fourths,* &c.; and an arc is said to be of as many degrees as it contains parts of the 360, into which the circumference is divided.

XI.

An ANGLE is the inclination of two lines which meet, but not in the same direction.

An angle is usually expressed by the letter placed at the angular point, as the angle A. But when two or more angles are at the same point, it is then necessary to express each by three letters, and the letter at the angular point is placed between the other two. Thus the angle formed by the lines AB, AC, is called the angle BAC, or CAB; and that formed by AB, AD, is called the angle BAD, or DAB.

An angle is measured by the arc of a circle comprehended between the two legs that form the angle; the centre of the circle being the angular point, and the whole circumference considered as equal to 360°.

Thus the angle A is measured by the arc BC described round the point A as a centre, and the angle is said to be of as many degrees as the arc is; that is, if the arc BC is 30°, then the angle BAC is said to be an angle of 30 degrees.

XII.

B

A

If a right line, AB, fall upon another, DC, so as to incline neither to the one side nor the other, but makes the angles ABC, ABD, equal to each other, then the line AB is said to be perpendicular to the line DC, and each of these angles is called a right angle, being each equal to a quadrant, or 90°; because the sum of the two angles, ABC, ABD, is measured by the semicircle DAC, described on the diameter DBC, and centre B.

XIII.

An ACUTE ANGLE is less than a right angle, as ABC.

XIV.

An OBTUSE ANGLE is greater than a right angle, as GEH. The least number of right lines that can include a space are three, which form a figure called a triangle, consisting of six parts, viz. three sides and three angles; it is distinguished into three sorts, viz. a right-angled triangle, an obtuse-angled triangle, and an acute-angled triangle.

XV.

A RIGHT-ANGLED triangle has one of its angles right; the side opposite the right angle is called the hypotenuse; and the other two sides are called legs; that which stands upright is called the perpendicular, and the other the base; thus BC is the hypotenuse, AC the perpendicular, and AB the base; the angles opposite the two legs are both acute.

B

B

E H

XVI.

An ACUTE-ANGLED TRIANGLE has each of its angles acute, as DEG.

XVII.

B

D

E

An OBTUSE-ANGLED TRIANGLE has one of its angles obtuse, or greater than a right angle, as BAF; the other two angles are

acute.

B

Note. All triangles that are not right-angled, whether they are acute or obtuse, are in general called oblique-angled triangles, without any other distinction.

A new division of the circumference of the circle has lately been adopted by several eminent French mathematicians, in which the quadrant is divided into 100°, each degree into 100', each minute into 100, &c., and tables of logarithms have been published conformable thereto. The general adoption of this division would tend greatly to facilitate most of the calculations of navigation and astronomy.

XVIII.

A QUADRILATERAL figure is one bounded by four sides, as ACDB. If the opposite sides are parallel, they are called PARALLELOGRAMS. Thus, if AC be parallel to BD, and AB parallel to CD, the figure ACDB is a parallelogram. A parallelogram having all its sides equal, and its angles right, is called a SQUARE, as B. When the angles are right, and the opposite sides only equal, it is called a RECTANGLE, as A,

XIX.

The SINE of an arc is a line drawn from one end of the arc perpendicular to a diameter drawn through the other end of the same arc; thus RS is the sine of the arc AS, RS being a line drawn from one end, S, of

that arc, perpendicular to DA, which is the diameter
passing through the other end, A, of the arc.

XX.

The COSINE of an arc is the sine of the complement of that arc, or of what that are wants of a quadrant; thus, AH being a quadrant, the arc SH is the complement of the arc AS; SZ is the sine of the arc SH, or the cosine of the arc AS.

XXI.

The VERSED SINE of an arc is that part of the diameter contained between the sine and the arc; thus RA is the versed sine of the arc AS and DCR is the versed sine of the arc DHS.

XXII.

The TANGENT of an arc is a right line drawn perpendicular to the diameter, passing through one end of the arc, and terminated by a line drawn from the centre through the other end of the arc; thus AT is the tangent of the arc AS.

XXIII.

The COTANGENT of an arc is the tangent of the complement of that arc to a quadrant; thus HG is the tangent of the arc HS, or the cotangent of the are AS.

XXIV.

The SECANT of an arc is a right line drawn from the centre through one end of the arc to meet the tangent drawn from the other end; thus CT is the secant of the arc AS.

XXV.

The COSECANT of an arc is the secant of the complement of that arc to a quadrant ; thus CG is the secant of the arc SH, or cosecant of the arc AS.

XXVI.

What an arc wants of a semicircle is called the SUPPLEMENT of the arc; thus the arc DHS is the supplement of the arc AS. The sine, tangent, or secant of an arc, is the same as the sine, tangent, or secant of its supplement; thus the sine of 80° = sine of 100°, and the sine of 70° = sine of 110°, &c.

XXVII.

If one line, AB, fall any way upon another, CD, the sum of the two angles, ABD, ABC, is always equal to two right angles.

