PROBLEM IX.

The Angles and Hypotenuse of a Right-angled Triangle given, to find either of the Legs.

Given the hypotenuse 250 leagues, the angle opposite the base 54° 30', consequently the other angle 35° 30; the base and perpendicular are required.

Draw the line CB, and at C make an angle 20. equal to 35° 30' by drawing the line CA, A34.30 take 250 from any convenient scale of equal C35.30 parts, and set it off from C to A; from A let fall the perpendicular AB, to cut the line CB, and it is done; for AB measured on the same scale gives 145, and CB 203.6 leagues.

C

250

35.30.

203.6.

A

NOTE. The two acute angles of a right-angled triangle make 90 degrees.

PROBLEM X.

The Angles and one Leg of a Right-angled Triangle being given, to find the Hypotenuse and the other Leg.

The angle ACB 33° 15', the leg AC 285 miles, to find the bypotenuse and the other leg AB.

B

90. C33.45 B56.45

D

341

33.45

+2.85

Draw the base AC, lay off on it 285 from your scale of equal parts, from A to C; on A erect the perpendicular AB: with the chord of 60° sweep the arch AD, and on it set off 3340, from your line of chords from A to D, through D and C; draw the right line BC, then BC will measure 341 nearly, and BA 187 nearly, on the same scale of equal parts that AC was taken from.

PROBLEM XI.

A

The Hypotenuse and one Leg given, to find the Angles and

the other Leg.

The leg AB 350, the hypotenuse 600 given, to find the angles,, and leg BC.

Draw the base CB, on B erect the perpendicular AB, on which set off 350 from B to A; on the point A, with an opening of 600, draw an arch to cut the line BC, in the point C; draw AC, and it is done; for the angle ACB will measure 35° 41' on the line of chords, and BC will measure 487 nearly, on the same scale of equal parts before used.

A

54.19

90. C 35.41 A54.19

PROBLEM XII.

The Legs given, to find the Angles and the Hypotenuse.

The leg AB 880 and BC 690 given, to find the angles A and C, and the hypotenuse AC.

Draw the base BC; on B erect the perpendicular AB, make BC equal to 690, and AB equal to 880; join AC, and it is done; for the angle C being measured as before, will be found as per figure, and the bypotenuse will measure 118.2.

PROBLEM XIII.

088

900 0* ZC 51 LA 38

54

6

4118.2.

38

06

B

51.54

-Two Angles and one Side of an Oblique-angled Triangle given, to find either of the other Legs.

The angle BDC 108° 30', and CBD 45° 15', and consequently the angle BCD 26° 15', and the leg BC 98 given, to find the sides CD and BD.

Draw the line BC, which make equal to 98, on the point B describe an angle of 45° 15', then add 45° 15' to 108° 30' and the sum 153° 45′ taken from 180, the remainder is the angle BCD=26° 15'; from the point Cdescribe an arch

B

46

45.15.

D

108.30

987

LB 45° 15.
D 108 30

153 45 180

LC 26 15

26.15.

with the chord of 60, and set off 26° 15', and it is done; for the side BD will be 46 nearly, and DC 73,4, as was required.

PROBLEM XIV.

Two sides and an Angle opposite to one of them given, to find the other Angle and the third Side.

The side BC 160, and BD 79, and the angle C 29° 9' given, to find the angle D, and the side CD.

Draw the line BC equal to 160, on C make the angle DCB equal to 29° 9' take 79 in your compasses, and with one foot on B, lay the other upon the line CD, draw the line BD, and it is done; for the angle D will be 99o 25', the angle B 51° 26', and the side DC 127 nearly.

PROBLEM XV.

Two Sides and their contained Angle given, to find either of the other Angles, and the third Side.

The side BC 109, BD 76,and angle CBD 101° 30' given, to find the angles BDC or BCD, and the side CD. Draw the line BC, which make equal to 109; on B describe an arch, on which set off from BC towards D 101° 30', then draw the line BD equal to 76, join DC, and it is done; for the angle BDC will be 47° 32', the angle BCD 30° 58', and the side DC will be 145, as was required.

PROBLEM XVI.

47°32

101.30

B

Three Sides given, to find the Angles.

30458 109

The sides BC 105, BD 85, and CD 50 miles given, to find the angles BDC, BCD, and CBD.

