Page:Encyclopædia Britannica, Ninth Edition, v. 23.djvu/587

Schmeisser's formulæ. If we use the values of $\sin {\frac {a}{2}}$, $\sin {\frac {b}{2}}$, $\sin {\frac {c}{2}}$, $\cos {\frac {a}{2}}$, $\cos {\frac {b}{2}}$, $\cos {\frac {c}{2}}$, given by (9), (10) and the analogous formulae obtained by interchanging the letters, we obtain by multiplication

$$\left.{\begin{aligned}\sin {\frac {a}{2}}\cos {\frac {b}{2}}\sin C&=\sin {\frac {c}{2}}\cos {\frac {B+C-A}{2}}\\\cos {\frac {a}{2}}\cos {\frac {b}{2}}\sin C&=\cos {\frac {c}{2}}\cos {\frac {B+C-A}{2}}\\\sin {\frac {a}{2}}\sin {\frac {b}{2}}\sin C&=\cos {\frac {c}{2}}\cos {\frac {B+C-A}{2}}\\\end{aligned}}\right\rbrace .$$

These formulae were given by Schmeisser in Crelle's Journ., vol. x.

Caguoli's formulæ. The relation $\sin b\sin c+\cos b\cos c\cos A=\sin B\sin C-\cos B\cos C\cos a$ was given by Cagnoli in his Trigonometry (1786), and was rediscovered by Cayley (Phil. Mag., 1859). It follows from (1), (2), and (3) thus: the right-hand side of the equation equals $\sin B\sin C+\cos a(\cos A-\sin B\sin C\cos a)=\sin B\sin C\sin ^{2}a+\cos a\cos A$, and this is equal to $\sin b\sin c+\cos A(\cos a-\sin b\sin c\cos A)$ or $\sin b\sin c+\cos b\cos c\cos A.$

Solution of triangles; The formulæ we have given are sufficient to determine three parts of a triangle when the other three parts are given; moreover such formulæ may always be chosen as are adapted to logarithmic calculation. The solutions will be unique except in the two cases (1) where two sides and the angle opposite one of them are the given parts, and (2) where two angles and the side opposite one of them are given.

Ambiguous cases. Suppose a, b, A are the given parts. We determine B from the formula $\sin B={\frac {\sin b}{\sin a}}\sin A$; this gives two supplementary values of B, one acute and the other obtuse. Then C and c are determined from the equations $\tan {\frac {C}{2}}={\frac {\sin {\frac {a-b}{2}}}{\sin {\frac {a+b}{2}}}}\cot {\frac {A-B}{2}}$, $\tan {\frac {c}{2}}={\frac {\sin {\frac {A+B}{2}}}{\sin {\frac {A-B}{2}}}}\tan {\frac {a-b}{2}}$. Now $\tan {\frac {C}{2}}$, $\tan {\frac {c}{2}}$ must both be positive; hence $A-B$ and $a-b$ must have the same sign. We shall distinguish three cases. First, suppose $\sin b<\sin a$; then we have $\sin B<\sin A$. Hence A lies between the two values of B, and therefore only one of these values is admissible, the acute or the obtuse value according as a is greater or less than b; there is therefore in this case always one solution. Secondly, if $\sin b>\sin a$, there is no solution when $\sin b\sin A>\sin a$; but if $\sin b\sin A<\sin a$ there are two values of B both greater or both less than A. If a is acute, $a-b$, and therefore $A-B$, is negative; hence there are two solutions if A is acute and none if A is obtuse. These two solutions fall together if $\sin b\sin A=\sin a$. If a is obtuse there is no solution unless A is obtuse, and in that case there are two, which coincide as before if $\sin b\sin A=\sin a$. Hence in this case there are two solutions if $\sin b\sin A\leq \sin a$ and the two parts A, a are both acute or both obtuse, these being coincident in case $\sin b\sin A=\sin a$; and there is no solution if one of the two A, a is acute and the other obtuse, or if $\sin b\sin A>\sin a$. Thirdly, if $\sin b=\sin a$ then $B=A$ or $\pi -A$. If a is acute, $a-b$ is zero or negative, hence $A-B$ is zero or negative; thus there is no solution unless A is acute, and then there is one. Similarly, if a is obtuse, A must be so too in order that there may be a solution. If $a=b={\frac {\pi }{2}}$, there is no solution unless $A={\frac {\pi }{2}}$, and then there are an infinite number of solutions, since the values of C and c become indeterminate.