For, on the point B as a centre, describe the circular arc CAD, cutting the line CD in C and D; then (by Art. 6), this arc is equal to a semicircle, but it is also equal to the sum of the arcs CA and AD, the measures of the two angles ABC, ABD; therefore the

A

sum of the two angles is equal to a semicircle, or two right angles. Hence it is evident that all the angles which can be made from a point in any line, towards one side of the line, are equal to two right angles, and that all the angles which can bẹ

XXVIII.

If a line, AC, cross another, BD, in the point E, the opposite angles will be equal, viz. BEA CED, and BEC=AED.

Upon the point E as a centre, describe the circle ABCD; then it is evident that ABC is a semicircle, as also BCD (by Art. 6); therefore the arc ABC=arc BCD; taking from both the common arc BC, there remains arc AB arc CD; that is, the angle BEA is equal to the angle CED. After the same manner we may prove that the angle BEC is equal to the angle AED.

XXIX.

B

If a line, GH, cross two parallel lines, AB, CD, it makes the external opposite angles equal to each other; viz. GEB-CFH, and AEG HFD.

For since AB and CD are parallel to each other, they may be considered as one broad line, and GH crossing it; then the vertical or opposite angles, GEB, CFH, are equal (by Art. 28), as also AEG HFD.

XXX.

E B

If a line, GH, cross two parallel lines, AB, CD (see the figure), the alternate angles, AEF and EFD, or CFE and FEB, are equal. For GEBAEF (Art. 28), as also CFH-EFD (by the same Art.), but GEBCFH by the last; therefore AEF is equal to EFD; in the same way may we prove FEB⇒CFE.

XXXI.

If a line, GH, cross two parallel lines, AB, CD (see the preceding figure), the external angle, GEB, is equal to the internal opposite one, EFD, or AEG equal to CFE.

For the angle AEF is equal to the angle EFD by the last, and AEF = GEB (by Art. 28); therefore GEB= ÉFD; in the same way we may prove AEG=CFE.

XXXII.

If a line, GH, cross two parallel lines, AB, CD (see the preceding figure), the sum of the two internal angles, BEF and DFE, or AEF and CFE, is equal to two right angles.

For since the angle GEB is equal to the angle EFD (by Art. 31), to both add the angle BEF, and we have GEB+BEF = BEF+EFD; but GEB+ BEF = two right angles (Art. 27). Hence, BEF + EFD two right angles; and in the same manner we may prove AEF+CFE=two right angles.

XXXIII.

In any triangle, ABC, one of its legs, as BC, being produced towards D, the external angle, ACD, is equal to the sum of the internal and opposite angles, ABC, BÁC.

To prove this, through C draw CE parallel to AB; then, since
CE is parallel to AB, and the lines AC, BD cross them, the angle
ECD=ABC (by Art. 31), and ACE=BAC (by Art. 30); adding
these together we have ECD + ACE = ABC + BAC; but B
ECD+ACE=ACD; therefore ACD = ABC+BAC.

XXXIV.

Hence it may be proved that if any two lines, AB and CD, be crossed by a third line, EF, and the alternate angles, AEF and EFD, be equal, the lines AB and CD will be parallel.

For, if they are not parallel, they must meet each other on one side of the line EF (suppose at G), and so form the triangle EGF, one of whose sides, GE, being produced to A, the exterior angle, AEF, must (by the preceding article) be equal to the sum of the two angles EFG and EGF; but by supposition it is equal to the angle EFG alone; therefore the angle AEF must be equal to the sum of the two angles EFG and EGF, and at the same time equal to EFG alone, which is absurd; therefore the lines AB, CD, cannot meet, and must be parallel.

A

E B

XXXV.

In any right-lined triangle, ABC, the sum of the three angles is equal to two right angles.

To prove this, you must produce BC (in the fig. Art. 33) towards D; then (by Art. 33), the external angle ACD ABC+BAC; to both add the angle ACB, and we have ACD+ACB ABC+BAC+ACB; but ACD+ACB=two right angles (by Art. 27). Hence, ABC+BAC+ACB two right angles; therefore the sum of the three angles of any plain triangle, ACB, is equal to two right angles.

XXXVI.

Hence in any plain triangle, if one of its angles be known, the sum of the other two will be also known.

For by the last article the sum of all three angles is equal to two right angles, or 180°; hence, by subtracting the given angle from 180°, the remainder will be the sum of the other two.

In any right-angled triangle, the two acute angles taken together are just equal to a right angle; for, all three angles being equal to two right angles, and one angle being right by supposition, the sum of the other two must be equal to a right angle; consequently, any one of the acute angles being given, the other one may be found by subtracting the given one from 90 degrees.

XXXVII.

