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

F

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

▲ DAE + ▲ AKL = 2A × T, (Prop. 15, Cor. 2). ▲ BEF + ABPK = 2B × T.

ACFL + A CDP = 20 × T.

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

whence,

That is,

AABC A.T+ B.T+ CT-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 denotes 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 polygoh 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

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

Enk

plane CBD is perpendicular to the plane DBA, CH will lie in the plane CBD, and be perpendicular to the line DB, and perpendicular to all lines that can be drawn in the plane DBA, from the point H (Def. 2, B. VI).

Draw HG perpendicular to DA, and draw GC; GC will lie wholly in the plane CDA, and CHG is a rightangled triangle, right-angled at H.

We will now demonstrate that the angle DGC is a right angle.

The right-angled ACHG, gives CH2+HG2-CG2 (1) The right-angled ADGH, gives DG2+HG2=DH2 (2) By subtraction, CH-DG2 CG- DH'

=

By transposition, CH + DH2 = CG2 + DG2

(3) (4)

But the first member of equation (4), is equal to CD', because CDH is a right-angled triangle;

Therefore,

=

CD2 CG2 + DG2

Hence, CD is the hypotenuse of the right-angled triangle DGC, (Th. 39, B. I).

From the point B, draw BE at right angles to DA, and BF at right angles to DB, in the plane CDB extended; the point F will be in the line DC. Draw EF, and as F is in the plane CDA, and E is in the same plane, the line EF is in the plane CDA. Now we are to prove that the triangle CHG is similar to the triangle BEF, and similarly situated.

As HG and BE are both at right angles to DA, they are parallel; and as HC and BF are both at right angles to DB, they are parallel; and by reason of the parallels, the angles GHC and EBF are equal; but GHC is a right angle; therefore, EBF is also a right angle.

Now, as GH and BE are parallel, and CH and BF are also parallel, we have,

DH : DB = HG : BE

And,

ᎠᎻ : ᎠᏴ

==

HC: BF

Therefore,

HG BE =
:

Or,

HG: HC =

HC: BF (Th. 6, B. II),
BE: BF.

Here, then, are two triangles, having an angle in the one equal to an angle in the other, and the sides about the equal angles proportional; the two triangles are therefore equiangular, (Cor. 2, Th. 17, B. II); and they are similarly situated, for their sides make equal angles at H and B with the same line, DB.

Hence the proposition.

SCHOLIUM. By the definition of sines, cosines, and tangents, wẻ perceive that CH is the sine of the arc BC, DH is its cosine, and BF its tangent; CG is the sine of the arc AC, and DG its cosine. Also, BE is the sine of the arc AB, and DE is the cosine of the same arc. With this figure we are prepared to demonstrate the following propositions.

PROPOSITION II.

In any right-angled spherical triangle, the sine of one side is to the tangent of the other side, as radius is to the tangent of the angle adjacent to the first-mentioned side.

Or, the sine of one side is to the tangent of the other side, as the cotangent of the angle adjacent to the first-mentioned side is to the radius.

For the sake of brevity, we will represent the angles of the triangle by A, B, C, and the sides or arcs opposite to these angles, by a, b, c, that is, a opposite A, etc. In the right-angled plane triangle EBF, we have, EB: BF = R: tan. BEF

That is, sin.c tan.a R tan.A,

=

which agrees with the first part of the enunciation. By reference to equation (5), Section I, Plane Trigonometry, we shall find that,