RULE I.

The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the tangents of the adjacent part.

RULE II.

The rectangle contained by the radius and the sine of the middle part, is equal to the rectangle contained by the cosines of the opposite parts.

These rules are demonstrated in the following manner: First, let either of the sides, as BA be the middle part, and therefore the complement of the angle B, and the side AC, will be adjacent extremes. And by Cor. 2. Prop. 17. of this, S, BA, is to the Co-T, B, as T, AC is to the radius, and therefore RXS, BA Co-T, В ХТ, АС.

B

AC

Fig.40.

A

The same side BA being the middle part, the complement of the hypothenuse, and the complement of the angle C, are opposite extremes: and by Prop. 18. S, BC is to the radius, as S, BA to S, C; therefore RXS, BAS, BC XS, C.

Secondly, let the complement of one of the angles, as B, be the middle part, and the complement of the hypothenuse, and the side BA will be adjacent extremes: and by Cor. Prop. 20. Co-S, B is to Co-T, BC as T, BA is to the radius, and therefore RX Co-S, B=Co-T, BCXТ, ВА.

Again, let the complement of the angle B be the middle part, and the complement of the angle C, and the side AC will be opposite extremes: and by Prop. 22. Co-S, AC is to the radius, as Co-S, B is to S, C; and therefore RX Co-S, B= Co-S, ACS, C.

Thirdly, let the complement of the hypothenuse be the middle part, and the complements of the angles B, C, will be adjacent extremes: but by Cor. 2. Prop. 19. Co-S, BC is to Co-T, C as Co-T, B to the radius: therefore RX Co-S, BC= Co-T, CX Co T, B.

:

Again, let the complement of the hypothenuse be the middle part, and the sides AB, AC will be opposite extremes: but by Prop. 21. Co-S, AC is to the radius, as Co-S, BC to Co-S, BA; therefore RX Co-S, BC Co-S, BAX Co-S, AC.

Q. E. D.

SOLUTION OF THE SIXTEEN CASES OF RIGHT-ANGLED
SPHERICAL TRIANGLES.

GENERAL PROPOSITION.

In a right-angled Spherical triangle, of the three sides and three angles, any two being given, besides the right angle, the other three may be found.

In the following Tables the solutions are derived from the preceding propositions. It is obvious that the same solutions may be derived from Baron Naper's two rules above demonstrated, which, as they are easily remembered are commonly used in practice.

Case. Given. Sought.

1

R:CoS, AC::S, C: CoS, B; and B is See fig. 40.

B of the same species with CA, by 22.

9

T, B: R:: T, CA: S, BA. 17.

R: CoS, C:: T, BC: T, CA. 20. If BC, CAC BC be less or greater than a quadrant, Cand B will be of the same or different affection. 15. 13.

The second, eighth, and thirteenth cases, which are commonly called ambiguous, admit of two solutions: for in these it is not determined, whether the side or measure of the angle sought be greater or less than a quadrant...

:

PROPOSITION XXIII.

In spherical triangles, whether right-angled or oblique-angled,

:

the sines of the sides are proportional to the sines of the an-
gles opposite to them.

First, let ABC be a right-angled triangle, having a right Seefig. 40.

angle at A; therefore, by Prop. 18. the sine of the hypothe-

nuse 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 the 11.5.
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.

1
i

C

Fig.

41.

Secondly, let BCD be an oblique-angled triangle, the sine
of either of the sides BC, will be to the sine of either of the
other two CD, as the sine of the angle D op-
posite to BC is to the sine of the angle B op-
posite to the side CD. Through the point C,
let there be drawn an arch of a great circle
CA perpendicular upon BD; and in the
right-angled triangle ABC * the sine of BC is
to the radius as the sine of AC to the sine of
the angle B; and in the triangle ADC (by 18.
of this): and, by inversion, the radius is to B
the sine of DC as the sine of the angle D to

the sine of AC: therefore, ex æquo pertur-

D

B

A
C

• 18. of this.

Fig.

42.

D

bato, the sine of BC is to the sine of DC, as the sine of the

angle D to the sine of the angle B.

Q. E. D.

PROPOSITION XXIV.

1

In oblique-angled spherical triangles, having drawn a perpen-
dicular arch from any of the angles upon the opposite side,
the cosines of the angles at the base are proportional to the
sines of the vertical angles.

:

- Let BCD be a triangle, and the arch CA perpendicular to See fig. 41.
the base BD; the cosine of the angle B will be to the cosine and 42.
of the angle D, as the sine of the angle BCA tto he sine of the
angle DCA.

For by 22. the cosine of the angle B is to the sine of the
angle BCA as (the cosine of the side AC is to the radius; that

is, by Prop. 22. as) the cosine of the angle D to the sine of the angle DCA; and, by permutation, the cosine of the angle B is to the cosine of the angle D, as the sine of the angle BCA to the sine of the angle DCA. Q. E. D.

PROPOSITION XXV.

The same things remaining, the cosines of the sides BC, CD, are proportional to the cosines of the bases BA, AD.

See fig. 41. and 42.

For by 21. the cosine of BC is to the cosine of BA, as (the cosine of AC to the radius; that is, by 21. as) the cosine of CD is to the cosine of AD: wherefore, by permutation, the cosines of the sides BC, CD are proportional to the cosines of the bases BA, AD. Q. E. D.

PROPOSITION XXVI.

See fig. 41. and 42.

The same construction remaining, the sines of the bases BA, AD are reciprocally proportional to the tangents of the angles B and D at the base.

For by 17. the sine of BA is to the radius, as the tangent of AC to the tangent of the angle B; and by 17. and inversion, the radius is to the sine of AD, as the tangent of D to the tangent of AC: therefore, ex æquo perturbato, the sine of BA is to the sine of AD, as the tangent of D to the tangent of

B. Q. E. D.

See fig. 41.

and 42.

PROPOSITION XXVII.

The cosines of the vertical angles are reciprocally proportional to the tangents of the sides.

For by Prop. 20. the cosine of the angle BCA is to the radius as the tangent of CA is to the tangent of BC; and by the same Prop. 20. and by inversion, the radius is to the cosine of the angle DCA, as the tangent of DC to the tangent of CA: therefore, ex æquo perturbato, the cosine of the angle BCA is to the cosine of the angle DCA, as the tangent of DC is to the tangent of BC. Q. E. D.

LEMMA.

In right-angled plane triangles, the hypothenuse is to the radius, as the excess of the hypothenuse above cither of the