PROPOSITION VIII.

In oblique-angled spherical triangles, a perpendicular arc being drawn from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle.

Let ABC be a triangle, and the arc CD perpendicular to the base BA; the cosine of the angle B will be to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD.

For (Sp. Tr. 6) the cosine of the angle B is to the sine of the angle BCD, as the cosine of the side CD is to the radius;

and also the cosine of the angle A to the sine of the angle ACD in the same ratio; therefore, by alternation, the cosine of the angle B is to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD.

PROPOSITION IX.

The same things remaining, the cosines of the sides BC, CA, are proportional to the cosines of BD, DA, the segments of the base.

For (Sp. Tr. 5) the cosine of BC is to the cosine of BD, as the cosine of DC to the radius, and the cosine of AC to the cosine of AD in the same ratio; wherefore, by alternation, the cosines of the sides BC, CA, are proportional to the cosines of the segments of the base BD, DA.

PROPOSITION X.

The same construction remaining, the sines of BD, DA, the segments of the base, are reciprocally proportional to the tangents of B and A, the angles at the base.

For (Sp. Tr. 1) the sine of BD is to the radius, as the tangent of DC to the tangent of the angle B; and also, the radius to the sine of AD, as the tangent of A to the tangent

of DC; therefore, by indirect equality, the sine of BD is to the sine of DA, as the tangent of A to the tangent of B.

PROPOSITION XI.

The same construction remaining, the cosines of the segments of the vertical angle are reciprocally proportional to the tangents of the sides.

For (Sp. Tr. 4) the cosine of the angle BCD is to the radius, as the tangent of CD is to the tangent of BC; and also (Sp. Tr. 4, by inversion), the radius is to the cosine of the angle ACD, as the tangent of AC to the tangent of CD; therefore, by indirect equality, the cosine of the angle BCD is to the cosine of the angle ACD, as the tangent of AC is to the tangent of BC.

PROPOSITION XII.

If, from the extremities of the base of any spherical triangle, arcs of great circles be described to meet the sides, and to cut off a part on each from the vertex, equal to the other side; the rectangle under the sines of half these arcs shall be equal to the rectangle under the sines of the excesses of the semiperimeter above the two sides.

Let ABC be a spherical triangle, having two unequal sides AB, BC. From BC, the greater, let BD be cut off equal to BA, and on BA produced make

BE BC; and let great circles
=
pass
through A, D, and E, C. Then if S
denote half the sum of the sides,
Sin AD sin EC-Sin (S-AB)· H
sin (SBC).

For, in BA produced, let AF= AC, and FG = AE; also, let the straight lines AG, GD, DA, EF, FC, CE, be drawn. The straight lines AD and EC are parallel; for, if a great circle bisect the angle at

F

B

B, it also bisects the arcs AD, EC, and is perpendicular to their planes; therefore the cords AD, EC, are perpendicular to the lines of common section, and consequently both are perpendicular to the plane of that great circle, and are

parallel. But AG and EF are also parallel; therefore the plane of the triangle DAG is parallel to the plane of the triangle CEF. Let the plane AFC, which cuts the latter in FC, cut the former in HK. Then HK is parallel to FC (So. Ge. I. 14). Therefore, since the cords AF, AC, are equal, HK is a tangent to the circle that passes through A, F, C, and consequently it is also a tangent to the circle that passes through A, D, G (Sp. Ge. 4, Cor. 5). Hence the angle ADG = GAH = EFC. But the angle DAG

CEF (So. Ge. I. 9). Therefore the triangles AGD, EFC, are equiangular, and AD: AG: = EF: EC, or AD: AG = EF: EC; or, considering the arcs, sin AD: sin & AG sin EF: sin EC; and hence Sin AD sin EC = Sin AG sin EF. But arc AG = AF +FG AC+AE AC+BC-AB, and AG (AC+BC+AB-2AB)=S-AB. Also EF AF

=

AE AC
=

=

=

(BC AB) = AC BC + AB (PI. Ge. Ad. II. 2), or EF (AC+ AB + BC-2 BC)= = (AC+AB+ S-BC. Hence, Sin AD sin EC Sin (SAB) · sin (S-BC).

PROPOSITION XIII.

In any spherical triangle, the rectangle under the sines of the two sides is to the square of the radius, as the rectangle under the excesses of the semiperimeter above these sides, to the square of the sine of half the vertical angle.

Let ABC be the triangle, BD =BA, BE=BC, and BFG bisecting the vertical angle; and AD, EC, drawn as in the preceding figure; then BFG bisects AD, EC, and cuts them at right angles, for it passes through their pole. In the right-angled triangles ABF, EBG,

D

Sin AB: R= sin AF: sin & B,
Sin BE: R= sin EG: sin B;

B

F

A

E

hence (Pl. Ge. VI. 23, Cor. 1) Sin AB sin BC: R2 = Sin AF sin EG: sin2 B. But (Sp. Tr. 12) Sin AF · sin EG Sin (SAB) sin (SBC); therefore,

Sin AB sin BC: R2-Sin (S-AB) sin (S-BC): sin2 B.

COR.-If the angles of the triangle be denoted by A, B, and C, and the sides opposite to them respectively by a, b, and c, and half the sum of the sides by s, then sin (sa). sin (8-c)

Sin2 B =

sin a. sin c

if R = 1.

By changing B into C, and c into b, a similar formula is found for sin C; and also, in the same manner, for sin2 A.

Solution of the Cases of Right-Angled Spherical Triangles.

PROBLEM.

In a right-angled spherical triangle, of the three sides and three angles, any two being given, besides the right angle to find the other three.

This problem has sixteen cases, the solutions of which are contained in the following table, where ABC is any spherical triangle right-angled at A.

AC and BC.

AB and AC.

B and C.

C.

AB.

B.

C.

BC.

B.

C.

tan B: tan AC-R : sin AB, (1)
sin B: sin ACR : sin BC, (2)
cos AC: cos B=R:sin C, (6)

456

789

cos AC: cos BC=R: cos AB, (5) |10 sin BC: sin AC=R : sin B, (2) │11 tan BC: tan ACR: cos C, (4) 12

R: cos AB=cos AC: cos BC, (5) 13
sin AB: R-tan AC: tan B, (1) 14
sin AC: R=tan AB: tan C, (1) 14

AB. sin B: cos C = R: cos AB, (6) 15
AC. sin C: cos BR: cos AC, (6) |15
BC. tan B: cot CR: cos BC, (3) |16]

B

A

Table for determining when the things found in the preceding are less than a Quadrant (Sp. Ge. 14 and 15).

The angle or arc found is less than 90°.

When B is less than 90°.

When BC and B are of the same affection.
When BC and B are of the same affection.
When C is less than 90°.

When AC and C are of the same affection.
When AC is less than 90°.

123

456

789

Ambiguous.

Ambiguous.

Ambiguous.

When AC and BC are of the same affection.
When AC is less than 90°.

10

11

When AC and BC are of the same affection.

12

When AB and AC are of the same affection.

13

When B and C are of the same affection.

16

Schol. The rules for the cases of right-angled spherical trigonometry may be reduced to two, called Napier's Rules of the Circular Parts.

In a right-angled spherical triangle, the right angle is neglected, and the hypotenuse, the two angles, and the complements of the two sides, are called circular parts; and any of these parts being called the middle part; the two