PROBLEMS.

142. A line of chords is made by transferring the divisions on the arc o. a quadrant to its chord. Thus, suppose ACB is a quadrant, and the right line BA the chord of its arc BDA. Let this arc be divided into 90

equal parts or degrees: then if one foot of a pair of compasses be kept on the point B, and arcs successively described with the other, from each of the 90 divisions in BDA to meet BA, those arcs will divide it into a line of chords.

143. To measure an angle with the line of chords.-Suppose the angle

АСВ. With the radius CD equal

to the extent of 60 degrees on the

G

line, about the angular piont C as a cen-
tre, describe the arc DG; then the ex-
tent from D to G measured on the
chords, gives the number of degrees,
&c. contained in the angle: which, in C
this example, is about 40.

D

144. Hence the method of laying down an angle which shall contain a proposed number of degrees is obvious. Suppose for example, it is required to make the angle ACB of 40 degrees, CB being a given line. With CD the chord of 60 degrees, describe an arc DO as before; then 40 degrees taken on the line of chords, will extend from D to the point G in the arc through which the line CA must be drawn to form the required angle.

When the angles are greater than 90 degrees, measure, or lay them off at twice. Or produce one side so as to form two angles at the angular point, and then measure the supplement to 180 degrees.

The chord of 60 degrees is taken for the radius, because the sum of the angles of a triangle being 180 degrees (41), each angle of

an equilateral triangle must therefore contain 60 degrees. R
Thus, if RCD is a quadrant, and the triangle BCD equi-
lateral, BD ( the radius CD) is the chord of the arc DB,
or of 60 degrees.

B

242

145. To measure the angle ACB with the sector. See the Fig. te Art. 143. About C with any radius CD, describe an intercepted arc DG, Open the sector till the distance between the brass points marked C, C, (the extremities of the chord-lines) is equal to the radius CD. Then if the distance DG be laid cross-ways on those chords, so that its extremities are equally distant from C, C, or from the centre of the instrument, the points of the compass will fall on the number of degrees in the angle. Thus it CO, CO, Be the chord-lines of 60 degrees (ach on the sector, (moveable about the centre Ond DO the chord of any other arc, 40 degrees for example: then by similar triangles CO (the radius): DO (the chord of 40 degrees) :: CC: DD; therefore i CC be made the radius of any arc, or circle, DD will be the chord of 40 degrees in that arc, or circle.

D

D

Hence it is, that the sector has frequently the advantage of the protractor, or common line of chords, because it may be set to different radii; the limits being the distance between the brass points C, C, when the instrument is shut, and their distance when it is quite open.

146. When it is proposed to trace an angle on the ground equal to another angle, the operation is similar to that in Art. 136. Thus, to lay down the angle ab equal to the angle ABN, the direction of bu being given. Measure equal distances BD, BG, and also the cross distance GD; then with those three distances lay down the triangle bdg (136), and the point g gives the direction of ba.

B

G

S

D

11

147. If the angle ahn (when laid down) is to contain a given number of degrees; first, make an angle DSR on paper equal to those degrees; then having measured the equal sides SD, SR, and the opposite side RD on some convenient scale of equal parts, let the triangle gbd be traced on the ground with three corresponding distances in feet or yards, &c. (136). Thus, suppose the angle RSD is 41 degrees, then if SR, SD are each 40 on a scale of equal parts, RD will be 28 on the same scale, nearly: consequently if the triangle ghd is traced on the ground, with 40, 40, and 28 feet, the angle abn will be 41 degrees.

་

148. And therefore to determine nearly, the angle P subtended by two distant objects A and B, measure equal distances PD, PG, and the crossdistance DG; then construct a triangle dpg on paper, similar to DPG, and measure the angle p with a protractor, or the chords. Thus if PD, PG, are each 30 feet, and DG = 284 feet, the triangle dog constructed with 30, 30, and 281 equal parts from any scale, will give the angle p (or P) = 563 degrees, nearly.

149. Through a given point P to draw a line CD parallel to a given line AB.

From P draw PG in any direction to meet the given line AB; then make the angle GPD equal to the angle AGP (140); and PD will be parallel to AB; because the alternate angles AGP, GPD are equal (40).

