PROP. XLIX.

The excess of the sum of the angles of a spherical triangle above two right angles, is to a right angle, as the triangle is to a tri-quadrantal triangle.

Let ABC be any spherical triangle; the excess of the sum of the angles A, B, and C above two right angles, is to a right angle, as ABC is to a tri-quadrantal triangle

Produce the sides AC and AB, but not so far as to make them intersect, and through E and F in

G

D

A

H

Ι

the parts produced, draw the circumference of a great circle EFGH; and let BA and CA produced meet this circumference again in D and G; also let BC produced meet it in H and I.

Then, by the two preceding propositions, the angle EAF, which is the angle A in the triangle BAC, is to four right angles, as the sum of the triangles EAF and DAG is to the spherical surface or eight tri-quadrantal triangles. If R denote a right angle and T a triquadrantal triangle, this proportion may be thus expressed,

A 4R:: EAF+DAG: 8T;

or, A R:: EAF+DAG: 2T.

:

To this may be added these two other like proportions;

B: R::DBI+EBH: 2T,

C: R:: FCI+GCH: 2T.

Hence (Euc. 5, 24),

A+B+C: R:: EAF + DAG + DBI + EBH+FCI+GCH : 2T Now by inspecting the figure it is seen that the third term of this proportion exceeds the convex surface of the hemisphere whose base is the circle EFGH, by twice the triangle ABC. It may therefore be replaced by the expression 4T+2ABC. The proportion then will be A+B+C R:: 4T+2ABC: 2T, A+B+C 2R:: 2T+ABC: 2T.

or

Hence (Euc. 5, 17),

A+B+C-2R: 2R:: ABC : 2T,

or A+B+C-2R: R:: ABC: T;

that is, the excess of the sum of the angles of the triangle ABC above two right angles, is to a right angle, as the triangle itself is to a tri-quadrantal triangle; which was to be proved.

PROP. L.

If the sum of the angles of a spherical polygon together with four right angles be diminished by twice as many right angles as there are sides of the polygon, the remainder will be to a right angle as the polygon is to a tri-quadrantal triangle.

Let ABCDE be a spherical polygon from A draw arcs to each of the other angular points of the polygon, except the adjacent ones, B and E; then will the polygon be divided into as as many triangles, wanting two, as it has sides.

C

3

Now by the last proposition, the sum of the angles of each triangle diminished by two right angles is to a right angle as the triangle is to a tri-quadrantal triangle. Hence (Euc. 5, 24), as the sum of all the angles of all the triangles diminished by twice as many right angles as there are triangles is to a right angle, so is the sum of all the triangles to a tri-quadrantal triangle. But the sum of all the triangles, is the polygon ABCDE, and the sum of the angles of the triangles is the same as that of the angles of the polygon. Therefore, the sum of the angles of the polygon diminished by twice as many right angles as there are triangles, is to a right angle, as the polygon is to a tri-quadrantal triangle. Now if the sum of the angles of the polygon, instead of being diminished by twice as many right angles as there are triangles, be first increased by four right angles, and then diminished by twice as many right angles as there are sides of the polygon, the result will not be changed. Then, the sum of the angles of a spherical polygon together with four right angles when diminished by twice as many right angles as there are sides of the polygon, is to a right angle, as the polygon is to a tri-quadrantal triangle.

SPHERICAL TRIGONOMETRY.

ART. 1. If two of the sides and angles of a spherical triangle be either two right angles, two quadrants, or a right angle and a quadrant adjacent or opposite, then is one vertex of the triangle a pole of the opposite side, and the angle at the vertex is measured by this side; the other two sides are quadrants, and the opposite angles right angles: and to be completely determined, such a triangle must have given its oblique angle or the opposite side which is a measure of this angle. These parts can neither of them be computed from the other parts of the triangle.

Triangles of this kind may be called bi-quadrantal, and in what follows they are left out of view.

Among such triangles are to be classed those that are tri-quadrantal.

of the sides and angles of a spherical triangle not biquadrantal, if any three be given, the other three may in general be determined from them. The cases in which this cannot be done, will be duly noticed hereafter.

It is the business of Spherical Trigonometry to furnish methods of computing the unknown from the known parts of a spherical triangle.

Spherical as well as Plane Trigonometry is divided into two branches; one relating to right and the other to oblique-angled triangles. It is convenient to treat first of triangles that are right-angled.

RIGHT-ANGLED SPHERICAL TRIANGLES.

2. If from the vertex of one of the oblique angles in a right-angled spherical triangle, as a pole, an arc of a great circle be drawn, meeting the sides that contain the other oblique angle (produced beyond the angular point if necessary), a second triangle will thus be formed, having one of its vertices at this angular point; which triangle is said to be complemental to the former triangle.

Thus, if ABC be a triangle rightangled at A, and from the vertex C as a pole an arc of a great circle DE be drawn, meeting AB and CB produced, the triangle DBE thus formed, having one of its vertices at B, is said to be complemental to ABC.

D

If from the vertex B as a pole, an arc of a great circle be drawn, meeting AC and BC produced, another triangle will be formed complemental to ABC. And thus every spherical triangle has two complemental triangles.

3. If one triangle be complemental to another, the latter is complemental to the former.

For in the preceding figure, the triangle BDE is right-angled at D, since C is the pole of DE: and E is