35. 3.) the rectangle CBG is equal to the rectangle EBF; that is, (16. 6.) BC is to BF, as BE is to BG, that is, the base is to the sum of the sides, as the difference of the sides is to the sum or difference of the segments of the base made by the perpendi cular from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Q. E. D.

PROP. VIII. PROB. FIG. 14.

The sum and difference of two magnitudes being given, to find them.

Half the given sum added to half the given difference, will be the greater, and half the difference subtracted from half the sum, will be the less.

For, let AB, be the given sum, AC the greater, and BC the less. Let AD be half the given sum; and to AD, DB, which are equal, let DC be added, then AC will be equal to BD, and DC together; that is, to BC, and twice DC; consequently twice DC is the difference, and DC half that difference, but AC the greater is equal to AD, DC; that is, to half the sum added to half the difference, and BC the less is equal to the excess of BD, half the sum about DC half the difference. Q. E. D.

SCHOLIUM.

Of the six parts of a plane triangle (the three sides and three angles) any three being given, to find the other three is the bu siness of plane trigonometry; and the several cases of that problem may be resolved by means of the preceding propositions, as in the two following, with the tables annexed. In these, the solution is expressed by a fourth proportional to three given lines; but if the given parts be expressed by numbers from trigonometrical tables, it may be obtained arithmetically by the common Rule of Three.

Note. In the tables the following abbreviations are used. R is put for the Radius; T for Tangent; and S for Sine. Degrees, minutes, seconds, &c. are written in this manner; 30° 25' 13", &c. which signifies 30 degrees, 25 minutes, 13 seconds, &c.

SOLUTION OF THE CASES OF RIGHT ANGLED TRIANGLES.

GENERAL PROPOSITION.

In a right angled triangle, of the three sides and three angles any two being given besides the right angle, the other three may be found, except when the two acute angles are given, in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them.

It is manifest from 47. 1. that of the two sides and hypothenuse any two be given, the third may also be found. It is also manifest from 32. 1. that if one of the acute angles of a rightangled triangle be given, the other is also given, for it is the complement of the former to a right angle.

If two angles of any triangle be given, the third is also given, being the supplement of the two given angles to two right angles.

The other cases may be resolved by help of the preceding propositions, as in the following table:

GIVEN.

Two sides, AB

1 AC.

2

3

AB, BC, a side and the hypothenuse.

SOUGHT.

The an

AB: AC: : R: T, B, of gles B, C. which C is the complement.

The an- BC: BA :: R: Ș, C, of gles B, C. which B is the complement.

AB, B, a side and The other R: T, B:: BA : AC. an angle.

5

side AC.

The hy- S, CR:: BA: BC. pothenuse

BC.

BC and B, the The side R: S, B:: BC : CA. hypothenuse and an AC.

Jangle.

These five cases are resolved by prop. 1.

SOLUTION OF THE CASES OF OBLIQUE ANGLED TRIANGLES.

GENERAL PROPOSITION.

In an oblique angled triangle, of the three sides and three angles, any three being given, the other three may be found except when the three angles are given; in which case the ratios of the sides are only given, being the same with the ratios of the sines of the angles opposite to them.

1 A, B, and there- [BC, AC, S, C : S, A :: AB: BC, and [fore C, and the side also, S, C: S, B, :: AB: AC. (2.) (Fig. 16. 17.)

2

3

AB.

Let AD be perpendicular to BC. 1. If ABq be less than ACq+CBq. Fig. 16. BC: BA+AC:: BA-AC : BD-DC, and BC the sum of BD, DC is given; therefore each of them is given. (7.)

2. If ABq be greater them ACq+CBq, Fig. 17. BC: BA+AC:: BA-AC: BD +DC; and BC the difference of BD, DC is given, therefore each of them is given. (7.)

and CA CD::R: Col S, C. (1) and C being found, A and B are found by case 2. or 3.

SPHERICAL TRIGONOMETRY.

DEFINITION'S,

I.

THE pole of a circle of the sphere is a point in the superficies of the sphere, from which all straight lines drawn to the circumference of the circle are equal.

II.

A great circle of the sphere is any whose plane passes through the centre of the sphere, and whose centre therefore is the same with that of the sphere.

III.

A spherical triangle is a figure upon the superficies of a sphere comprehended by three arches of three great circles, each of which is less than a semicircle.

IV.

A spherical angle is that which on the superficies of a sphere is contained by two arches of great circles, and is the same with the inclination of the planes of these great circles.

PROP. I.

GREAT circles bisect one another.

As they have a common centre their common section will be a diameter of each which will bisect them.

PROP. II. FIG. 1.

THE arch of a great circle betwixt the pole and the circumference of another is a quadrant.

Let ABC be a great circle, and D its pole; if a great circle DC pass through D, and meet ABC in C, the arch DC will be a quadrant.

Let the great circle CD meet ABC again in A, and let AC be the common section of the great circles which will pass