To trace the parallel CD on the ground; Fix on any convenient point G in AB, and measure an isosceles triangle RGK; then at the point Play down the triangle OPQ equal to RGK; and PQ will be parallel to GK.

150. By means of this last problem we can bisect an inaccessible angle. Let it be required to determine the direction of the capital OP of a bastion. At any points B, S, in the directions of the faces DP. AP, set up two marks; and from B trace BR parallel to PS; measure equal distances BC, BR, and mark the point K in the direction CR; then find G the middle of CK: and the prolongation of GP will bisect the angle APD.

P

D

For the triangle CPK being similar to the isosceles triangle CBR, the line GP from the middle of the base CK bisects the opposite angle (46, corol. 1).

Corol. Hence if we measure CB, CR, CK, the distances CP, KP, are

found by similar triangles. For CR: CB :: CK: CP. And a perpen dicular from B on CR will give the distance GP at another proportion.

151. When it is proposed to trace a line through a given point P parallel to an inaccessible line AB, set up marks at any convenient points C, R, in the directions AP, BP; next, by means of three equal isosceles triangles Caa, Paa, Oaa, trace PO parallel to CB, and OD parallel to PC; then the direction DP is parallel to AB.

D:

R

a

B

For by construction OD is parallel to CA, and OP to CB; therefore the triangles ORD, CRA; and OPR, CBR, are respectively similar;

And RO RC :: OP: CB; therefore by equality of

ratios, OD: OP :: CA: CB.

Now the sides about the equal angles DOP, ACB of the triangles DOP, ACB being proportional, those triangles are therefore similar (94, corol. 1); and since the homologous sides are respectively parallel and like situated, the third sides DP, AB must also be parallel.

Corol. Because the quadrilaterals RDPO, RABC are similar, if we measure the sides RO, DP, RC, the inaccessible distance AB may be found at one proportion; for RO : DP :: RC : AB.

151". In castramentation it is sometimes necessary to change the di rection instead of continuing the fronts of all the battalions or divisions in the same line. Let QR be two divisions of the encampment, the fronts being in the same line OT, and IL the distance between them; and let it be required to place the other divisions GC, &c. that the fronts SG, &c. may be in a given direction or parallel to a given line BA, the distance between the divisions remaining as before or RS = IL, and (as is usual) the two pro

B

P

C

R

I

longations RO, SO of the fronts equal to each other:

In BQ and BA take two equal distances BP, BD, and measure the siden of the isosceles triangle DBP; then the point O is found thus, DP: PB ::

RS (or IL): RO. Suppose BP = BD= 30, DP = 50, and RS IL= 20 feet; then 50: 30 :: 20 : 12 feet = RO. Therefore if RO be made = 12 feet, a string or tape OS 12 feet, and another RS 20, when stretched from O and R will give the point S, and the new direction OSG. For the triangles DBP, SOR being similar (94), and RO parallel to PB, the angles SOR, DBP are equal, and consequently OS is parallel to BD.

152. From a given point P to let fall a perpendicular PG upon a given line AB.

About P as a centre with any radius PD greater then the distance of P from AB, describe an arc DC; and from D and C with a radius greater than half DC, describe arcs. intersecting each other in R; join PR : then PG is the perpendicular required.

A

P

D

B

Draw the radii RC, RD. Then RC, CP being equal to RD, DP, respectively, and the side RP common to both the triangles RCP, RDP, those triangles are therefore identical, consequently the angles CPG, DPG are equal, and the triangle CPD isosceles; therefore PG is perpendicular to CD (46, corol. 1).

When the point is nearly opposite the end of the line. From any point Cin AB, describe an arc PDR; take DR equal to DP; then join RP: and PG will be perpendicular to AB.

B

For by construction CD bisects the arc PDR in D ; therefore PG is perpendicular to CD (65).

When a perpendicular is to be traced on the ground: First trace the line CPD parallel to AB (by 149); then 2 perpendicular to CD at the point P (137) will also be perpendicular to AB,

C P D