the angle A 40°, and consequently the angle C 50°, with BC 170, given, to find AC and AB. 1st. By Construction. Make an angle CAB of 40° in blank lines (by prob: 16, sect. 4); with BC 170 from a line of equal parts draw the lines EF parallel to AB (by prob. 8, sect. 4), the lower line of the angle, and from the point where it cuts the other line in C let fall a perpendicular BC (by prob. 7, sect. 4), and the triangle is constructed: the measures of AC and AB, from the same scale that BC was taken, will answer the question. What has been said in the two foregoing cases is sufficient to render the operations in this, both by calculation, Gunter's scale, and natural sines, so obvious, that it is needless to insert them; however, for the sake of the learner, we give for Answers, AC 264.5, and AB 202.6. CASE IV. The base and hypothenuse given, to find the angles and perpendicular. Pl. 5. fig. 7. In the triangle ABC, there is given AB 300 and AC 500; the angles A and C and the perpendicular BC are required. 1st. By Construction. From a scale of equal parts lay 300 from A to B; on B erect an indefinite blank perpendicular line ; with AC 500 from the same scale, and one foot of the compasses in A, cross the perpendicular line in C; and the triangle is constructed. By prob. 17, sect. 4, measure the angle A, and let BC be measured from the same scale of equal parts that AC and AB were taken from; and the answers are obtained. 2d. By Calculation. 1. Making AC the radius. AC: R::AB: S.C. R: AC:: S.A : BC. is to radius, =90° =300 2.698970 10.000000 2.477121 12.477121 to the sine of C,= 36° 52' 9.778151 By cor. 2, theo. 5, 90°—36° 52'=53° 08', the angle A. As radius =90° 10.000000 2.698970 9.903108 to BC, 2.602111 Or BC may be found from cor. 2, theo. 14, sect 4. 3d. By Gunter's Scale. 1. Making AC the radius. Extend from 500 to 300, on the line of numbers ; that ex. tent will reach from 90°, on the line of sines, to 36° 52' for the angle C. Again, extend from 90° to 53° 08', on the line of sines, that extent will reach from 500 to 400, on the line of numbers, for BC. 2. Making AC the radius, the second stating is thus performed. Extend from radius, or the tangent of 45°, to 53° 08', that extent will reach from 300 to 400, for BC. The perpendicular and hypothenuse given, to find the angles and base. Pl. 5. fig. 8. In the triangle ABC there is BC 306 and AC 370 given, to find the angles A and C and the base AB. 1st. By Construction. Draw a blank line from any point, in which at B erect a perpendicular, on which lay BC 306, from a scale of equal parts : from the same scale, with AC 370 in the compasses * For finding the natural sines and co-sines, the reader is referred to table 3. draw the first drawn blank line in A, and the triangle ABC is constructed. Measure the angle A (by prob. 17, sect. 4), and also AB, from the same scale of equal parts the other sides were taken from, and the answers are now found. The operations by calculation, the square root, Gunter's scale, and natural sines are here omitted, as they have been heretofore fully explained: the statings, or proportions, must also be obvious, from what has already been said. Answers. The angle A 55° 48'; therefore the angle C 34° 12', and AB 208. CASE VI. The base and perpendicular given, to find the angles and hypothenuse.' Pl. 5. fig. 9. In the triangle ABC, there is AB 225 and BC 272 given, to find the angles A and C and the hypothenuse AC. 1st. By Construction. Draw a blank line, on which lay AB 225, from a scale of equal parts; at B erect a perpendicular; on which lay BC 272 from the same scale; join A and C, and the triangle is constructed. As before, let the angle A and the hypothenuse AC be measured, in order to find the answers. 2d. By Calculation. 1. Making AB the radius. AB: R::BC: T.A." R: AB :: sec. A: AC. 2. Making BC the radius. BC:R:: AB: T.C. R:BC:: sec. C: AC. By calculation, the answers from the foregoing proportions are easily obtained as before. But because AC, by either of the said proportions, is found by means of a secant, and since there is no line of secants on Gunter's scale, after having found the angles as before, let us suppose AC the radius, and then 1. S.A : BC::R: AC or 2. S.C: AB::R: AC. Given, the base AB 162 chains, LA Ans. BC 216 These proportions may be easily resolved, either by calculation or Gunter's scale, as before; and thus the hypothenuse AC may be found without a secant. From the two given sides the hypothenuse may be easily obtained, from cor. 1, theo. 14, sect. Thus, the square of AB=50625 124609(353=AC. 9 65)346 325 703)2109 2109 From what has been said on logarithms, it is plain, 1. That half the logarithm of the sum of the squares of the two sides will be the logarithm of the hypothenuse. Thus,* The sum of squares, as before, is 124609; its log. is 5.095549, the half of which is 2.547774; and the corresponding number to this in the tables will be 353, for AC. 2. And that half of the logarithm of the difference of the squares of AC and AB, or of AC and BC, will be the logarithm of BC, or of AB. The following examples are inserted for the exercise of the learner. Ex. 1. In the right-angled triangle ABC, Given, { the hypothenuse AC 540 perches, Ans. { { AB 449 To find the other two sides. Ex. 2. In the right-angled triangle ABC, AC 270 33° 45' To find the other two sides. { * Demonstration. The square of the hypothenuse of a right-angled tri. angle is equal to the sum of the squares of the sides (theo. 14); hence the log. of (AC2AB4)=the log. of BC?, and by the nature of logarithms the log. of BC is equal to the log. of BC? divided by 2; and in like manner the log. of (AB+BC)= the log. of AC%, hence dividing the log. of AC? by 2 gives the log.of AC. Q. E. D. |