Fig. 15.

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 hypothe nuse 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:

the hypothenuse. gles B, C. which B is the complement.

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.

3

The an-j

S, CS, A :: AB : BC. and also, S, C: S, B, : : AB : AC. (2.)

AC: AB::S, B : S, C.

two sides and an an-gles A and (2.) This case admits of

gle opposite to one C.

of them.

two solutions; for C may be

greater or less than a quad-
rant. (Cor. to def. 4.)

The an-AB+AC: AB-AC: : T,

two sides and the gles B and C+B: T, C-B : (3.)

AB, AC, and A,

included angle. IC.

Fig. 16,

17.

SPHERICAL TRIGONOMETRY.

DEFINITIONS.

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 cir cumference 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. I.

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 through

E the centre of the sphere: join DE, DA, DC: by def. 1. DA, DC are equal, and AE, EC are also equal, and DE is common; therefore (8. 1.) the angles DEA, DEC are equal; wherefore the arches DA, DC are equal, and consequently each of them is a quadrant. Q. E. D.

PROP. III. FIG. 2.

IF a great circle be described meeting two great circles AB, AC passing through its pole A in B, C, the angle at the centre of the sphere upon the circumference BC, is the same with the spherical angle BAC, and the arch BC is called the measure of the spherical angle BAC.

Let the planes of the great circles AB, AC intersect one another in the straight line AD passing through D their common centre; join DB, DC.

Since A is the pole of BC, AB, AC will be quadrants, and the angles ADB, ADC right angles; therefore (6. def. 11.) the angle CBD is the inclination of the planes of the circles AB, AC; that is, (def. 4.) the spherical angle BAC. Q. E. D.

COR. If through the point A, two quadrants AB, AC be drawn, the point A will be the pole of the great circle BC, passing through their extremities B, C.

Join AC, and draw AE a straight line to any other point E in BC; join DE: since AC, AB are quadrants, the angles ADB, ADC are right angles, and AD will be perpendicular to the plane of BC therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE, each to each; therefore AE, AC are equal and A is the pole of BC, by def. 1. Q. E. D.

PROP. IV. FIG. 3.

IN isosceles spherical triangles, the angles at the base are equal.

Let ABC be an isosceles triangle, and AC, CB the equal sides; the angles BAC, ABC, at the base AB, are equal.

Let D be the centre of the sphere, and join DA, DB, DC; in DA take any point E, from which draw, in the plane ADC, the straight line EF at right angles to ED meeting CD