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It often happens, however, that only one side of a polygon can be measured, and the distant angles be determined by intersection; in this case the area may be found, independent of construction, by the following problem.


Given the Length of One of the Sides of a Polygon, and the
Angles made at its two extremities by that Side and Lines
drawn to all the Other Angles of the Polygon; to find an
Expression for the Surface of that Polygon.

Here we suppose known PQ; also
AQPa", BQP = b", cap=c", DQP=d".
Now, sin PAQ = sin (a+a"); sin PBQ =
sin (b'+b").

Therefore, sin (a' + a"): PQ :: sin a": PA =

And,. . . sin (b' + b′′) : PQ :: sin b′′ : P3 =


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sin a"


sin (a'+a")
sin "


sin (+b)

But, triangle APB=AP. PB. 1⁄2 sin APB=÷AP. PB.sin (a ́—b'). .

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▲ DPQ = QP. PD. sin DPQ = PQ.

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sin d'. sin d"
sin (d'+d') *


Surface PABCDQ = PQ2.

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PQ. sind


sin d"
sin (d+d")

sin a". sin " . sin (a'-b')
sin (a+a"). sin (l' + b′′)
sin b". sin c", sin (b'-c')
sin (b' + b). sin (c' + c')
sin c". sin d'. sin (c'-d')

+ sin (c′ + c′′). sin (d'+d′′)

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The same method manifestly applies to polygons of any number of sides: and all the terms except the last are so perfectly symmetrical, while that last term is of so obvious a form, that there cannot be the least difficulty in extending the formula to any polygon whatever.



Given, in a Polygon, All the Sides and Angles, except three; to find the Unknown Parts.

This problem may be divided into three general cases, as shown at the beginning of this chapter: but the analytical solution of all of them depends on the same principles; and these are analogous to those pursued in the analytical investigations of plane trigonometry. In polygonometry, as well as trigonometry, when three unknown quantities are to be found, it must be by means of three independent equations, involving the known and unknown parts. These equations may be deduced from either theorem 1, or 3, as may be most suited to the case in hand; and then the unknown parts may each be found by the usual rules of extermination.

For an example, let it be supposed that

in an irregular hexagon ABCDEF, there
are given all the sides except AB, BC, and F
all the angles except B; to determine
those three quantities.




The angle B is evidently equal to (2n-4) right angles (A+C+D+E+F); n being the number of sides, and the angles being here supposed the interior ones.

Let AB = x, BC=y: then, by th. 1,

y = x

x = y. cos B + DC. COS AB CD + DE. COS ABAED + EF. COS ABAEF+ AF. COS ABAF; COS BAF. COS BCAAF + FE. COS BCAFE DE. COS BCADE + DC. COS BCACD. In the first of the above equations, let the sum of all the terms after y. cos B, be denoted by c; and in the second the sum of all those which fall after r. cos B, by d; both sums being manifestly constituted of known terms: and let the known coefficients of x and y be m and n respectively. Then will the preceding equations become

x = ny + c . . . y = mx + d. Substituting for y, in the first of the two latter equations, its value in the second, we obtain x = mnx + nd + c. Whence there will readily be found

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-- mn

Thus AB and BC are determined. Like expressions will serve for the determination of any other two sides, whether contiguous or not: the coefficients of r and y being designated by different letters for that express purpose; which would have been otherwise unnecessary in the solution of the individual case proposed.


Remark. Though the algebraic investigations commonly lead to results which are apparently simple, yet they are often, especially in polygons of many sides, inferior in practice to the methods suggested by subdividing the figures. The following examples are added for the purpose of explaining those methods: the operations however are merely indicated; the detail being omitted to save room.


