SCHOLIU M. The foregoing propofitions contain all that is neceffary to the folution of the different queftions in plane trigonometry. It is, however, useful in fome of the cafes of oblique angled triangles, to be provided with other methods of calculation; on which account the following propofitions are added. I PRO P. VI. F, as the greater of any two fides of a triangle to the lefs, fo the radius to the tangent of a certain angle; then will the radius be to the tangent of the difference between. that angle and half a right angle, as the tangent of half the fum of the angles at the base of the triangle to the tangent of half their difference. C Let ABC be a triangle, the fides of which are BC and CA, and the bafe AB, and let BC be greater than CA. Let DC be drawn at right angles to BC, and equal to AC; join BD, and because (Prop. 1.) in the right angled triangle BCD, BC: CD::R: D tan. CBD, GBD is the angle, of which the tangent is to the radius as CD to BC, that` is, as CA to BC, or as the A leaft of the two fides of the triangle to the greatest. But BC+CD: BC-CD :: tan. (CDB+CBD): tan. (CDB-CBD), (Prop. 5.); and and alfo, BC+CA: BC-CA:: tan. (CAB+CBA): tan. (CAB-CBA). Therefore, fince CD=CA, tan. tan. ཀཱ (CDB+CBD): tan: (CAB+CBA) : tan. † (CAB—CBA). But, becaufe the angles CDB+CBD 90°, tan. ÷ (CDB+CBD): tan. (CDB-CBD) :: R: tan. (45°-CBD), (2. cor. Prop. 3.); therefore, R : tan. (45°-CBD): : tan. (CAB+CBA): tan. (CAB-CBA); and CBD 1 14 was already fhewn to be fuch an angle that BC: CA:: R: tan. CBD. Therefore, &c. Q. E. D. FOO PROP. VII. OUR times the rectangle contained by any two fides of a triangle, is to the rectangle contained by two straight lines, of which one is the base or third fide of the triangle increased by the difference of the two fides, and the other the bafe diminished by the difference of the fame fides, as the fquare of the radius to the fquare of the fine of half the angle included between the two fides of the triangle. Let ABC be a triangle, of which AB is the bafe, and CB the greater of the two fides. From B draw BD at right angles to AC. Cafe I. When ACB is an acute angle, fo that BD falls within the triangle. From the centre C, with the distance CB, defcribe the circle EBF, meeting AC produced in E and F. Join BE, BF; and it is evident, that EBF is a right angle, and also, that the angle EFB is half the angle ECB. Then Then, (13. 2.) AC2+CB2, that is, AC2+CÈ2=AB2+ 2AC.CD. But, (7.2.) AC2+CE2 2AC.CE+AE2, therefore, 2AC.CE+AE2— AB2+ 2AC.CD; and if the rectangle 2AC.CD be taken from both, 2AC.CE-2AC.CD+AE2 AB2, and therefore, because 2AC.CE-2AC.CD= 2AC.ED, 2AC.ED+ AE-AB', or 2AC.ED AB'-AE'. But (5. 2.) AB'—AE'-(AB+AE)(AB—AE), therefore, 2AC.ED=(AB+AE)(ÁB—AE). Again, AC:CE:: 2AC.DE: 2CE.DE, because rectangles that have equal altitudes are as their bases; and, for the fame reafon, 4AC.CE 4CE: 2AC.DE: 2CE.DE, or alternately, 4AC.CE: 2AC.DE:: 4CE: 2CE.DE. Now, 4CE EF, and 2CE.DE-FE.DE-EB2 (cor. 8. 6.), therefore, 4AC.CE: 2AC.DE:: EF2: EB2. But EF: EB:: R: fin. EFB :: R: fin. C, whence, 2 4AC.CE: 2AC.DE:: R2: (fin. C). And it was proved, that 2AC.ED= (AB+AE)(AB-AE), therefore 4AC.CE : (AB+AE)(AB—AE): : R2 : (fin. 1⁄2 C)2. Cafe 2: Cafe 2. Let C be an obtufe angle, so that the perpendicu lar DB falls without the triangle. Then, because AC2+ CB2+ 2AC.CD=AB2 (12. 2.), or AC2+CE2+ 2AC.CD-AB2; and because AC2+CE2- 2AC.CE+AE2, 2AC.CE+ 2AC.CD+ AE AB2. Now, 2AC.CE+ 2AC.CD= 2AC.ED, therefore 2AC.ED+AE AB. Therefore, it will be demonftrated as before, that 4AC.CE: (AB+AE)(AB—AE) : : R2: (fin. 1 C)2. B E Now, 4AC.CE, or 4AC.CB is four times the rectangle under the fides, and AE being the difference of the fides, (AB+AE)(AB-AE) is the rectangle which has for one of its fides the base increased by the difference of the two fides of the triangle, and for its other fide, the base diminished by the fame difference. Therefore, in any triangle, &c. Q.E. D. now, COR. Because BDC is a right angle, BC: CD:: R: fin.CBD ; fin. CBD cof. C, therefore, BC: CD:: R: cof. C, and 2AC.CB 2AC.CDR cof. C. But, (13.2.) 2AC.CD=AC2+CB2-AB2, therefore, : 2AC.CB: AC2+CB2-AB2 :: R: cof. C; that is, twice the rectangle AC.CB, is to the excess of the fquares of AC and CB above the fquare of AB, as radius to the co-fine of the included angle ACB. PROP. Then, (13. 2.) AC2+CB2, that is, AC2+CE2-AB2+ 2AC.CD. But, (7.2.) AC2+CE2= 2AC.CE+AE2, therefore, 2AC.CE+AE2 AB2+ 2AC.CD; and if the rectangle 2AC.CD be taken from both, 2AC.CE-2AC.CD+AE2 =AB2, and therefore, because 2AC.CE-2AC.CD= 2AC.ED, 2AC.ED+ AE'=AB', or 2AC.ED AB'-AE'. But (5. 2.) AB-AE'-(AB+AE)(AB-AE), therefore, 2AC.ED=(AB+AE)(AB—AE). 1 Again, AC:CE:: 2AC.DE: 2CE.DE, because rectangles that have equal altitudes are as their bases; and, for the fame reafon, 4AC.CE : 4CE2::2AC.DE: 2CE.DE, or alternately, 4AC.CE: 2AC.DE::4CE2:2CE.DE. Now, 4CE2-EF2, and 2CE.DE-FE.DE=EB2 (cor. 8. 6.), therefore, 4AC.CE: 2AC.DE:: EF2: EB2. But EF: EB:: R: fin. EFB :: R: fin. C, whence, EF2 : EB2 : : R2 : (fin. C)2, and therefore, 2 4AC.CE: 2AC.DE:: R2: (fin. C). And it was proved, that 2AC.ED= (AB+AE)(AB-AE), therefore 4AC.CE: (AB+AE)(AB-AE) :: R2: (fin. C)2. Cafe 2: |