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Similar equations may be deduced for each of the other sides. Hence, generally,
R2 cos a=R cos b cos c+sin b sin c cos A.
That is, radius square into the cosine of either side of a spherical triangle is equal to radius into the rectangle of the cosines of the two other sides plus the rectangle of the sines of those sides into the cosine of their included angle. +
V. Each of the formulas designated (2) involves the three sides of the triangle together with one of the angles. These formulas are used to determine the angles when the three sides are known. It is necessary, however, to put them under another form to adapt them to logarithmic computation. Taking the first equation, we have
Rcos am -R cos b cos c
sin b sin c Adding R to each member, we have
Rcos a+R sin b sin c-R cos b cos c
sin b sin c
2 cos A But, R+cos A
(Art. XXIII.), and R sin b sin c-R cos b.cos c=- -R2 cos (b+c) (Art. XIX.),
2 cosʻlA_R? (cos a—cos (b+c))_
sin b sin c
(Art. XXIII). Putting s=a+b+c, we shall have
is=}(a+b+c) and ismari (6+c-a): hence
sin } ( s )sin (1 s—a)
sin b sin c
1 cos 4 O=RA
sin a sin b
Had we subtracted each member of the first equation from R, instead of adding, we should, by making similar reductions have found
sin (a+b-c) sin(a+-6) sin A=RV
sin b sin c
sin f(a+b- c) sin } (6+c-a) sin ; B=RV
sin a sin c
sin j (a+c—b) sin } (b+cma) sin } CERV
sin a sin b
Putting s=a+b+c, we shall have Is—a=}(6+c-a), is—b=} (a+c—b), and js-c=f(a+bc) hence,
sin șA=Rsin (1s—c) sin (fs—b)
sin b sin c
sin a sin c sin ?C=
Rsin (fs—b) sin (tsa) ?
sin u sin b VI. We may deduce the value of the side of a triangle in terms of the three angles by applying equations (4.), to the polar triangle. Thus, if a', u', c', A, B, C, represent the sides and angles
of the polar triangle, we shall have
_C (Book IX. Prop. VII.): hence, omitting the ', since the equations are applicable to any triangle, we shall have cos La=RA
cos (A+B-C) cos } (A+C-B)
sin B sin C cos }b=RV = cos } (A+B-C) cos } (B+C—A) (6.)
sin A sin C cos įc=Rcos (A+C—B) cos } (B+C—A)
sin A sin B. 32
Putting S=A+B+C, we shall have
fS-A=(C+B—A), 1S--B=} (A+C—B) and
sin B sin C
VII. If we apply equations (2.) to the polar triangle, we shall have
-R2 cos A'=R cos B' cos C'-sin B’ sin C' cos a'. Or, omitting the ', since the equation is applicable to any triangle, we have the three symmetrical equations,
R?.cos A=sin B sin C cos d-R cos B cos C
R?.cos C=sin A sin B cos C-R cos A cos B That is, radius square into the cosine of either angle of a spherical triangle, is equal to the rectangle of the sines of the two other angles into the cosine of their included side, minus radius into the rectangle of their cosines.
VIII. All the formulas necessary for the solution of spherical triangles, may be deduced from equations marked (2.). If we substitute for cos b in the third equation, its value taken from the second, and substitute for cos a its value R2-sino a, and then divide by the common factor R.sin a, we shall have R.cos c sin a=sin c cos a cos B+R.sin b cos C.
sin B sin c But equation (1.) gives sin b
sin C hence, by substitution,
sin B cos C sin R cos c sin a=sin c cos a cos B +.R.
sin C Dividing by sin c, we have
sin B cos C R sin a=cos a cos B+R
cot a sin b=cos b cos C+cot A sin C
cot c sin a=cou a cos B+cot C sin B That is, in every spherical triangle, the cotangent of one of the sides into the sine of a second side, is equal to the cosine of the se
' cond side into the cosine of the included angle, plus the cotangent of the angle opposite the first side into the sine of the included
IX. We shall terminate these formulas by demonstrating Napier's Analogies, which serve to simplify several cases in the solution of spherical triangles.
If from the first equations (2.) cos c be eliminated, there will result, after a little reduction,
Rcos A sin c=R cos a sin bccs C sin a cos b. By a simple permutation, this gives
R cos B sin c=R cos b sin a-cos C sin b cos a. Hence by adding these two equations, and reducing, we shall have sin c (cos A+cos B)=(R
=(R-cos C) sin (a+b)
we shall have
sin c (sin A-sin B)=sin C (sin a-sin b). Dividing these two equations successively by the preceding one; we shall have
sin A +sin B sin C sin a+sin b
sin (a + b
And reducing these by the formulas in Articles XXIII. and XXIV., there will result
cos } (a−b) tang I (A+B)=cotic.
'cos } (a+b)
sin } (a−b) tang $ (A–B)=cot C.
sin } (a+b) Hence, two sides a and b with the included angle C being given, the two other angles A and B may be found by the analogies,
cos } (a+b) : cos} (a—b) : : cot 1C : tang} (A+B)
sin } (a+b) : sin } (a—b) : : cot 1C : tang } (A-B). If these same analogies are applied to the polar triangle of ABC, we shall have to put 180°—A', 180°—B',180°—a',180°46', 180°—c', instead of a, b, A, B, C, respectively; and for the result,
, we shall have after omitting the ', these two analogies,
cos }(A+B) : cos }(A—B) : : tang ?c : tang /(a+b)
sin } (A+B) : sin }(AB) : : tang lc : tang 1 (a−b), by means of which, when a side c and the two adjacent angles A and B are given, we are enabled to find the two other sides a and b. These four proportions are known by the name of Napier's Analogies.
X. In the case in which there are given two sides and an angle opposite one of them, there will in general be two solutions corresponding to the two results in Case II. of rectilineal triangles. It is also plain that this ambiguity will extend itselt to the corresponding case of the polar triangle, that is, to the case in which there are given tv'o angles and a side opposite one of them. In every case we sl.all avoid all false solutions by recollecting,
1st. That every angle, and every side of a spherical triangle is less than 180°.
2d. That the greater angle lies opposite the greater side, and the least angle opposite the least side, and reciprocally.
NAPIER'S CIRCULAR PARTS.
XI. Besides the analogies of Napier already demonstrated, that Geometer also invented rules for the solution of all the cases of right angled spherical triangles.