Er. 1. In a hexagon ABCDEF, all the sides except AF, and all the angles except A and F, are known. Required the unknown parts. Suppose we have

1284 Ext. ang.

BC 1782


B 32°

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Then, by cor. 3 th. 2, tan BAF =

BC. sin B+CD. sin (B+C) + DE . sin (B+C+D) + EF . Sin (B + C + D + £) AB + BC. COS B + CD.COS (B+C) + DE. COS (B+C+D) + EF. COS (B + C + D + E ) BC sin 32° + CD sin 80+ DE, sin 1320+ EF. sin 1980 AB+BC.COS 32°+ CD. cos 80°+ BE. cos 132° + EF . cos 1980 BC. sin 32°+ CD sin 80+ DE . sin 48° EF. sin 180 AB+ BC.cos 32° + CD. cos 80°-DE, cos 480




EF.COS 130*

Whence BAF is found 106731′38′′; and the other angle AFE= 91°28′22′′. So that the exterior angles and F are 73°28′22′′, and 88°31′38′′ respectively: all the exterior angles making 4 right angles, as they ought to do. Then, all the angles being known, the side AF is found by th. 1 = 4621.5.

If one of the angles had been a re-entering one, it would have made no other difference in the computation than what would arise from its being considered as subtractive.

Er. 2. In a hexagon ABCDEF, all the sides except AF, and all the angles except C and D, are known: viz,

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The second of these will give for c, a re-entering angle; the second will give exterior angle c = 33° 23′26′′, and then will D 14° 36′ 34". Lastly,

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Ex. 3. In a hexagon ABCDEF, are known, all the sides except AF, and all the angles except в and E; to find the rest. 1200 Exterior angles a = 64°

Given AB

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F = 84°.

Suppose the diagonal BE drawn, dividing the figure into two trapeziums. Then, in the trapezium BCDE, the sides except BE, and the angles except в and E, will be known; and these may be determined as in exam. 1. Again, in the trapezium ABEF, there will be known the sides except AF, and the angles except the adjacent ones в and E. Hence, first for BCDE: (cor. 3 th. 2),

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CD. sin 720 + DB. Siu 147°



CD. sin 72+ DE . sin 330 BC+ CD.COS 72°+DE. cos 1470 BC+ CD. Cus 72°-DE. cos 33°° Whence CBE = 79° 2′1′′; and therefore DEB = 67° 57′59′′.

Then EB


BC.cos 79° 2′ 1

+ CD. cos 7° 2′
+ DE . cos 67°57′59′′.

Secondly, in the trapezium ABEF,

AB. sin A+ BE

sin (A+B)

sin (A+B)


sin F AB


=2548 581.

=EF. sin F: whence

sin B
= sin

S 20°55'54", 159° 4' 6". Taking the lower of these, to avoid re-entering angles, we have B (exterior ang.) = 95°4′6′′; ABE=84°55′54′′; FEB= 63°4′6′′; therefore ABC= 163° 57'55"; and FED=131°2′5′′: and consequently the exterior angles at B and E are 16°2′5′′ and 48° 5755" respectively.

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Lastly, AF= - AB.COS A — BE. COS (A+B) - EF. COS F—— AB. COS 64° + BE. cos 20° 55′ 54′′- EF cos 841645.292. Note. The preceding three examples comprehend all the varieties which can occur in Polygonometry, when all the sides except one, and all the angles but two, are known. The unknown angles may be about the unknown side; or they may


be adjacent to each other, though distant from the unknown side; and they may be remote from each other, as well as from the unknown side.

Er. 4. In a hexagon ABCDEF, are known all the angles, and all the sides except AF and CD: to find those sides. Given AB 2200 Ext. Ang. A =



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B =


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Here, reasoning from the principle of cor. th. 2, we have

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Ex. 5. In the nonagon ABCDEFGHI, all the sides are known, and all the angles except A, D, G: it is required to find those angles.

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Whence BAD 39° 30′42, CDA = 32° 29′18′′.

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2dly. In the quadrilateral DEFG, where DG and the angles about it are unknown; we have


EF. Sin E+FG. sin (E+F)

EF. Sin 36° + FG. sin 81° DEEF. Cos 36"+ FG.cos810


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