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.

Ext. ang:

B

C

.

EXAMPLES. 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 AB = 1284

Whence
BC = 1782

= 32
Btc

80°
CD = 2400

48°

B+c+D = 132°
DE = 2700

52°

B+c+D+E = 1989
EF = 2860
E = 66° A +

162o.
Then, by cor. 3 th. 2, tan BAF =
BC. sin B + CD. sin ( B+ c)+ De . sin (B+C+D)+ ep. sin (B+C+D+E)
AB + BC.COS B+ CD.cos (B+ c) +DE.cos (8+c+D) + Er.cos (B+C+D+ E)

BC ,sin 32° + CD sin 80° + DE , sin 1320 + er , sin 1989
AB + BC.cos 32° + CD .cos 80° + DE, cos 132° + EF . cos 1980

BC , sin 32° + CD, sin 80° + DE, sin 48° EF, sin 180
AB + BC. cos 32° + CD cos 80o-- DE, CO3 450

EF.cos 130 Whence bafis found 106°31'38"; and the other angle Are= 91°28°22". So that the exterior angles A and Fare 73428'22”, 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.

Ex. 2. In a hexagon ABCDEF, all the sides except AF, and all the angles except c and D, are known: viz, AB=2400

We shall have, by th. 4 cor. 1, BC=2700 A=54°

AB . sin a CD=3200 B=62°

DE.sin (e+F) DE=3500

+ BC .sin(A+B) E=640

+EF. sin f. +cD.sin(A+B+c) EF=3750 F=72°

AB.

. sin 54° Therefore, cd. sin (116° + c) =

- BC. sin 116° + DE . sin 136

ter, sin 72°. Or, 116° +2=

149°23'26". + 33°36'34".

The

Ex. Ang

.

<

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,

AB. cos 54°

AF

+BC. Cos 64°

+ CD. cos 30°36′34′′ =3885.905.

+ DE COS 44°

- EF. cos 72°

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. Given AB 1200 Exterior angles A = 64°

BC 1500

CD = 1600

DE 1800
EF 2000

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),

B

C = 72°

D

75°

=

CD. sin C+ DE . sin (C+D) tan CBE = BC+ CD. COS C+ DE . COS (C + D) CD. sin 72° + DE. Siu 1470 CD sin 72+ DE . Sin 33 BC+ CD.COs 72°+ DE. cos 1470 BC+ CD. COS 72°-DE. COS 330* Whence CBE = 79° 2′-1′′; and therefore DEB = 67° 57′59′′.

Then EB

BC. cos 79° 2′ 1′′ +CD.cos 7° 2′ 1′′ +DE . cos 67°57′59′′ Secondly, in the trapezium ABef, AB. sin A + BE. sin (A + B)

EF. Sin F AB.

=2548 581.

=EF. sin F: whence
S 20°55′54′′,

sin B

sin

BE

sin (A + B) = 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° 57′55′′ respectively.

AB.COS A

AB

Lastly, AF- BE COS (A+B)- EF. COS F=— cos 64° BE. cos 20° 55' 54"- EF. COS 84°1645.292. 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

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.

Given AB

BC

Er. 4. In a hexagon ABCDEF, are known all the angles, and all the sides except AF and CD: to find those sides. 2200 Ext. Ang. A = 2400

B =

DE

EF

=

{

4800

5200

DE. sin 166° +EF. Sin 148°.

Whence S DE. sin 14. cosec 10°

CD=+EF. sin 32°. cosec 10°

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

AB. sin 96° + BC. sin 150° + CD. sin 170°

Or AB. sin 84°
+ BC. sin 30°.
+ CD. sin 10°
- AB. sin 84°. cosec 10°

BC. Sin 30°. COSCC 10°=3045-58.

CB. sin 20° } =14374.98.

And S DE. sin 24°. cosec 10°

AFEF. sin 42°, côsec 10°

And AD =

96°

C=

D=

E=

F = 148°.

54°

20°

24°

Suppose diagonals drawn to join the unknown angles, and dividing the polygon into three trapeziums and a triangle; as in the marginal figure. Then,

1st. In the trapezium ABCD, where AD and the angles about it are unknown; we have (cor. 3 th. 2)

tan BAD

BC. sin B+ CD. sin (B+ c)
AB+ BC. COS B+ CD. COS (B+C)
Whence BAD

18°

EF. Sin E+ FG. Sin (£ + F)
DE+ EF. COS E+FG. COS(E+ F)

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.

Given AB 2400 FG = 3800 Ext. Ang. в = 40°

BC

2700 GH 4000

CD

2800 HI 4200 DE 3200 IA 4500. EF 3500

{

S DE. Sin 14

H

+EF.Sin 32.

