CASE 4th. All the fides of an oblique triangle given, to find the angles. Solution, by letting fall a perpendicular from the greateft ang e to the longeft fide or bafe, dividing that bafe into two unequal fegments or parts. Take the fum of the two leaft fides; and half their difference, then state. As the base or longeft fide: the fum of the other two fides :: half their difference: half difference of the fegments of the base. Then half the base + half difference of fegments of base, is the greater fegment; and half the bafe half the difference of the fegments, is the less segment of the bafe. Again as the longer fide: radius :: the greater fegment: co-fine of the included between the bafe and longer leg Or one angle may be found by logarithms. Thus, from half the fum of three fides, fubtract each fide feverally, but first the fide oppofite to the required angle, then. As rectangle of half the fum of the fides and first remainder: fquare of radius. :: The rectangle of the other two remainders. : fquare of the tangent of half the angle fought. Which may be thus done by logarithms. Take the compliment arithmetic of the logarithms of half the fum of fides, and firft remainder feparately. And the logarithms of the other two remainders, then collect these two logarithms and the two arithmetic compliments into one fum, and half it, then that half found in the table of artificial tangents, and the degrees and min. Correfponding doubled, is the angle equired. Whence the other angles may be found, by ftating oppofite fides to the fines of their oppofite angles, and need not be repeated. So fhall conclude, with two or three examples as follows, wrought out both by natural fines, tangents, &c. and by logarithms, &c. First, in case the first of right angled triangles. If the at the bafe be 52d. 25m. and the hypotheaufe 156 yards required the prependicular? ift. Sol. For perpendicular, hypothenufe being radius. Log. 2d. term + Log. 3d. — Log. 1ft. = Log. 4th. By fcale and compaffes, extend from the firft term to the third, and that extent will reach from the second to the fourth or answer, carefully minding the lines refered to in the proportion. 2d. Solution, the bafe being radius, for perpendicular. As fec. of at base 52.25 nat. fec. 1.639 10.214731 : The hypothenufe 156yds. 156 2.193125 :: Tang. ofat base 52.25 nat. tang 1.299 10.113712 123.6 2.092106 : The perp. 123.6yds. By natural fecants and tangents. 156×1.229 1.639 123.6 anfwe r. By logarithms. Logarithm of fecond term + logarithm thirdlogarithm firft logarithm fourth as above. = 3d. Solution, the perpendicular being radius, to find the perpendicular. Thus As As co-fec. at bafe 52.25 nat.co-f. 1.262 log. 10.101010 : The hyp. 156yds. :: The tang. of rad.45 00 nat. tang.I 156yds. 2.193125 10 000000 123.6yds. 123.6yds. 2.092115 : the perp. 156×1 By natural fecants and tangents. 1.262 123.6 the perpendicular. By logarithms. Logarithm fecond + logarithm of third-logarithm of first = logarithm fourth. Anf. The application of Trigonometry to Longimetry and Altimetry, acceffable and inacceffable. Admit a monument ftanding upon a plain (as for inftance, that erected in memory of King EDWARD, on Brugh-marfh, not far from Carlifle) the distance of which is required from a fhip in Solway firth. The practical method, measure from the monument in any convenient direction, but not directly towards the fhip 4 or 5 hundred yards, or more to a station; then with a good inftrument as a theodolite, take the angle at that station, between the monument and ship, alfo at the monument take the angle between the ship and station. As for example let the former angle be 115d. 40m. and the latter 36 50; diftance of the ftation and monument 700 yards, to find the diftance of the ship from the monument. = 180 degrees-115d. 40m. +36d.50m. 27d. 30m. the angle at the ship. Then, as fine of at fhip 27 30 : The measured distance 700 yds. :: Sine of fup. of Lat ftation 115 40 Co. Ar. log. 0.33560 2.84509 9.95488 : Distance required — 1366yds. answer 3.13557 . Second, admit the angle of altitude of faid monument at 50 yards diftance therefrom be 381d. quere the height of the monument? per Trigo. by Log. Co. Ar. As co-fine of altitude obferved 38d. 30m. 0.10646 The ground distance :: Sine of the observed : The height required 50 yards 1.69897 Third, admit the angle of altitude between the horizontal line, and apex of the monument, at a certain distance unknown be 381d. and at the distance of 50yd. direaly farther from the monument the faid of altitude be 21d. 40m. required its height? per Trigo. by Log. As fine of the diff. of ob. angles 16d. 50m. : The measured distance :: Sine of lefs L : A fourth number Again as radius god. oom. Co. Ar. 0.53822 N. B. The fmall difference in the heights in the fecond and third viz. 1 tenth of a yard and a half +, arifes from the inacuracy of the angles. On the fea coaft of Cumberland ftands Dubmill, and Bankend, on oppofite fides of Alonby bay, from each. of which appear two white houfes, which for diftinction we call A, and B, on the Scotch fhore, whofe distance from Dubmill, and Bankend, and from one another are required from the data fubjoined ? Suppofe at Dubmill, the line to houfe B, bifects the angle there between Bankend, and house A, then admit 80 chains meafured from Dubmill in the line from house A, fo that a line to Bankend was perpendicular thereto, and there a fignal put up, or a fire made, next at Bankend, the following angles were obferved. ft. Between the fignal or fire and Dubmill 12d. 30m. zd. Between Dubmill and house A 3d. Between houfe A and B 54 30 61 30 The answers are put down, and the work is left for exercife to the young ftudent. 义 THE |