The other case of ambiguity may be discussed in a similar manner, or the different cases may be deduced from the above by the use of the polar triangle transformation. The method of classification according to the three cases $\sin b{\underset {<}{\overset {>}{=}}}\sin a$ was given by Professor Lloyd Tanner (Messenger of Math., vol. xiv. ).

Radii of circles related to triangles. If r is the angular radius of the small circle inscribed in the triangle ABC, we have at once tan r=tan-2-sin (s-a), where triangles. 2s=a + b + c ; from this we can derive the formulas tanr=i cosecs= sec sec ^-sec^ =sin sin sin sec (21), 2t 2&gt; t Z B SI Z where n, N denote the expressions !sin s sin (* - a) sin (s - b) sin (s - c} &, { - cos Scos (S- A) cos (S- B) cos (S- 0)} J. The escribed circles are the small circles inscribed in three of the associated triangles ; thus, applying the above formulae to the triangle (a, v - b, ir - c, A, IT - B, w - C), we have for r v the radius of the escribed circle opposite to the angle A, the following formulae A . . . N A B C tan r^ = tan - 9 - sm s= n cosec (s - a) = ^-sec -^- cosec -^- cosec ^-

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The pole of the circle circumscribing a triangle is that of the circle inscribed in the polar triangle, and the radii of the two circles are complementary ; hence, if R be the radius of the circum scribed circle of the triangle, and R v R& R^ the radii of the circles circumscribing the associated triangles, we have by writing - - R 2 for r, - - RI for r v ir-a for A, &c., in the above formulae cot R= cot cos (S- A) = - cosec - cosec ^ cosec |= - JVsec S = sin A cos g cos = cosec = a v n a b c cot BI-- cot ^ cos S= ~ cosec ^ sec ~ sec ^ = a . . . b . c a = sm .4 sin - sm - cosec ^. . . (23) ; (S-A) .(24). The following relations follow from the formulae just given : 2 tan R = cot r t + cot r z + cot r 3 - cot r, 2 tan .Kj = cot r + cot r 2 + cot r s - cot r lt tan r tan r x tan r 2 tan r 3 = n 2 , sin 2 s=cotrtanrj tan r 2 tan 7*3, sin* (s - a) = tan r cot r t tan r 2 tan r 3.

Formulæ for spherical excess. If E=A+B + C-ir, it may be shown that E multiplied by the square of the radius is the area of the triangle. We give some of the more important expressions for the quantity E, which is called the spherical excess.

A + B cos 2 We have - -^ . O sin a + b A + B a-b c cos and (J cos . C SU1 2 and a-b 5 ~2~ hence therefore Similarly cos -= c " a + b COS 2~ COS ~2~ . G . sm - + sm COS jj + COS II a + b tan -7 4 tan 2 -^ E , n C-E . tan tan- - = tan ^ tan s-b i 22 ^ f E (, s. s-a. s-b. s-c~ I ,.... therefore tan - = -j tan - tan -- tan -^- tan - - Y (2o). This formula was given by L Huillier. a + b Also C E C . E cos s a-b C E . cos - cos -fsin- E .(26). whence, solving for cos , we get E l + cosa + cosi + cosc cos - = - 2 a b c 4 COS jr COS x COS 5 2&gt; 2t

This formula was given by Euler (Nova acta, vol. x.). If we find siii from this formula, we obtain after reduction I . E n sm -r = ^ , A _ M/ V If 2 cos- cos - cos - m jm- m a formula given by Lescell (Acta Petrop., 1782).

From the equations (21), (22), (23), (24) we obtain the following formulae for the spherical excess : sin 2 -= = tan R cot RI cot R. 2 cot R 3 _ 4(cot rj + cot r 2 + cot ?,,) (cot r - cot r x + cot r + cot r 3 ) (cot r + cot r x - cot r 2 + cot r 3 ) (cot r + cot T-J + cot r a - cot r 3 ) The formula (26) may be expressed geometrically. Let M, NM the middle points of the sides AB, AC. Then we find cos MN 1 + cos a + cos b + cos c ~~b ^ 4 cos - cos 5 2, 2t hence cos - = cos MN sec . A geometrical construction has been given for E by Gudermann (in Crelle s Journ., vi. and viii. ). It has been shown by Cornelius Keogh that the volume of the parallelepiped of which the radii of