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Let ABCDEF be a polygon: then will AF AB. COS A + BC. COS CBA FA + CD COS CDA AF + DE. COS DE AAR + EF. COS EFA AF*.

For, drawing lines from the several angles, respectively parallel. and perpendicular to AF; it will be


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вр = BC. cos CBB = BC. COS CBAAF,
cd=&D CD. cos CD

de = εE = DE. COS DEƐ


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CD. cos CD ^ AF,



But AF bc + cd + de + eF Ab; and ab, as expressed above, is in effect subtractive, because the cosine of the obtuse angle BAF is negative. Consequently,

AF AC + cd+de+ eF = AB. CUS BAF + BC. COS CBAAF + &c, - as in the proposition. A like demonstration will apply, mutatis mutandis, to any other polygon.

Cor. When the sides of the polygon are reduced to three, this theorem becomes the same as the fundamental theorem in chap. ii, from which the whole doctrine of Plane Trigonometry is made to flow.


The Perpendicular let fall from the Highest Point or Summit of a Polygon, upon the Opposite Side or Base, is Equal to the Sum of the Products of the Sides Comprised between that Summit and the Base, into the Sines of their Respective Inclinations to that Base.

Thus, in the preceding figure, cc=CB. Sin CBAFA+BA.Sin A; or cc CD. sin CDA AF + DE . sin DE^ AF + EF. Sin F. This is evident from an inspection of the figure.

Cor. 1. In like manner nd = DE. sin DE A AF + EF. Sin F, or Dd CB. sin CBFA + BA. Sin A

A CD. sin CD AF. Cor. 2. Hence, the sum of the products of each side, into the sine of the sum of the exterior angles, (or into the sine of the sum of the supplements of the interior angles), comprised between those sides and a determinate side, is = + perp. perp. or 0. That is to say, in the preceding figure, AB. sin A + BC. sin (A+B) + CD. sin (A+B+c)+DE. sin (A+B+C+D) + EF. Sin (A+B+C+D+E) = 0.

* When a caret is put between two letters or pairs of letters denoting lines, the expression altogether denotes the angle which would be made by those two lines if they were produced till they met: thus CRAFA denotes the inclination of the line ca to FA.


Here it is to be observed, that the sines of angles greater than 180° are negative (ch. ii equa. vII).

Cor. 3. Hence again, by putting for sin (A+B), sin (A+B+C), their values sin A. COS B+ sin B. Cos A, Sin A. COS (B+C) + sin (BC). Cos A, &c (ch. ii equa. v), and recollecting that tang = (ch. ii p. 55), we shall have,



sin A.(AB+BC.COSB+CD.Cos(B+c)+DE.COS(B+C+D)+&C) +cos A. (BC.sin


and thence finally, tan 180° - A, or tan BAF =

BC sin B+ CD in (B+C+DE. sin (B+C+D) + EF. Sin (B+C+D+E) AB+ BC. COS B + CD. Cos (B+ c) + DE. COS (B + C + D) + E F · COS (B+C+D+ E)

A similar expression will manifestly apply to any polygon; and when the number of sides exceeds four, it is highly useful in practice.

Cor. 4. In a triangle ABC, where the sides AB, BC, and the angle ABC, or its supplement B, are known, we have

tan CAB


sin B

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in both which expressions, the second term of the denominator will become subtractive whenever the angle ABC is acute, or B obtuse.


The Square of Any Side of a Polygon, is Equal to the Sum of
the Squares of All the Other Sides, Minus Twice the Sum
of the Products of All the Other Sides Multiplied two and
two, and by the Cosines of the Angles they Include.
For the sake of brevity, let the sides
be represented by the small letters which
stand against them in the annexed figure:
then, from theor. 1, we shall have the
subjoined equations, viz.

α = b. cos ab c. cos a^c + ♪
b = a. cos a^b + c COS bac + 8

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cos and, cos bad,

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c = a. cos a^c + b,. cos b^c + d. cos c^d,

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d = a cos and b. cos b^d + c. cos cad. Multiplying the first of these equations by a, the second by b, the third by c, the fourth by d; subtracting the three latter products from the first, and transposing b, c, d, there will


a2 = b2 + c2 + d2 — 2 (bc.cos b^c+bo.cos bad+co.cos c^d). In like manner,

c2=a2 + b2 + d'—2(ab. cos a^b+as. cosand+bo.cos bad).

&c. &c.



Or, since bec, bad = c + D 180°, CAD, we have a2 = b2 + c2 + 2(bc.cos c-b. cos (c+D)+co. cos D), · c2 = a2 + b2 + ♪2-2(ab. cos B − bò.cos (A + B) +að. cos A). &c. &c.

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The same method applied to the pentagon ABCDE, will give bd. cos (c+D) + be. cos (C + D + E) ? b2 + c2 + d2 + e2 — 2 ) bc. cos c -+ cd. cos Dce . cos (D + E) + de . cos E And a like process is obviously applicable to any number of sides; whence the truth of the theorem is manifest.

