BG:GC::BF:FE.
But DC is the difference of the sides t A and n c ; and because D F : F A G C; but D F and FA are equal, whence Da and G c are equal ; therefore, n o or a c, is half the difference of the sides ; now half the difference of the sides, added to the lesser side, is half the stun of the sides. It is evident, that BA D is half the sum of the angles c AB and B CA, and D A C is half their difference; now, by con sidering At as radius, BF will be the tangent of the angle B A F, or B A D, and F E the tangent of the angle F A E, or DA C ; therefore, as B a, half the sum of the two sides, is to c, half their difference, so is B F, the tangent of the half sum of. the opposite angles, to F E, the tangent of half their difference ; and, consequently, as the sum of the two sides is to their difference, so is the tangent of the half sum of the opposite angles, to the tangent of half their difference.
Theorem any plane triangle, if a perpendicular be let fall upon the longest side from the opposite angle ; then, as the sum of the segments of the longest side, or base, is to the sum of the other two sides, so is the difference of the sides to the difference of the segments of the base.
Figure 3.—In the triangle A B c, left fall the perpendicular A E upon BC; about A as a centre with the distance, A c, of the shortest side, describe a circle : produce BA to meet the circumference in a ; then will B o be the sum of the two sides, and n F their difference ; also, BE and EC are the segments of the base, and B D their difference; now, by the property of the circle, na=pc X n n; whence, BC : B G : : B F : BD; that is, as the longest side, n c, is to R a, the sum of the other two sides, so is n F the difference of the sides, to BD, the dikrence of the segments of the base.
When two angles of a triangle are given, the third angle found by subtracting the sum of the two given angles from I SO°, or two right angles ; and, consequently, if one of the acute angles of a right-angled triangle be given, the remain ing angle will be fimnd, by subtracting the acute angle from 90 ; for, in this case also, two angles are given, viz., one of
the acute angles and the right angle • and these being sub tracted from two right angles, or 180, the other acute angle will be found.
Various propositions might be given, but the preceding are sufficient for every case of plane trigonometry.
The analogies of plane trigonometry may easily be deduced from those ofspherical trigonometry ; the former being par ticular eases of the latter.
Proposition.— The sum, s, and difference, d, of two quan tities, x and y, being given ; to find the quantities themselves. Let .r y s x — y = d add these equations together, and 2x=s+d or subtract these equations from each other, and 2y=s — d, — or y = s whence x, the greater of the two quantities, is half the sum and difference of these quantities ; and y, the lesser, is half of the difference between the sum and differ ence ; or, in other words, the half sum added to the half difference, gives the greater quantity ; and the half sum sub tracted from the half difference, gives the lesser quantity.
Solutions of the three cases of triangles.
Every plane triangle consists of six parts, the three sides and the three angles ; three of these parts must always be given, and of these given parts one at least must be a side ; to find the remaining three parts.
Case 1.—Two angles and a side being given, to find the remaining sides.
As the sine of the angle opposite the given side is to the sine of the angle opposite the required side, so is the given side to the required side.
In the oblique-angled triangle, A B c, given the angle A 59°, the angle c 52°15', and the side AB 276.5, to find A and n c.
We shall find angle B = 180° — (52° 15' 59°) = .45, as the sine < c 52° 15' 9.89801 is to sine of < n 68°45'.
SO is the side A 76.5 2.•4170 12.411 I 2 to the side AC 325.9 2.51311 To find B c.
As the sine of < c 52° 15' 9.89801 is to the sine of < A 59° 9.93307 so is A u 276.5 2.44170 12.37477 to B c 299 S 2.47676