DEHAD; hence the sum of the A's, BDC and ADE, 18 equivalent to the same lune.

By the same course of reasoning, we prove that the sum of the opposite A's, DAC and DBE, is equivalent to the lune DCHAD, whose angle is ADC.

PROPOSITION X V.

The surface of a lune is to the whole surface of the sphere, as the angle of the lune is to four right angles; or, as the arc which measures that angle is to the circumference of a great circle.

Let ABFCA be a lune on the surface of a sphere, and BCE an arc of a great circle, whose poles are A and F, the vertices of the angles of the lune. The arc, BC, will then measure the angles of the lune. Take any arc, as BD, that will be contained an exact number of times in BC, and in the whole circum

E

ference, BCEB, and, beginning at B, divide the arc and the circumference into parts equal to BD, and join the points of division and the poles, by arcs of great circles. We shall thus divide the whole surface of the sphere into a number of equal lunes. Now, if the arc BC contains the arc BD m times, and the whole circumference contains this arc n times, the surface of the lune will contain m of these partial lunes, and the surface of the sphere will contain n of the same; and we shall have, Surf. lune surf. sphere :: m : n.

But, mn: BC: circumference great circle; Lence, surf. lune surf. sphere :: BC cir. great circle; surf.lune : surf. sphere :: BOC: 4 right angles.

or,

This demonstration assumes that BD is a common measure of the arc, BC, and the whole circumference. It may happen that no finite common measure can be found; but our reasoning would remain the same, even though this common measure were to become indefinitely small.

Hence the proposition.

Cor. 1. Any two lunes on the same sphere, or on equal spheres, are to each other as their respective angles.

SCHOLIUM.-Spherical triangles, formed by joining the pole ɔf an arc of a great circle with the extremities of this arc by the arcs of great circles, are isosceles, and contain two right angles. For this reason they are called bi-rectangular. If the base is also a quadrant, the vertex of either angle becomes the pole of the opposite side, and each angle is measured by its opposite side. The three angles are then right angles, and the triangle is for this reason called tri-rectangular. It is evident that the surface of a sphere contains eight of its trirectangular triangles.

Cor. 2. Taking the right angle as the unit of angles, and denoting the angle of a lune by A, and the surface of a tri-rectangular triangle by T, we have,

surf. of lune : 8T :: A: 4;

whence, surf. of lune = 2A × T.

Cor. 3. A spherical ungula bears the same relation to the entire sphere, that the lune, which is the base of the ungula, bears to the surface of the sphere; and hence, any two spherical ungulas in the same sphere, or in equal spheres, are to each other as the angles of their respective lunes.

PROPOSITION XVI.

The area of a spherical triangle is measured by the excess of the sum of its angles over two right angles, multiplied by the tri-rectangular triangle.

Let ABC be a spherical triangle, and DEFLK the eir cumference of the base of the hemisphere on which this triangle is situated.

Produce the sides of the triangle until they meet this circumference in the points, D, E, F, L, K, and P, thus forming the sets of opposite triangles, DAE, AKL; BEF, BPK; CFL, CDP.

Now, the triangles of each of these sets are together equal to a lune, whose angle is the cor

P

B

K

F

responding angle of the triangle, (Prop. 14); hence we

If the first members of these equations be added, it is evident that their sum will exceed the surface of the hemisphere by twice the triangle ABC; hence, adding these equations member to member, and substituting for the first member of the result its value, 4T+ 2AABC, we have

or,

4T+2AABC2A.T+ 2B.T+2C.T

2T+ AABC A.T+ B.T+ C.T

whence,

That is,

=

AABC A.T+ B.T+ C.T-2T.

=

^ABC = (A + B + C − 2) T.

But A + B + C · 2 is the excess of the sum of the angles of the triangle over two right angles, and T do notes the area of a tri-rectangular triangle.

Hence the proposition; the area, etc.

PROPOSITION XVII.

The area of any spherical polygon is measured by the excess of the sum of all its angles over two right angles, taken as many times, less two, as the polygon has sides, multiplied by the tri-rectangular triangle.

Let ABCDE be a spherical polygon; then will its area be measured by the excess of the sum of the angles, A, B, C, D, and E, over two right angles taken a number of times which is two less than the number of sides, multiplied by T, the tri-rectangular triangle. Through the vertex of any of the angles, as E, and the vertices of

B

the opposite angles, pass arcs of great circles, thus dividing the polygon into as many triangles, less two, as the polygon has sides. The sum of the angles of the several triangles will be equal to the sum of the angles of the polygon.

Now, the area of each triangle is measured by the excess of the sum of its angles over two right angles, multiplied by the tri-rectangular triangle. Hence the sum of the areas of all the triangles, or the area of the polygon, is measured by the excess of the sum of all the angles of the triangles over two right angles, taken as many times as there are triangles, multiplied by the trirectangular triangle. But there are as many triangles as the polygon has sides, less two.

Hence the proposition; the area of any spherical polygon, etc.

Cor. If S denote the sum of the angles of any spherical polygon, n the number of sides, and T the tri-rectangular triangle, the right angle being the unit of angles; the area of the polygon will be expressed by

[S-2 (n-2)] × T = (S— 2n + 4) T.

SECTION II.

SPHERICAL TRIGONOMETRY.

A Spherical Triangle contains six parts-three sides and three angles-any three of which being given, the other three may be determined.

Spherical Trigonometry has for its object to explain the different methods of computing three of the six parts of a spherical triangle, when the other three are given. It may be divided into Right-angled Spherical Trigonometry, and Oblique-angled Spherical Trigonometry; the first treating of the solution of right-angled, and the second of oblique-angled spherical triangles.

RIGHT-ANGLED SPHERICAL TRIGONOMETRY.
PROPOSITION I.

With the sines of the sides, and the tangent of ONE SIDE of any right-angled spherical triangle, two plane triangles can be formed that will be similar, and similarly situated.

Let ABC be a spherical triangle, right-angled at B; and let D be the center of the sphere. Because the angle CBA is a right angle, the plane CBD is perpendicular to the plane D DBA. From C let fall CH, perpendicular to the plane DBA; and as the

EA

F

B