It is easily seen that a is less than A when both are acute (they must be of the same affection), and greater than A when both are obtuse; and when the data do not satisfy these conditions the solution is impossible.

It appears at once from the figure that there are in general two solutions, for the colunar triangle A'BC has also the given parts A' and a, and therefore satisfies the given conditions.

I sin b 9.6417030 25° 59' 27.8", or 154° 0′ 32.2′.

=

=

I sin c = 9.7400177 33° 20′ 13.4", or 146° 39′ 46′6′′.

=

L sin B 9.9016853 52° 23′ 2.8", or 127° 36′ 57.2′.

=

SECTION III.

Theorems.

53. In this section we propose to investigate some theorems which apply to spherical triangles in general, but which are immediately deducible from a knowledge of the properties of right-angled triangles.

54. The arcs drawn from the vertices of a spherical triangle perpendicular to the opposite sides are concurrent. Let AP, BQ (fig. 26), be arcs drawn from A, B, per

pendicular to BC and CA, respectively, and let O be their point of intersection.

(formula (6), Art. 44); and it is obvious that the sines of the perpendiculars from any other point of AP on the sides b and c bear the same ratio to each other.

and hence, on dividing this equation by the former, we

and therefore the arc joining C to O is perpendicular to the side AB. Thus the perpendiculars meet in a point; and if a, ß, y be the lengths intercepted on them between their point of concurrence and the sides a, b, and c, respectively, we have

sin a sin ẞ sin γ = sec A: sec B: sec C.

In the same manner, it may be shown that the arcs joining the vertices to the middle points of the opposite sides are concurrent; and if a', B', y' be the perpendiculars from their point of concurrence on the sides a, b, c, we have

sin a sin B': sin y = cosec A: cosec B: cosec C.

:

Similarly for point of intersection of perpendicular at middle points of sides

sin a": sin ẞ": sin y"= cos (S-A): cos (S-B): cos (S-C).

55. If from three points A, B, C on a great circle perpendiculars AA', BB, CC' be drawn to another great circle

(1) sin BC sin AA' + sin CA sin BB' + sin AB sin CC' = 0, (2) sin B'C'tan AA'+ sin C'A' tan BB'+ sin A'B' tan CC'= 0, (3) sin BC cot A + sin CA cot B + sin AB cot C (4) sin B'C' cos A+ sin C'A' cos B

+ sin A'B' cos C

=

= 0,

=

= 0,

where A, B, C are the angles the perpendiculars AA, BB', CC' make with the great circle ABC.

But since A, B, C are concyclic (p. 11, Ex. 15),

sin BC sin AD + sin CA sin BD + sin AB sin CD = 0.

Substituting in this expression the values of sin AD, sin BD, and sin CD, obtained from the equations (a), we obtain (1).

sin B'C' sin A'D + sin C ́A′ sin B′D + sin A′B′ sin C′D = 0.

Substituting in this expression the values of sin A’D, sin BD, and sin C'D, obtained from the equations (3), we obtain (2).

sin BC cos AD + sin CA cos BD + sin AB cos CD = 0;

sin B'C' cos A'D + sin C'A' cos B'D + sin A'B' cos C'D = 0; therefore, &c.

By observing that AA', BB', and CC' meet at P, the pole of A'B'C', we are enabled to write down the following corollaries:

Cor. 1.—If B be a point on the base AC of a triangle APC, we have

sin BC cos AP + sin CA cos BP + sin AB cos CP = 0.

Cor. 2.

[Cf. Art. 40.]

cot AP sin BPC+ cot BP sin CPA + cot CP sin APB = 0.

[Cf. p. 58, Ex. 10.]