D

If in any two triangles, ABC, DEF, two legs of the one, AB, AC, be equal to two legs of the other, DE, DF, each to each respectively, that is, AB=DE, and AC = DF, and the angles BAC, EDF, included between the equal legs be equal; then the remaining leg of the one will be equal to the remaining leg of the other, and the angles opposite to the equal legs will be equal; that is, BC= EF, ABCDEF, and ACB=DFE. For if the triangle ABC be supposed to be lifted up and put upon the triangle DEF, with the point A on the point D, and the line AB upon DE, it is plain, since AB DE, that the point B will fall upon E; and since the angles BAC, EDF are equal, the line AC will fall upon DF; and these lines being of equal length, the point C will fall upon F; consequently the line BC will fall exactly upon the line EF, and the triangle ABC will in all respects be exactly equal to the triangle DEF, and the angle ABC will be equal to the angle DEF, also the angle ACB will be equal to the angle DFE.

XXXVIII.

Δ.Δ.

B

C E

After the same manner it may be proved that if in any two triangles, ABC, DEF (see the preceding figure), two angles, ABC and ACB, of the one be equal to two angles, DEF, DFE, of the other, and the included side, BC, be equal to EF, the remaining sides and included angles will also be equal to each other respectively; that is, AB=DE, AC=DF, and the angle BAC the angle EDF.

For if the triangle ABC be supposed to be lifted up and laid upon the triangle DEF, the point B being upon the point E, and the line BC upon the line EF, then, since BCEF, the point C will fall upon the point F; and, as the angle ACB=the angle DFE, the line CA will fall upon the line FD; by the same way of reasoning, the line BA will fall upon the line ED; therefore the point of intersection, A, of the two lines, BA, CA, will fall upon D, the point of intersection of the lines ED, FD; consequently AB DE, AC=DF, and the angle BAC the angle EDF.

XXXIX.

If two sides of a triangle are equal, the angles opposite these sides will also be equal; that is, if ABAC, the angles ABC, ACB, will also be equal.

For, draw the line AD, bisecting the angle BAC, and meeting the line BC in D, dividing the triangle BAC into two triangles, ABD, ACD, in which the side AB AC, the side AD is common to both triangles, and the angle BAD=the angle DAC; consequently (by Art. 37), the angle ABD must be equal to the angle ACD.

B

A

C

The converse of this proposition is also true; that is, if two angles of a triangle are equal, the opposite sides are also equal. This is demonstrated nearly in the same

XL.

Any angle at the circumference of a circle is equal to half the angle at the centre, standing upon the same arc.

Thus the angle BAD is half the angle BCD, standing upon the same arc, BD, of the circle BEDA whose centre is Č. To demonstrate this, draw through A and the centre C, the right line ACE; then (by Art. 33) the angle CAD+angle CDA = angle ECD; but AC=CD (being two radii of the same circle); therefore (by Art. 39), the angle CAD: the angle CDA, and the sum of these two angles is the double of either of them; that is, CAD +CDA = twice CAD; therefore ECD=twice CAD; in the same manner it may be proved that BCE twice BAC, and by adding these together, we have ECD+BCE=twice CAD+twice BAC; that is, BCD=twice BAD, or BAD equal to half of BCD. The demonstration is similar when B, D, fall on the same side of E.

XLI.

An angle at the circumference is measured by half the arc it subtends. For an angle at the centre, standing on the same arc, is measured by the whole arc (by Art. 11); but since an angle at the centre is double that at the circumference (Art. 40), it is evident that an angle at the circumference must be measured by half the arc it stands upon. Hence all angles, ACB, ADB, AEB, &c., at the circumference of a circle standing on the same chord, AB, are equal to each other; for they are all measured by the same arc, viz. half the arc AB.

XLII.

An angle in a segment greater than a semicircle is less than a right angle.

Thus, if ABC be a segment greater than a semicircle, the arc AC on which it stands must be less than a semicircle, and the half of it less than a quadrant or a right angle; but the angle ABC in the segment is measured by the half of the arc AC; therefore it is less than a right angle.

An angle in a semicircle is a right angle.

For since DEF is a semicircle, the arc DKF must also be a semicircle; but the angle DEF is measured by half the arc DKF, D that is, by half a semicircle or by a quadrant; therefore the angle DEF is a right one.

An angle in a segment less than a semicircle is greater than a right angle.

Thus, if GHI be a segment less than a semi-circle, the arc GKI on which it stands must be greater than a semicircle, and its half greater than a quadrant or right angle; therefore the angle GHI, which is measured by half the arc GKI is greater than a right angle.

XLIII.

G

B

E

K

H

K

B

If from the centre, C, of the circle ABE there be let fall the perpendicular CD on the chord AB, it will bisect the chord in the point D.

Draw the radii CA, CB; then (by Art. 39) the angle CBA=the angle CAB, and as the angles at D are right, the angle ACD must be equal to the angle BCD (by Art. 36), Hence in the triangles ACD, BCD, we have the angle ACD equal to the angle BCD, CA CB, and CD common to both triangles, consequently (by Art. 37) AD=DB; that is, AB is bisected at D.

XLIV.

E C

D

If from the centre, C, of the circle ABE there be drawn a perpendicular, CD, to the chord AB, and it be continued to meet the circle in F, it will bisect the arc AFB in F (See the preceding figure.)

=

For in the last article it was proved that the angle ACD the angle BCD; hence (by Art. 11) the arc AF the arc FB.