(228.) Der. If a given right-angled spherical triangle have two angles, that are not right angles, and from the summit of either of them, as a pole, a great circle be described, cutting the opposite side and the hypotenuse produced, if necessary, the triangle contained by the segments of the circumference so described, and of the two sides which it cuts, is called the Complemental Triangle of the given triangle.

Thus, let the angle C, and no other angle, of the spherical triangle ABC, be a right angle; and from either

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of the oblique angles, B, as a pole, let there be described the great circle FG, cutting the opposite side AC, and the hypotenuse BA, produced, if necessary. ", in Fand G: the triangle FAG is one of the complemental triangles of ABC; and by a similar construction, if a great circle be described from A, as a pole, the other complemental triangle will be found.

* In the figure, the two sides of the given spherical triangle are supposed to be of the same species, and to be, each of them, less than a quadrant: there are, therefore, two other cases, which might be illustrated by separate figures: but, as the reasoning is general and very easy to be understood, it seems unnecessary to exhibit more than one

of the cases.

(229.) Cor. Let the circle FG meet the side BC, produced, if necessary, in H.

Then (Art. 36.) BG and BH are quadrants ; so that AG is the complement of BA, and CH of BC. Again, since the angle C is a right angle, the pole of BCH is (Art. 50.) in CF; and, by the construction, BF is a quadrant; wherefore, (Art. 36. 50.) F is the pole of BCH; and, consequently, FA is the complement of AC: also (Art. 54.) CH, the complement of BC, measures the angle AFG; and GH, the complement of FG, measures the angle ABC; and FG, likewise, measures the angle FBG, the complement of ABC.

It is manifest, also, that similar conclusions may be drawn, from the other complemental triangle.


(230.) Theorem. The cosine of any one of the sides, of a spherical triangle, is equal to the product of the cosines of the other two sides, together with the continued product of the sines of those two sides, and the cosine of the angle contained by them; unity being put, as well for the tabular radius, as for the radius of the sphere.

Let AC be any one of the sides of the spherical triangle ABC; then, unity being put for the tabular radius,






and also for the radius of the sphere, cos AC=cos AB cos BC + sin AB sin BC cos LABC.

For, let K be the sphere's center, and KA, KB, KC, each of them a radius of the sphere; also, let BD and BE be two straight lines, which touch the arches AB and BC, in their common point B, and meet KA and KC, produced, in D and E: and let the two points D, E be supposed to be joined, by the straight line DE.

Wherefore, (Introd. 25.)

KD+KE'—2KD. KE.cos _DKENDE: also, BDP + BE-2BD.BE. cos _DBE=DE.

Whence, by subtraction, if unity be put for (KDBD%), and for (KE - BE), which are each of them (E. 18. 3. and 47. 1.) equal to the square of the radius, that is, to unity, if, also, the spherical angle ABC be put (Art. 41.) for the plane angle DBE, and the arch AC (E. 33. 6.) for the angle DKE,

1+1+2BD.BE.cos z ABC-2KD.KE.cos AC=0, that is, 2+2BD.BE.cos - ABC - 2.KD.KE.cos AC=0; .:. 1+BD. BE.cos - ABC-KD.KE.cos AC=0.

sin AB But, (Introd. 17.) BD =tan AB=

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and, if these values be substituted in the equation last found, it will become, when cleared of fractions, COS AB cos BC+sin AB sin B C cos 4 ABC-COS AC=0; .. cos AC = cos AB cos BC + sin AB sin BC cos ABC.

And, in the same manner, may the proposition be proved to be true of the cosines of the other sides of the triangle.

If, therefore, A, A', A", be put for the three angles of a spherical triangle, and S, S', $", for the three sides respectively opposite to them, (I.) (1.)'cos S=cos S' cos S" + sin S' sin S" cos A*.

(2.) cos S' = cos S" cos S + sin S" sin S cos A'.
(3.) cos S" = cos S cos S' + sin S sin S'cos A".

* According to the system of notation here adopted, any one of the variations of the same Form may be deduced from any other, if the Form itself be perfectly general, by placing an additional accent over every letter which has not two accents, and wherever there are two accents, by suppressing them.

(231.) Cor. 1. If a, d', a" be put for the angles, and s, s', s" for the opposite sides of the polar triangle of ABC, then

cos s = cos s' cos s" + sin s' sin s" cos a; that is, (Art. 78. and Introd. 19.) cos A = cos A' x - cos A" - sin A' sin A" cos S.

cos A' cos A" - sin A' sin A" cos S; .. (II.) (1.) cos A = sin A' sin A" cos S-cos A' cos A".

(2.) cos A' = sin A" sin A cos S' - cos A" cos A. (3.) cos A" =sin A sin A' cos $" -cos A cos A'.

(232.) Cor. 2. Hence, in a right-angled spherical triangle, having the angle A for a right angle, since cos A = 0,

cos S = cos S'cos S" (Art. 230. I. 1.)

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Again, the angle A being a right angle, and therefore, COS A = 0; i. (Art. 231. II.) sin A' sin A" cos S-cos A' cos A" =0;

cos A' cos A" ... cos S = sin A' sin A"=(Introd. 17.) cot A' cot A" ;

cot A" :(IV.) cos S=cot A' cot A"=


tan A' tan A'tan A" (Introd. 17.)

* In this, and the subsequent Forms, fractions are equated, rather than rectangles; because the quantities, under comparison, are thus more completely separated; and their mutual relation is the more readily and the more distinctly perceived. The implied theorems, also, are most easily read off, in words, as proportions, when the Forms, which represent them, are thus exhibited.

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