Sidebilder
PDF
ePub

F

B В

A

PROPOSITION IV, In right-angled spherical triangles, the cosine of an angle is to the radius, as the tangent of the side adjacent to that angle is to the tangent of the hypotenuse.

The same construction remaining. In the triangle CEF (Sp. Tr. 1), the sine of the side EF is to the radius, as the tangent of the other side CE is to the tangent of the angle CFE opposite to it; that is, in the triangle ABC, the cosine of the angle ABC is to the radius as the cotangent of the hypotenuse BC to the cotangent of the side AB, adjacent to ABC, or as the tangent of the side AB to the tangent of the hypotenuse, since the tangents of two arcs are reciprocally proportional to their cotangents (Pl. Tr. 4, Cor. to Def.) Cor.–And since by this proposition the cosine of the

angle ABC is to the radius, as the tangent of the side AB is to the tangent of the hypotenuse BC; and as the radius is to the cotangent of BC, so is the tangent of BC to the radius; by equality, the cosine of the angle ABC will be to the cotangent of the hypotenuse BČ, as the tangent of the side AB, adjacent to the angle ABC, to the radius.

PROPOSITION V. In right-angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the hypotenuse is to the cosine of the other side.

The same construction remaining. In the triangle CEF, the sine of the hypotenuse CF is to the radius, as the sine of the side CE to the sine of the opposite angle CFE (Sp. Tr. 2); that is, in the triangle ABC, the cosine of the side CA is to the radius as the cosine of the hypotenuse BC to the cosine of the other side BA.

PROPOSITION VI. In right-angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle opposite to that side is to the sine of the other angle.

The same construction remaining. In the triangle CEF, the sine of the hypotenuse CF is to the radius as the sine of the side EF is to the sine of the angle ECF opposite to it; that is, in the triangle ABC, the cosine of the side CA is to the radius, as the cosine of the angle ABC, opposite to it, is to the sine of the other angle ACB.

PROPOSITION VII. In spherical triangles, whether right-angled or obliqueangled, the sines of the sides are proportional to the sines of the angles opposite to them.

First, let ABC be a right-angled triangle, having a right angle at A ; therefore (Sp. Tr. 2) the sine of the hypotenuse BC is to the radius (or the sine of the right angle at A) as the sine of the side AC to the sine of the angle B. And in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (Pl. Ge. V. 11) the sine of the side AC is to the sine of the angle B, as the sine of AB to the sine of the angle C.

Secondly, let ABC be an oblique-angled triangle, the sine of any

of the sides BC, will be to the sine of any of the other two AC, as the sine of the angle A, opposite to BC, is

B

A

[ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small]

to the sine of the angle B, opposite to AC. Through the point C, let there be drawn an arc of a great circle CĎ perpendicular upon AB; and in the right-angled triangle BCD (Sp. Tr. 2), the sine of BC is to the radius, as the sine of CD to the sine of the angle B; and in the triangle ADC, by inversion, the radius is to the sine of AC as the sine of the angle A to the sine of DC; therefore, by indirect equality, the sine of BC is to the sine of AC, as the sine of the angle A to the sine of the angle B.

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 ;

[blocks in formation]

D

А 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 È, C. Then if s denote half the sum of the sides, Sin . AD sin EC=Sin (S_AB): ! sin (S-BC).

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 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

B

E

= AF

[ocr errors]

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= 1 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 + FG = AC + AE= AC + BC --- AB, and AG=; (AC+ BC + AB-2 AB) = S-- AB. Also EF = AR

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

A and cuts them at right angles, for it passes through their pole. In the right-angled triangles ABF, EBG,

B

D

[ocr errors]

E

B;

Sin BE:R=sin EG: sin hence (Pl. Ge. VI. 23, Cor. 1) Sin AB sin BC: R2 = Sin AF.sin EG:sin? į B. But (Sp. Tr. 12) Sin AF.sin EG =Sin (S-AB) sin (S-BC); therefore, Sin AB sin BC:R=Sin (S-AB).sin (S-BC):sinB.

« ForrigeFortsett »