divide by 2 to take v, quot. = 1.759459* add 10 to multiply by R, log. prod. = 9.759459 log. sin 35° 4′ 49′′ = P. Hence p or the hour angle is 70° 9′38′′. This converted into time, allowing 15° to the hour, is 4 hours 40′ 38′′. In the same manner, and by the same formula, each of the other angles of the given spherical triangle might be found if required, taking care to use the two sides adjacent the angle about to be found each time. The result is not ambiguous, for sine of half the required angle must be acute since the whole angle cannot exceed 180° (Geom. B. 9. Prop. 16. Schol.) 85. It is proved in Geometry, (B. 9, Prop. 15,) that the three angles of a triangle being given the triangle is determined. A formula for calculating either of the sides when the three angles are given, may be easily derived from that of the last article, by means of the polar triangles. It is necessary first to premise that the polar triangles of the whole range of triangles will include all possible triangles; for as each side of a triangle passes through all values from 180° to 0°, the opposite angle of the polar triangle will pass through all values from 0° to 180°. Wherefore whatever can be proved of the polar triangles of all possible triangles, may be considered as proved for all triangles. Roşume equation (4) of the preceding article. sinA= R sin (a + b c) sin (a + c - b) sin b sin c For the parts of the triangle in this formula, substitute their equivalents in the polar triangle. It will be remembered (Geom. B. 9, Prop. 8,) that each angle of a spherical triangle is measured by a semicircumference or 180°, minus the side • To accomplish this division add -1 to the characteristic and +1 to the decimal part of the logarithm, which additions balance each other and leave the value of the logarithm the same as before; the characteristic thus becomes-2; then say 2 into 2= 1; 2 into 15 = 7; and so on. opposite in the polar triangle, and vice versa; hence if a', b', c', represent the sides, and A', B', c', the angles of the polar triangle, we have A=180°-a',a=180°-a', b=180° - в' and c=180°-c putting these values of the letters a, a, and e in their places in the formula above, it becomes sin (180° - a') R 'sin(180°-(A'+'-c'))sin(180°-(A+C-B')) sin (180° — в') sin (180° - c') But (180°-a')=(90° a') and sin (90°- a')=cos a' (Art. 23); also (180°-(Α'+Β'-c')=90°-1(A'+B'C') and sin (90°-1 (Α ́+B'C'))=cos (A'B'C'); also (180o(A'+c'-B')=90°-1('+c'-в') and sin, (90°-1 (A'+C'-B')) =COS(A'+C'-в'); also sin 180°-в'= sin в' (Art. 15) and sin (180°-c')= sin c'. Making these substitutions, the formula becomes COS (A' + B' - c') COS (A' + C' — в') sin B' sin c' a formula for the cosine of half a side in terms of the three angles of a triangle. We may leave out the accents over the letters, which we have employed only to distinguish the polar from the triangle to which it corresponds, and which are superfluous in a general formula. This formula will undergo a similar modification to that made in formula (4) of the last article, and (A+B-C) = (A+B+C) — с (A+C-B) = (A+B+C) - в Represent A + B + C by s, and the second members of these two equations may be written s-candS-B substituting them in the place of the first members in the formula, it becomes cos (s - B) COS (S-с COS a=R sin B sin c or, the cosine of half either side of any spherical triangle is equal to radius into the square root of the cosine of half the sum of the three angles minus one adjacent angle into the cosine of half the sum minus the other adjacent angle divided by the rectangle of the sines of the adjacent angles. 86. We shall next derive two sets of proportions applicable to the solution of a spherical triangle, the first set when two sides and the included angle are given, and the second when two angles and the included side. For this purpose it will be necessary to establish some additional general relations of trigonometrical lines. Resume from (Art. 70) the formulæ for the cosine of the sum, and for the cosine of the difference of two arcs. Add these two equations together, cancelling sin a sin b in the second member, and substituting p for a+b, and q for a--b, (p+q) for a, and (p-q) for b, as in Art. 74; there results cos p + cos q= 2 R cos(p+q) cos(p-q) (1) ... By this last equation divide equation (2) of Art. 74. 2 sin p+sin q sin(p+q)cos(p-q) R 2 cos p + cos q cos(p+q)cos(p-q) R 2 R striking out the common factors and cos (p-q) in the numerator and the denominator of the second member, and Again, divide equation (4) of Art. 74, by equation (1) of this article, cos(p-q) sin(pq)_tan(pq) the last equation becomes R sin p - sin q tan(p-q) cos p + cos q R (3) Again, referring to Art. 71, we find a formula (3) for the sine of an arc, in terms of half the arc, viz. substituting (p+q) for a in this formula, it becomes 2 sin (p+q) = sin(p+q)cos(p+q) R By this last equation divide equation (2) Art. 74, 2 sin p + sin q -sin(p+q) cos (p-q) R 2 sin (p+q) R sin 1⁄2 (p+q) cos + (p+q) 2 (4) striking out the common factors - and sin (p+q) in the R Again, divide equation (4) of Art. 74, by equation (4) of sin (p+q) cos (p+q) striking out the common factors, as in the last case, this be Making q=0 in equation (1) of Art. 83, and observing that COS OR, we have Dividing the latter by the former, and striking out the cosp 87. Resume now the formula for the cosine of an angle in terms of the three sides of a spherical triangle, (Art 82.) clearing both these equations of fractions, they become and cos c sin a sin b = r2 cos c R COS a cos b Eliminate cos c from these last two equations by the method of addition and subtraction. The cc-efficients of cos c will be rendered the same by multiplying the first equation |