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increased very nearly in the duplicate ratio of the sine of the latitude.

Ex. 15. If be the measure of a degree of a great circle perpendicular to a meridian at a certain point, m that of the corresponding degree on the meridian itself, and d the length of a degree on an oblique arc, that arc making an angle a with the meridian, then is d =

demonstration of this theorem.


p+(m-p)sin2 a

Required a



THE theorems and problems in Polygonometry bear an intimate connection and close analogy to those in plane trigonometry; and are in great measure deducible from the same common principles. Each comprises three general cases.

1. A triangle is determined by means of two sides and an angle; or, which amounts to the same, by its sides except one, and its angles except two. In like manner, any rectilinear polygon is determinable when all its sides except one, and all its angles except two, are known.

2. A triangle is determined by one side and two angles; that is, by its sides except two, and all its angles. So likewise, any rectilinear figure is determinable when all its sides except two, and all its angles, are known.

3. A triangle is determinable by its three sides; that is, when all its sides are known, and all its angles, but three. In like manner, any rectilinear figure is determinable by means of all its sides, and all its angles except three.

In each of these cases, the three unknown quantities may be determined by means of three independent equations; the manner of deducing which may be easily explained, after the following theorems are duly understood.


In Any Polygon, any One Side is Equal to the Sum of all The Rectangles of Each of the Other Sides drawn into the Cosine of the Angle made by that Side and the Proposed Side*.

*This theorem and the following one, were announced by Mr. Lexel of Petersburg, in Phil. Trans. vol. 65, p. 282: but they were first demonstrated by Dr. Hutton, in Phil. Trans, vol. 66, pa. 600.


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bc ==
cd ᎴᎠ


de = E DE. COS DES

eF.... EF. COS EFE

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But AF bc + cd + de + CF 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 + er = AB. COS BAF + BC

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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 CD4AF + DE . Sin DE^AF+EF. Sin F. This is evident from an inspection of the figure

Cor. 1. In like manner Dd = DE. sin DEA AFEF. sin F, or Dd CB. sin CBAFA+BA. sin A - -CD. Sin CDAAF.


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 CBAFA denotes the inclination of the line CB 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 в . COS A, Sin A .COS (B+C) + sin (BC. COS A, &c (ch ii equa. v), and recollecting that hii

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sin A.(AB+BC.COSB+CD.COS(B+C)+DE. COS(B+C+D) + &C) +cos A.(BC.Sin B+CD.sin(B+C)+DE.COS(B+C+D)+&c) = 0; and thence finally, tan 180o — A, or tan BAF =

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

AB+BC.COSB+CD.COS(B+C)+DE.COS(B+C+D)+EF.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 в, are known, we have

tan CAB

BC.sin B


tan BCA

AB. sin B ; BC+AB. COS B

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



The Squre 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.

a = b. cos arb + c

b = a. cos a^b + c

c = a. cos a^c + b





a B

cos a^c +d. cos and,

cos bac + cos bad,

cos bac +d. cos cad,

cos bad+c. cos cad.

d = a.cos asd + b. Multiplying the first of these equations by a, the second by b, the third by c, the fourth by ; subtracting the three latter products from the first, and transposing 62, c2, d2, there will


a2=b2+c2+d2—2(bc.cos bac+bo. cos bad+co. cos cad). In like manner,

c2=a2+b2+d2 = 2(ab. cos arb+ad. cos and +od. cos bad).

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Or, since bac = c, bad = c + D

180°, cad = D, we have a2 = b2 +c2 + No2 —2(bc. cos c-bd. cos (c+D) + co. cos D), · c2 = a2 +¿ 3 + No2 −2(ab.cos B-bd. cos (A+B) + að. cosa).

&c. &c

The same method applied to the pentagon ABCDE, will give a2=62+c2+a2+e2-2 Sbc.cos c-bd.cos(c+D)+be.cos(C+D+E)) +cd.cos D-ce.cos (D+E) +de. cos E.S 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. 1 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 triangle PAD is (Mens. of Planes pr. 2 rule 2) AP. PD (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

sin P BC. sin c

sin P

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sin P =

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

Twice surface
of trapezium.

Cor. Since AB. BC

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

}-{ {

sin B =

twice triangle ABC, it follows that twice triangle ACD is equal to the remaining two terms,viz,

twice area ACD =

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

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

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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. 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} = {4P, DE, sin (P+D) or sin (B+C+D)

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sin (B+ c)'

two terms become



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BC. DE . Sin C. sin(B+C+D) BC.DE.SinB sinn
sin (B+C)
sin (B+C)

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

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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 corallary, we have,

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