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Schol.—The preceding propositions are sufficient for the solution of all the cases of trigonometry; but, in particular cases, some of the following propositions may be employed with advantage.
PROPOSITION VIII. THEOREM.
biB In any triangle, twice the rectangle contained by any two of its sides, is to the difference between the sum of their squares, and the square of the other side, as radius to the cosine of the angle contained by these two sides.
Given ABC any triangle; to prove that 2AB · BC is to the difference between AB? + BC and AC as radius to cos. B. spa
(Const.) For from C draw CD perpendicular to AB or AB produced. Then BC : BD = R : cos. B. But (VI. 1)
..be BD Å
SA Р В Р А
(18 2AB · BC : 2AB · BD = BC: BD; therefore 2AB · BC: 2AB • BD = R:cos. B. But (II. 12, 13, otherwise), 2AB. BD is the difference between AB? + BC2 and AC?; and hence the proposition is proved. COR.-If the sides opposite to the angles A, B, and C be called
a, b, and c respectively, and radius = 1, then 2ac:a? + c –
R_a2 + c 62
11 the numeri. cal value of the cosine calculated from this expression be positive, the angle B is acute, but if negative, the angle B is obtuse; · but the cosine of an obtuse angle is negative,
hence, in either case, the above expression gives the true cosine of the angle B. From the above expression, it is evident that the cosine of an angle of a triangle is equal to the sum of the squares of the two sides that contain the angle, diminished by the square of the side opposite the angle, divided by twice the product of the sides which contain the
angle, Hence also
By multiplying both sides of these three values for the three cosines of the angles of a triangle by the denominators of the second side, and, transposing, they become
a’ = 12 + c* — 2bc cos. A,
PROPOSITION XI. THEOREM. In any triangle, the rectangle under the semiperimeter, and its excess above the base, is to the rectangle under its excesses above the two sides, as the square of the radius to the square of the tangent of half the vertical angle.
For (Prop. X.) AC-CB: S(S - AB) = R2 : cos.4C, and (Prop. IX.) AC CB:(S – ÀC)(S – BC) = RP. sin.4C, therefore, by direct equality, S(S – AB): (S – AC)(S – BC) = cos. ¿C:sin.. But (Cor. 7 to Definitions, and 'VI. 22, Cor.)
PROPOSITION XII. THEOREM. Let AB, AC, and AD be three such arcs, that BC the difference of the first and second is equal to CD the difference of the second and third; the radius is to the sine of AC, the middle arc, as the cosine of the common difference BC to half the sum of the sines of AB and AD, the extreme arcs.
(Const.) Draw CE to the centre, and BF, CG, and DH perpendicular to AE,
then BF, CG, and DH are the sines of the arcs AB, AC, and AD.
Join BD, and let it meet CE in
also draw IM and BL perpendicular to DH.
(Dem.) Then, because the arc BD is bisected in C, EC is at right angles to BD, and bisects it in I; also BI is the sine, and El the cosine of the arc BC or CD. And, since BD is bisected in I, and IM is parallel to BL, LD is bisected in M (2. VI.) and LB is bisected in N. Now BF is equal to LH, therefore BF + DH = DH + HL = DL + 2LH = 2LM + 2LH = 2HM = 2IK; and therefore IK is half the sum of BF and DH. But because the triangles ECG and EIK are equiangular, having the angles at G and K right angles and the angle at É common, EC: CG :: EI : IK; but it has been shewn that CG is the sine of the arc AC, EI the cosine of the arc BC;
therefore R: sin. AC :: cos. BC: (sin. AB + sin. AD). COR. 1.-Radius is to the cosine of AC, the middle arc, as the
cosine of the common difference is to half the sum of the cosines of the extreme arcs.
(Dem.) For EH is the cosine of AD, and EF is the cosine of AB, but EH + EF = 2EH + HF = 2EH + 2HK = 2EK, therefore EK = half the sum of the cosines of the extreme arcs.
Now the triangles ECG and EIK are equiangular, hence EC:EG :: EI: EK, that is, R:cos. AC :: cos. BC: (cos. AD + cos. AB). Cor. 2.-Radius is to the sine of AC, the middle arc, as the
sine of CD, the common difference, is to half the difference of cosines of the extreme arcs.
(Dem.) The difference of the cosines of the extreme arcs is evidently HF = LB = 2LN = 2MI; therefore, MI = half the difference of the cosines of the extreme arcs. Now the triangles BCG and DMI are equiangular, for the angles at G and M are right angles, and the angles DIM and ECG are equal, for they are the complements of the equal alternate angles MIE and IEĞ, therefore EC: CG :: DI:IM, or R: sin. AC :: sin. CD: (cos. AB – cos. AD).
COR. 3.-Radius is to the cosine of AC, the middle arc, as the sine of DC, the common difference, is to half the difference of the sines of the extreme arcs.
(Dem.) The difference of the sines of AD and AB is the difference of DH and BF, which is DH - LH = DL = 2DM;
therefore DM is equal to half the difference of the sines of AD and AB. And since it was proved in last corollary that the triangles CEG and IDM are equiangular, EC: EG :: ID : DM,
or R: cos. AC :: sin. CD: }(sin. AD – sin. AB).