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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.

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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.

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Given ABC any triangle; to prove that 2AB BC is to the difference between AB2 + BC2 and AC2 as radius to cos. B.

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(Const.) For from C draw CD perpendicular to AB or AB produced. Then BC BD = R: cos. B. But (VI. 1)

44D

2AB

B

BC: 2AB BD = BC: BD;

2AB BD = R: cos. B.

B

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therefore 2AB BC:

But (II. 12, 13, otherwise), 2AB · BD is the difference between AB2 + BC2 and AC2; and hence the proposition is proved.

=

= 1, then 2ac: a2 + c2· a2 + c2 b2

If the numeri

COR.-If the sides opposite to the angles A, B, and C be called a, b, and c respectively, and radius b21: cos. B, or cos. B = = 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,

2ac

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

and

cos. A =

b2 + c2
2bc

a

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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

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In any triangle, the rectangle under the two sides is to that under the excesses of the semiperimeter above these sides, as the square of the radius to the square of the sine of half the vertical angle.

(Let ABC be any triangle, produce CB till CD = CA; join AD, and draw CE bisecting the angle C, then (I. 4) CE is perpendicular to AD. Draw BF parallel to DE, and BG to CE. Then AC: AE = Rad.: sin. C (Prop. IV.) if AC be radius; also BC BF or EG Rad.: sin. 1C; hence (VI. 23, Cor.) AC CB: AE EG R2: sin.2 C.

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But (II. c. Cor. 1) AE EG = (AB +
BD)(AB - BD) in the triangle ABD.
But AB + BD = AB + CD – CB
= AB+ AC — CB = AB + AC + CB

if S = semiperimeter.

= AB (AC - CB)

— 2AC = 2(SAC);

or

or

R2: sin.21C.

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B

- 2CB = 2(SCB), = AB (CD - CB) ABAC + CB

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And AB BD AB AC + CB therefore ACCB: (S - AC) (S — BC)

COR.-If the numerical values of the sides opposite to the angles A, B, and C be called a, b, and c respectively, and if s = the semiperimeter and R = 1, then

ab: (8 - a)(s — b) = 12: sin.2 C.

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and

sin.24B =

(s — a) (s — c)

ac

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In any triangle, the rectangle under its two sides is to that under the semiperimeter, and its excess above the base, as the square of the radius to the square of the cosine of half the vertical angle.

and draw the figure ACDH as

Let ABC be any triangle, in the last; produce BF to meet AC produced in E, and draw CG parallel to AD.

Then angle ACH =

CBE, and CE = CB.

E, and DCH

=

= CBE, therefore E Also since CG is parallel to AD, the

angles at G are right angles. And HG is a rectangle, therefore CH= GF. Now, AC: CH or GF = : R: cos. C, and CE or CB: EG = R: cos. E or cos. C; therefore (VI. 23, Cor.) AC CB: EG· GF = R2: cos.21C. But EG GF = EB (EF+FB) (II. c. Cor. 1) (EA+ AB) × (EA AB) =

=

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(BC + CA + AB) × (BC + CA +

AB - 2AB) = S(SAB); since

B

H

F

G

EF and FB are the segments of the base of the triangle EAB made by the perpendicular AF from the vertex; therefore AC CB: S(SAB) R2: cos.2 C.

=

COR.-Take a, b, c, &c., as in Cor. to last proposition, and

ab: s(sc)
s(sc), or

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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.

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For (Prop. X.) AC CB: S(SAB) = R2: cos.2C, (Prop. IX.) AC CB: (SAC)(S therefore, by direct cos.2 C: sin.2 C.

and

BC) = R2. sin.2 4C, equality, S(S — AB): (S — AC)(S — BC) = But (Cor. 7 to Definitions, and VI. 22, Cor.)

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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 I; draw IK perpendicular to EA, also draw IM and BL perpendicular to DH.

M

L

E.

KG FA

also BI is the sine,

(Dem.) Then, because the arc BD is bisected in C, EC is at right angles to BD, and bisects it in I; and EI 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 E common, EC: CG:: EI: IK; but it has been shewn that EI the cosine of the arc BC; therefore R : sin. AC:: cos. BC: (sin. AB + sin. AD).

CG is the sine of the arc AC,

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 EHEF = 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 IEG, 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).

COR. 4.-If AC be considered as one arc, and CB or CD as another, then AD is their sum and AB is their difference; and if AC be called A, and CB or CD be called B, then the arc AD = (A + B), and AB = (A — B); also, if radius = 1, then the proposition becomes

1: sin. A :: cos. B: {sin. (A + B) + sin. (A — B)};

.. {sin. (A + B) + sin. (A – B)} = sin. A cos. B;

similarly, from Cor. 3,

{sin. (A + B) — sin. (A — B)} = cos. A sin. B ;

adding and subtracting these, gives

sin. (A + B) = sin. A cos. B + cos. A sin. B

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In (A), let B = A, then sin. 2A = 2 sin. A cos. A

(C).

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