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the same sign with tang c. Hence in every right-angled spherical triangle, an oblique angle and the side opposite to it are always of the same species; in other words, are both greater or both less than 100°.


LXVII. A spherical triangle may have three right angles, and then its three sides are each 100°; it may have only two right angles, in which case, the opposite sides are both 100° each, and there remains an angle with its opposite side, both of which are measured by the same number of degrees. These two kinds of triangles can evidently give rise to no problem; we may, therefore, leave them out of view entirely, and limit our attention to such triangles as have only one right angle.

Let A be the right angle, B and C the other two angles which are called oblique ; let a be the hypotenuse opposite the angle A; b and c the sides opposite the angles B and C. Two of the five quanties B, C, a, b, c, being given, the solution of the triangle will always be reducible to one of the six following



LXVIII. Given the hypotenuse a, and a side b, the two angles B and C with the third side c may be found by the equations,

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The angle C and the side c have in them no uncertainty as to their signs; the angle B must be of the same species with the side b.


LXIX. Given b and c the two sides containing the right angle, the hypotenuse a and the angles B and C may be found by the equations,

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There is no ambiguity in any of these values.


LXX. The hypotenuse a and an angle B being given, we shall obtain the two sides b and c, with the other angle C, by the equations,

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cos a tang B

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The elements c and C are determined without ambiguity by these formulas; the side 6 will be of the same species with the angle B.


LXXI. Given b a side of the right angle with the opposite angle B, we shall find the three other elements a, c, and C, by the formulas,

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In this case, the three unknown elements being determined by means of their sines, the question is susceptible of two solutions. It is evident, accordingly,

that the triangle ABC and the tri

angle AB'C, are both right-angled at B
A, and have both the same side AC=
b, and the same opposite angle B=B'.
It only remains for the double values
to combine so that c and C shall be of
the same species; then the species of
c and b will determine that of a, ac-
cording to the formula cos b cos c=
R cos a



The value of a may also be derived immediately

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LXXII. Given b a side of the right angle and the adjacent angle C, the other three elements a, c, B, may be found by the formulas,

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There is no uncertainty, in this case, with regard to the species of the unknown elements.


LXXIII. The oblique angles B and C being given, the three sides a, b, c will result from the formulas,

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LXXIV. The spherical triangle, whose angles are A, B, C, the sides opposite them being a, b, c, always corresponds to another polar triangle whose angles are supplements of the sides a, b, c, while its sides are supplements of the angles A, B, C; so that, calling the angles of the polar triangle A', B', C', and the sides opposite them a', b', c', we shall have

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This being settled, if a spherical triangle has one of its sides a equal to a quadrant, the corresponding angle A' of the polar triangle will evidently be right, and thus that triangle will be right-angled. Hence the two data which, in addition to the side of 100°, we must have before solving the proposed triangle, will likewise serve for solving the polar triangle, and consequently for solving the proposed triangle From this property, we derive formulas similar to the foregoing, for the direct solution of spherical triangles which have one side of 100°.

An isosceles triangle divides itself into two right-angled triangles which are equal in all their parts: hence the solution of isosceles spherical triangles likewise depends on that of rightangled spherical triangles.

Let ABC be a spherical triangle such that the two sides AB, BC are supplements of each other; the 4 sides AB, AC being produced till they meet, it is evident that BC




and BD will be equal, since they are supplements of the same side AB; also it is plain that the parts of the triangle BCD being known, those of the triangle ABC, which is the remainder of the lune AD, are likewise known, and vice versa. Hence the solution of the triangle ABC, whereof two sides together make 200°, is reducible to that of the isosceles triangle BCD, or to that of the right-angled triangle BDE, which is the half of BCD.

When the two sides AB, BC are supplements of each other, the opposite angle ACB, BAC must also be supplements of each other; for BCD is the supplement of BCA, and BCD=D=A. Hence we cannot have a+c=200° without at the same time having A+C=200°, which is a reciprocal property.

Thus it appears that the solution of the right-angled spherical triangles includes, first, that of spherical triangles having a side equal to a quadrant; secondly, that of isosceles spherical.

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triangles; thirdly, that of spherical triangles in which the sum of two sides and also of the opposite angle is 200°.


LXXV. In every spherical triangle the sines of the angles are as the sines of the opposite sides.

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angled triangles ABD, ACD will give the proportions,

sin BR:: sin AD: sin AB

R sin C:: sin AC: sin AD

Multiplying together the terms of these two proportions, omitting the common factors; we shall have

sin B: sin C:: sin AC: sin AB.
If the perpendicular AD falls
without the triangle we shall
have the same two proportions,
in one of which sin C will de-
signate sin ACD; but the angles
ACD and ACB being supple- B
mental to each other, their sines
are equal, and we shall still
have sin B sin C:: sin AC:
sin AB.


Let a, b, c be the sides respectively equal to the angles A, B, C; by this proposition, we shall have sin A: sin a :: sin B: sin b:: sin C: sin c; which gives the double equation

sin A sin B sin C.

sin a sin b sin c

LXXVI. In every spherical triangle the cosine of an angle is equal to the square of the radius multiplied by the cosine of the opposite side, minus the product of the radius by the cosines of the adjacent sides, the whole divided by the product of the sines of those sides: in other words, we shall have, for an angle C, R2 cos c-R cos a cos b

cos C

sin a sin b

; for the two other angles, we

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Let ABC be the proposed triangle, in which BC-a, AC=b, AB=c. From O, the centre of the sphere, draw the indefinite straight lines OA, OB, OC; assume OD at will; and through D draw DE in the plane OCA, and DF in the plane OCB, both perpendicular to OD, and meeting the radii OA, OB produced in E and F; lastly join EF.

The angle D of the triangle EDF is, by construction, the measure of the angle which is formed between the planes OCA,



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OCB; hence the angle EDF is equal to the angle C of the spherical triangle ACB. Now (Art. 45.) in the triangles DEF, OEF, we have

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Taking the value of EF2 in the second, and substituting it in the first, we shall have

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But OE-DE-OD2, and OF2-DF2=OD2, hence we have

cos EDF

OE.OF.cos EOF-OD2. R.


We have now only to substitute the values which relate to the spherical triangle: but here

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