and F, and the base C B in G: also draw the perpendicular AD.

Then, because AE, AF are each equal to ac, it is manifest that BE is the sum of the sides A B, AC, and BF their difference.

And because the perpendicular AD, from the centre, bisects CG in D, it is also plain that BG is the difference of the segments of the base CD, DB, or DG, DB.

But since the lines CG, EF in the circle, cut each other in B, the rectangle of EB, BF is equal to the rectangle of CB, BG.

Hence, CB is to BE as BF is to BG; that is, the base BC is to the sum of the sides AB, AC, as the difference of those sides is to the difference of the segments of the base CD, db.

Q. E. D.

SCHOLIUM. When the perpendicular AD falls without the triangle, the segments of the base must be both reckoned the same way, from the angle c or B, to the foot of that perpendicular.

To these three propositions, which furnish solutions to all the problems that can occur in Plane Trigonoinetry, it may be proper to add the following, which, when applied to each of the angles, is alone sufficient for that purpose.

THEOREM IV.

96. As twice the rectangle of any two sides of a plane triangle radius: sum of the squares of the same two sides-the square of the other side: cosine of the angle included by those sides.

For let AD be perpendicular to the base BC, falling either within or without the triangle, as in the figures. Then, in the case in which the perpendicular falls within the triangle, we shall have Ac2= AB2+ BC2—

A B2+B C A C3

2 BC X BD, or BD = 2 BC

But ABD being a right-angled triangle, it will be as

rad: AB :: sin BAD or COS B: BD; or BD =

A B COS B

And if this value be substituted in the first equation,

Again, if the perpendicular AD falls without the triangle, we shall have, AC2= AB2+ BC2+ 2 BC X BD,

2

And since ABD is a right-angled triangle, rad: AB ::

sin BAD or cos ABD: BD; or BD =

AB COS A BD

T

But ABD being the supplement of ABC, COS ABD=

And if this value be substituted in the first equation,

=rx

AB+BC-AC

2 AB X BC

; whence, if these equations be

turned into analogies, we shall have 2 AB X BC: rad :: AB+BC-AC2: COS ABC.

2

SCHOLIUM. If the three angles of the triangle be denoted by A, B, C, and their opposite or corresponding sides by a, b, c, the four theorems here demonstrated may be exhibited in general terms, as follows:

IN SPHERICAL TRIGONOMETRY.

In treating of the practical part of Plane Trigonometry, no distinction was made between right-angled and oblique-angled triangles, on account of the three rules, which are there given, being sufficient to solve every problem that can occur in this branch of the subject, whatever may be the species of the triangle.

But in Spherical Trigonometry, where, from the nature of the subject, the rules of practice become more numerous, it was judged proper to class the various species of these kind of triangles under the three heads of right-angled, quadrantal, and oblique-angled triangles, and to give the rule for each case separately.

The common rules, however, for the various cases in this branch of the science, as well as in the former, may

be all derived from the three following theorems, which have the same generality with those mentioned above; and are here demonstrated, from geometrical principles, in a manner equally simple and perspicuous.

THEOREM I.

97. In any right-angled spherical triangle ABC, the sine of either of the legs is to radius, as the tangent of the other leg is to the tangent of its opposite angle.

For let o be the centre of the sphere, and having joined OA, OC, OB, draw CE perpendicular to OB: also, in the plane AOB, draw FE perpendicular to the same line OB, meeting OA, produced, in F; and join FC.

Then, since OE is at right angles to both EC and EF, it will be at right angles to the plane EFC.

And because the plane coв passes through OE, it will also be perpendicular to the plane EFC; or, which is the same thing, the plane FFC will be perpendicular to the plane COB.

But the angle ACB being a right angle (by hyp.) the plane COA or COF, will be perpendicular to the same plane COB.

Hence the planes EFC, COF being each perpendicular to the plane coв, their common section Fc will, also, be perpendicular toc o B.

And since EC, EF, which lie in the planes COB, Aob, are each perpendicular to OB, the angle FEC will be the measure of their inclination, or of the spherical angle CBA.

Also, FCO, CEO, being right angles, FC will be the tangent of the arc AC, and CE the sine of the arc CB, to the radius of the sphere o c.

Hence, FCE being a right-angled plane triangle, rightangled at c, we shall have EC: rad :: CF: tan FEC; or sin CB rad :: tan AC: tan ABC.

Q. E. D.

SCHOLIUM. By using the same notation as in the former theorem the present one becomes r tan b = sin a tan B; and if

дов

cot B

be put for tan B, we shall have

r sin a tan b cot B,

which is the formula for the second case of rightangled spherical triangles.

THEOREM II.

98. In any spherical triangle ABC, whether rightangled or oblique-angled, the sines of the sides are as the sines of their opposite angles; and conversely,

E

D

F

B

For, let o be the centre of the sphere, and having joined o A, OC, OB, draw AD perpendicular to the plane OBC; also make DE perpendicular to OB, and DF to oc; and join AE, AF.