Page 74 Let us take on as radius and draw the perpendiculars AD and AE at its extremity and in the planes AOC and AOB; produce these perpendiculars till they meet the lines oc and OB in D and E. AD will be the tangent and on the secant of the side b of the spherical triangle, and AE the tangent and OE the secant of the side c. This being premised, let us take the value of DE in terms of the other sides and one angle of each of the two plane triangles DAE and DOE, to both of which it belongs. This b2 + c2- a2 may be done by means of the formula cos a = R 2 bc `(Art. 68.) from which taking the value of the square of the side opposite the angle in the formula, we have and in the triangle EOD 2 DO X EO Cos a ED2 = DO2 + EO2 R subtracting the former from the latter of these two equations and observing that in the right angled Do2 AD2012 and E02 there results 020 A2 triangles OAD and OAE AE2 = 0A2 2 DOXEO cos a 2 ADXA E COS A R + R Taking the value of cos a from this equation, we have substituting for o A its value R for Do its value sec b R2 = cos b R sin b for A D its value tan b = (Art. 32) and for A E its value tan c = pression becomes COS C COS A Rsinb Rsine cos b COS C R2 sin b sin c or striking out R2 COS A= R2 COS a R cos b cos c sin b sin c But the angle A of the plane triangle D A E is the same as the angle of the spherical triangle (Geom. B. 9, Prop. 6,) hence, translating the above formula, The cosine of either * angle of a spherical triangle is equal to radius square into the cosine of the side opposite, minus radius into the rectangle of the cosines of the adjacent sides, divided by the rectangle of the sines of the adjacent sides. The above formula will serve to calculate one of the angles of a spherical triangle when the three sides are given if we * We say either angle because in the above demonstration no particular angle was selected. employ the table of natural sines and cosines; but is unsuitable for the application of logarithms, in consequence of the sign — in the numerator requiring a subtraction to be performed which operation is impracticable by means of logarithms. We shall therefore derive from this another formula, involving only multiplications, divisions, &c. of the trigonometrical lines contained in it, to which operations logarithms apply. It will be necessary first to establish another general relation of trigonometrical lines which we shall require in the course of the demonstration. 83. Resume the formulæ for the cosine of the sum and difference of two arcs (Art. 70.) subtract the first from the second, substituting p for a + b, That is, the difference of the cosines of any two arcs is equal to 2 divided by radius into the sine of half their sum into the sine of half their difference. 84. Resume now the formula (Art. 72) sinA = √ R2 R COS A substitute in this the value of cos a in terms of the three sides of the spherical triangle given in the last article but one, (Art. 82) and there results sin A= 1 R2 - R R2 cos a .(3) reducing the two terms under the radical to a common deno minator they become R2 sin b sin c R3 cos a + R2 cos b cos c 2 sin b sin c collecting together the first and third terms of the numerator of this expression, the two may be written thus but R2 (cos b cos c + sin b sin c) cos b cos c + sin b sin c = R cos (b (Art.69.... (8)) hence the whole numerator of the fraction becomes R3 (cos (bc) cos a) but by (1) above substituting bc for q and a for p cos (b — c) — cos a = 2 R sin (a+b—c) sin 1 (a − b + c) Multiplying the second member by R3, writing the denominator 2 sin b sin c under it, striking out the common factors 2 and R from the numerator and denominator, placing back the fraction thus reduced under the radical sign in equation (3), and extracting the square root of R2 that equation becomes sin (a + b — c) sin † (a + c — b) sin b sin c sin A = R and 1/2 (a + c—b) = = 1 (a + b + c) - b representing a + b + c by s, these two expressions become (4) and equation (4) becomes sin A = R √sin (3 s b) sin (s sin b sin c read thus; the sine of half either angle of a spherical triangle is equal to radius into the square root of the sine of half the sum of the three sides of the triangle minus one of the adjacent sides, into the sine of half the sum minus the other adjacent side, divided by the rectangle of the sines of the adjacent sides. This formula, for the solution of a spherical triangle when the three sides are given, is very convenient for calculation by logarithms. Applied to each of the angles separately, it will serve to determine them all. Take the following EXAMPLE. N Let p be the pole of the equator, z the zenith of the place of observation, and s the place of the sun. Then ZP will be the colatitude, Ps the codeclination, zs the P coaltitude, and the angle ZPS the hour angle, or time of s day, (apparent astronomical time). Given the latitude of a place of observation 30° 43' 37" the sun's declination 19° 55' 42", his altitude 26° 38', 33”. Required the hour angle. colatitude the latitude = 59° 16' 23" codeclination=900 the declination 700 4' 18" coaltitude 90° - the altitude = 63° 21' 27" = 900 The colatitude and codeclination are the sides adjacent the hour angle p. sum of the 3 sides = 1920 42 811 half sum = 960 21' 4" subtract one of the adj. sides = 590 16' 23" remainder = 37° 4'41" subtract from s the other adjacent side = 70° 4' 18" remainder = 26° 16' 46" log. sin 37° 4' 41' = 9.780247 log. sin 26° 16' 46" = 9.646158 ar. comp. of log. sin of one adj. side 0.065697 ar. comp. of log. sin of the other adj. side = 0.026817 sum rejecting twice 10 = 1.518919 |