Cor. The property of a plane triangle expressed in equa. I ch. ii, is only a particular case of this general theorem.


Twice the Surface of Any Polygon, is Equal to the Sum of the Rectangles of its Sides, except one, taken two and two, by the Sines of the Sums of the Exterior* Angles Contained by those Sides.

1. For a trapezium, or polygon of four sides. Let two of the sides AB, DC, be produced till they meet at P. Then the trapezium ABCD is manifestly equal to the difference between the triangles PAD and PBC. But twice the surface of the tri

angle PAD is (Mens. of Planes pr. 2 rule 2) AP. PD. sin P = (AB + BP). (DC + CP). sin P; and twice the surface of the triangle PBC is BP. PC. sin P: therefore their difference, or twice the area of the trapezium, is = (AB. DC+AB.CP + DC. BP). sin P. Now, in ▲ PBC,

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BC. sin B

BC. sin c

sin P

Substituting these values of PB, PC, for them in the above. equation, and observing that sin P = sin (PEC+ PCB) = sin sum of exterior angles B and C, there results at length,

Twice surface

of trapezium. S


it follows

AB. BC. sin B +AB. DC. sin (B+C) + BC. DC. sin c. twice triangle ABC, that twice triangle ACD is equal to the remaining two terms, viz, AB. DC. sin (B+ c) + BC. DC. sin

Cor. Since AB. BC . sin B =

twice area ACD =



* The exterior angles here meant, are those formed by producing the sides in the same manner as in th. 20 Geometry, and in cors. 1, 2, th. 2, of this chap.

2. For

2. For a pentagon, as ABCDE. Its area is obviously equal to the sum of the areas of the trapezium ABCD, and of the triangle ADE. Let the sides AB, DC, as before, meet when produced at P. Then, from the above, we have

Twice area of

the trapezium



AB. BC . sin B


+AB. DC. sin (B+c)
+ BC DC. sin c.

And, by the preceding corollary,

Twice triangle


That is, twice triangle DAE


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BC. sin c

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AP. DE. Sin (P+D) or sin (B+C+D) + DP. DE. sin D.

AB DE. sin (B+C+D)

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+ DC. DE. sin D

+BP. DE. sin (B+C+D)
+CP. DE. sin D.

BC. sin B

sin (B+c) BC. DE. sin c. sin (B+C+D) sin (B+c)

Now, BP =

and CP =

sin (B+ c)' two terms become

=BC.DE ·

sin B. sin D + sin c. sin (B+C+D)

sin (B+ c)

:therefore the last


BC. DE. Sin B. sind sin (B+C)

: and this expression,

by means of the formula for 4 arcs (art. 30 ch. iii), becomes BC. DE. sin (c+D). Hence, collecting the terms, and arranging them in the order of the sides, they become

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Cor. Taking away from this expression, the 1st, 2d, and 4th terms, which together make double the trapezium ABCD, there will remain

Twice area of
the triangle =



AB. DE. sin (B+C+D) +BC. DE. sin (C+D)

DC. DE. sin D.

3. For a hexagon, as ABCDEF. The double area will be found, by supposing it divided into the pentagon ABCDE, and the triangle AEF. For, by the last rule, and its corollary, we have,


Twice area of
the pentagon


Twice area of)
the triangle=


Or, twice area the triangle



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BP. EF. sin (B+C+D+E) +CP. EF . sin (D+E).

Now, writing for BP, CP, their respective values,

BC.sin c


BC. sin B sin (B+c)'

the sum of the last two expressions,

sin (B+ c)
in the double areas of AEF, will become
sin (B+C+D+ E) + sin B . sin (D+ E)
sin (B+c)


sin c


and this, by means of the formula for 5 arcs (art. 30 ch. iii) becomes BC. EF . sin (c+D+E). Hence, collecting and properly arranging the several terms as before, we shall obtain AB. BC. sin B

Twice the area

of the hexa-


+AB. CD. sin (B+C)

+AB. DE. sin (B+C+D)
+AB. EF . sin (B+C+D+E)
+BC. CD. sin c

+BC. DE. sin (C+D)
+BC. EF. sin (c+D+E)
+CD. DE. sin D

+CD. EF. sin (D+E)

+DE. EF. sin E.

4. In a similar manner may the area of a heptagon be determined, by finding the sum of the areas of the hexagon and the adjacent triangle; and thence the area of the octagon, nonagon, or of any other polygon, may be inferred; the law of continuation being sufficiently obvious from what is done above, and the number of terms=".2, when the number of sides of the polygon is n: for the number of terms is evidently the same as the number of ways in which n - 1 quantities can be taken, two and two; that is, (by the nature of Permutations) = 1.2-2




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