Draw the line BC equal to 105, take CD equal to 50 in your compasses, and with one foot in C, describe an arch as at D, then take BD $5 in your compasses, and with one foot in B cut the former arch in D, join BD and DC, and it is done; for the angle B being measured,

ZB 28°
LC 53

81

180

LD 98

8

12

48

50

85

28

B

will be found 28° 4', the angle C 53° 8', which being added together is 81° 12' their sum subtracted from 180°, leaves angle D 98 48' as was required.

PROBLEM XVII.

To find the Centre to a given Circle. With any radius, and one foot in the circumference as at A, describe an arch of a circle, as CBD, then removing the foot from A to whence it cuts the given circle, as at B, on B describe another arch, cutting or crossing the former, as CAD, and through the points of intersection draw the right line CD, which will give one right line passing through the centre; in like manner may another right line be drawn, as EFG, which will cross the first right line at the centre required, for any two diameters will always cut or cross one another in the central point.

PROBLEM XVIII..

To divide a Circle into any Number of equal even Parts, as 4, 8, 16, 32,

First draw the diameter through the centre, which will divide it into two equal parts; bisect the diameter with another right liné perpendicular thereto, and the circle will be divided into four equal parts or quadrants; bisect each of these quadrants again by right lines drawn through the centre, and it will be divided into eight equal parts, and so may you continue on your bisections any number of times, that is 4, 8, 16, 32, &c. doubling the number of

even parts.

X

+

This problem is useful in constructing the Mariner's Compass. I. A chord or subtense of an arch, is a right line that divides the circle into two unequal parts, and is

a chord to them both as FH, TI.

II. A right sine of an arch is a line drawn from the end or termination of an arch, perpendicular to the radius, or is half the chord of twice the arch, so that TV is the sine of the arch TG, and of the arch TF, the sum of F which arches together make 180°, or a semi-circle.

III. The versed sine of an arch is part of the diameter intercepted between the right sine and the arch, as VG.

H

R

IV. The tangent of an arch is a line drawn perpendicular to the end of the radius, or diameter, just touching the arch, as DG. V. The secant of an arch is a right line drawn from the centre though the circumference, meeting the end of the tangent line to the same arch, as OD is the secant of the arch TG, to which DG is tangent; also OR is the secant of the arch CT, to which CR is the tangent.

NOTE. Sines, Tangents, Secants, are said to be the measure of so many degrees as the arch contains parts of 360, so that radius being the sine of a quadrant, or a fourth part of the circumference, contains 90 degrees; thus the radius is always equal to the sine of 909, as is also the tangent of 45o, and the chord of 60°.

PROJECTION

OF THE

LINES OF SINES, TANGENTS, AND SECANTS,

ON THE PLANE SCALE.

1st. WITH the radius you intend for your scale, describe a semi-circle ADBC, and upon the centre C raise the perpendicular CD, (which will divide the semi-circle into two quadrants, AD, BD), continue CD directly to S, and upon B raise the perpendicular BT, then draw the right lines BD and AD.

2dly. Divide the quadrant BD into 9 equal parts, then will each of these be 10 degrees. Again, you may subdivide each of these parts into single degrees; and these again, if your radius admits it, into minutes, or some aliquot parts of a degree greater than minutes.

3dly. Set one foot of the compasses in B, and transfer each of the divisions in the quadrant BD to the right line BD, then

is BD a line of chords.

4thly. From the points 10, 20, 30, &c. in the quadrant BD, draw right lines parallel to CD, till they cut the radius CB, then is the line CB divided into a line of sines, which must be numbered from C towards B.

5thly. If the same line of right sines be numbered from B towards C, it will become a line of versed sines, which may be continued to 180°, if the same divisions be transferred on the same line on the other side of the centre C.

6thly. From the centre C, through the several divisions in the quadrant BD, draw right lines till they cut the tangent BT, so will the line BT become a line of tangents.

7thly. Setting one foot of the compasses in C, extend the other to the several divisions 10, 20, 30, &c. on the tangent line BT, and transfer these extents severally into the right line CS, then will the line CS, be a line of secants.

8thly. Right lines drawn from A to the several divisions, 10, 20, 30, &c. in the quadrant BD, will divide the radius CD into a line of semi-tangents.

9thly. Divide the quadrant AD into eight equal parts, and from A transfer these divisions severally into the line AD, then is AD a line of rhumbs, each division answering to 11° 15' upon the line of chords.