I A

BC. sin 400+ CD. sin 72° AB+BC.COS 40°+ CD.cos72°* 39° 30′42, CDA = 32° 29′18′′.

cos 39°30′42′′

AB.
+ BC. COS 0 29 18
32 29 18

=6913.292.

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

tan EDGE

EF. Sin 360+ FG. sin 81° DEEF. COS 360+ FG.cos81°*

Whence

Whence edt = 41° 14' 53", Fgv = 39° 45' 7".

DE . cos 41°14'53"
And DG = + EF. cos 5° 14'53" = 8812.803,

+ FG. cos 39° 45' 7" 3dly. In the trapezium GHIA, an exactly similar process gives HGA = 50° 46' 53", IAG = 47° 13' 7", and ag = 9780.591.

4thly. In the triangle ADG, the three sides are now known, to find the angles: viz, DAG=60°53' 26", AGD=43° 15' 54", ADG = 75° 50' 40". Hence there results, lastly,

, IAB = 47° 13' 7" +60°53'26" + 39° 30'42" = 147° 37'15", cde = 32° 29' 18" +70° 50' 40" +41° 14' 53" = 149° 34'51",

FGA= 39° 45' 7" +43° 15' 54" + 50° 46' 53" = 133° 47'54". Consequently, the required exterior angles are a=32° 22' 45", D = 30° 25'9", G = 46° 12'6". Ex. 6. Required the area of the hexagon in ex. 1.

Ans. 16530191. Ex. 7. In a quadrilateral abcd, are given AB=24, bc=30, CD = 34; angle abc = 92°18', BCD = 97°23'. Required the side AD, and the area.

Ex. 8. In prob. 1, suppose PQ = 2538 links, and the angles as below; what is the area of the field ABCDQP?

APQ=89° 14', BPQ=68°11', cPQ=36° 24', DPQ= 19°57'; AQP=25°18', BQP=69° 24', cap=94° 6', DQP=121°18'.

CHAPTER VII.

PROBLEMS RELATIVE TO THE DIVISION OF FIELDS OR

OTHER SURFACES.

PROBLEM I.

To Divide a Triangle into Two Parts having a Given Ratio,

m: n.

1st. By a line drawn from one angle of the triangle. Make Ad : AB::m:m+n; draw cd.

2; So shall adC, BDC, be the parts required. Here, evidently, AD=

m

n

2dly. By a line parallel to one of the sides of the triangle. Let ABC be the given triangle, to be divided into two parts, in the ratio of m to n, by a line parallel to the base AB. Make CE to EB as m to n: erect ED perpendicularly to CB, till it meet the semicircle described on CB, as a diameter, in

D. Make CF CD: and draw through F, GF || AB. So shall GF divide the triangle ABC in the given ratio.

CD2

CE

For, CE: CB= :: CD2(=CF2) : CB2. But CE: EB :: m:n, or CE : CB :: mm+n, by the construction: therefore, CF2: CB2 :: M:m+n. And since ▲ CGF : ▲ Cab::cf2 : CB2; it follows that CGF: CAB :: m : m+n, as required.

Computation. Since CB2: CF :: m+n: m, therefore, (m + n)cF2 = m. CB2; whence CF√ (m + n) = CB √√m, or

m

m

CFCB

-

In like manner, CG CA

m+ n

m+n

Sdly. By a line parallel to a given line. Let HI be the line parallel to which a line is to be drawn, so as to divide the triangle ABC in the ratio of m

to n.

CP =

A

By case 2d draw GF parallel to AB, so as to divide ABC in the given ratio. Through F draw FE parallel to HI. On CE as a diameter describe a semicircle; draw GD perp. to AC, to cut the semicircle in D. Make CP = CD : through P, parallel to EF, draw PQ, the line required.

The demonstration of this follows at once from case 2; because it is only to divide FCE, by a line parallel to FE, into two triangles having the ratio of FCE to FCG, that is, of CE to CG,

Computation. Co and CF being computed, as in case 1, the distances CH, CI being given, and CP being to co as CH to CI: the triangles CGF, GPQ, also having a common vertical angle, are to each other, as CG.CF to ca.cr. These products therefore are equal; and since the factors of the former are known, the latter product is known. We have hence given the ratio of the two lines CP(=x) to ca(=y) as CH to CI; say, as p to 9; and their product = CF. CG, say ab: to find x and y.

abp

Here we find x =

y = √ aba

CF

D

✔abq. That is,

CF.CG. CH

CG CI
CH

CI

·; CQ = √⋅ N.B. If the line of division were to be perpendicular to one of the sides, as to CA, the construction would be similar: VOL. III.

M

CP

B

J.