OBLIQUE ANGLED TRIGONOMETRY, THE folution of all plane triangles, may be deduced from, the three following theorems. THEOREM I. In any plane triangle, the fides are in the fame proportion, as the fines of the oppofite angles. Fig. 53. plate 4. Dem. From the angles A and B, draw BE and AD perpen dicular, to the oppofite fides, BC and AC produced if neceffary. Because the triangles ADB, AEB, are right angled triangles, the fide AD becomes the fine of the angle ABD, and BE the fine of the angle BAE; if AB the hypothenufe, common to both the triangles, be made the radius; but the two triangles ADC, BEC, have each a right angle at D and E, likewise the common angle ACB, therefore, they are fimilar, and confequently, BC is to CA, as BE is to AD; that is, the fides are in the fame proportion as the fines of the oppofite angles THEO THEOREM II. In any plane triangle, the fum of any two fides is to their difference as the tangent of half the fum of the angles at the bafe, is to the tangent of half their difference. Fig 54. plate 4. Dem. LET ABC be a plane triangle, AB+BC: AB—BC : : tan. ang. A+Ang. C: tan. ang. C-ang. A, upon A as cen 2 tre with AB the longeft fide for a radius, defcribe a circle, meeting AC produced in E and F ; produce BC to D, join DA, FB, EB, and draw FG parallel to BC, meeting EB in G. The angle EAB is equal to the fum of the angles at the base, and the angle EAB at the centre is double the angle EFB at the circumference, therefore, EFB is half the fum of the angles at the bafe; but the angle ACB, is equal to the angles CAD, and ADC, or ABC together, therefore, FAD is the difference of the angles at the base, and FBD is half that difference, but FBD is equal to the alternate angle BFG; fince the angle FBE in a femi-circle, is a right angle, FB being radius, BE, BG will be tangents of the angles EFB, BFG; but it is plain, that EC is the fum of the fides, BA and AC, also that CF is their difference; and fince EG and BC are parallel, EC: CF as EB: BG, that is, the fum of the fides is to their difference: as the tangent of half the fum of the angles at the base, is to the tan gent of half their difference. THEO THEOREM III. In a plane triangle, the base is to the sum of the fides, as the difference of the fides, is to the difference of the fegments of the base, made by the perpendicular upon it from the vertex. Fig. 55. plate 4. LET ABC be a plane triangle, if from B the vertex a perpendicular BD be dropped on the bafe, AC: AB+BC:: BC—AB: DC-AD. Upon B as centre with BC, the greater fide for a radius, describe a circle meeting BA and CA, produced in F and E. It is manifeft, that AF is the difference of the fides, and that EA is the difference of the fegments of the bafe, for ED and DC are equal, and AG is the sum of AB and BC; but, because FG and EC cut each other within a circle in the point A, the rectangle contained by the fegments of the one, is equal to the rectangle contained by the fegments of the other that is, FAXAG=EA×AC, and by Euclid vi. 16. AC: AG::FA: EA. Wherefore, in any plane triangle, the base is to, &c. Note, 'The fum and difference of two magnitudes being given, to find each of them. Rule, To half the fum, add half the difference, the fum will be the greater, and from half the fum, fubtract half the difference, the remainder will be the lefs. In plane triangles may be given, The three angles and one side. Two fides and the angle oppofite to one of them. to find the other parts. PRO PROBLEM I. The angles and one fide given, to find the two remaining fides. As fine ang. C52° 15' 9.89801 | As fine ang. C 52° 15′ 9.89801 2.44170 is to AB 276.5 2.44170 is to A B 276.5 So is fine an. B68° 45′ 9.96942 | So is fine ang. A 59° 9.93307 CAB=2600 Given Ang. A 47° 30′ Req. AC BC. Ang. C41°15' 180--47°30+41°15'=91°15'ang.B.. The fine, tangent, fecant, &c. of any arch, is the fine, tangent, fecant, &c. of its fupplement. Hence the fine of 91° 15′ may be obtained thus, 180°-91° 15′=88' 45 the fupplement of 91° 15'. 3. |