The last figures cannot be expected to agree. The logarithms are correct to tenths of seconds, though seconds only are inserted in the second column. (18.) A small angle cannot be correctly found from its cosine (Tr. 90.), hence R, becomes useless when the angle to be found is small. But- sin. (c (19.) In nearly a similar way may be proved (which we leave to the student) which, though in an impossible form, is not really so, for, as we shall see, A+B must be greater than a right angle; so that cos. (A + B) is negative. This formula may be used when c is nearly a right angle. When R, is to be used to find a, and a is nearly a right angle, proceed as follows: (20.) We give the following as exercises for the learner: sin. a cos. b sin. c cos. B. ✅ tau. (45° — x) . R. R10 CHAPTER III. ON OBLIQUE-ANGLED TRIANGLES. (21.) Most of the cases of oblique-angled triangles may be reduced to those of right-angled triangles, as we shall afterwards see. But we shall first supply the necessary formulæ for completing the subject, and afterwards proceed to each particular case. (22.) Let ABC be any spherical triangle, and let O A, O B, O C, contain the corresponding solid oangle at the centre of the sphere. From any point Rin OC draw RP and R Q respectively perpendicular to OC in the planes O CA, OC B. We have then a pyramid, having the right-angled tri R angles OR P, OR Q, and the oblique-angled triangles OPQ, RPQ; and in each triangle of which we find one of the angles of the solid angle, or one of the parts of the spherical triangle; namely,— . QRP the angle C, because RP and R Q are perpendicular to O C. (Tr. 55.) PQ PR2 + R Q - 2 PR. RQ cos. PR Q = PO2 + O Q2 0 PO PR2 + QO2 - Q R 2 PO. OQ cos. c +2 PR. RQ cos. C OROR-2 PO.OQ cos. c+2 PR. RQ cos. C. + cos. C OR, OR RQ PR OQ OP OQ PO cos. a cos. b + sin. a sin. b cos. C. (23.) This formula may also be deduced as follows: B C D x; whence, from R1, Draw BD perpendicular to A C, and let C D x, BD = p. Then DA = b cos. x cos. p= cos. a x) cos. p cos c cos. (b cos. ccos. b cos. a + sin. b. cos, a tan. x. =cos. b cos. a cos. x + sin. b cos. a sin. x But, (R1), or cos. c an. a cos. C = tan. x sin. a cos. C = tan. x cos a, cos. b cos. a + sin. b sin. a cos. C. (24.) From the preceding, by treating the remaining angles in a similar manner, we get the following formulæ : cos. a cos. b cos. c + sin. b sin. c cos. A) cos. b = cos. c cos. a + sin. c sin. a coș. B the preceding formulæ, with the ones corresponding to the other angles, then become from which formulæ may be obtained for tan. 2 A &c. . sin. (s sin. 2 B = a) sin. (s - b) sin. 2 C = (25.) From O, by simple multiplication and extraction of the square From O, we find the following, by simple division, cos.c sin. (a+b) sin.c in which 1-cos. b is put for sin. 2b. From this we shall find, by reduction, (28.) In the first formula of O, substitute the value of cos. b given by the second, which gives cos. a= (cos. a cos. c + sin. a sin. c cos. B) cos. c + sin. b sin. c cos. A Remove the term cos. a cos. 2c to the left side, substitute sin. c for cos. 2c, and divide by sin. c, which gives sin. a cos. c cos. B + sin. b cos. A. cos. a sin. c sin. a which substitute, giving sin. a sin. A cos. a sin. c Divide by sin. a, and by this, and similar processes, we have the following formula: (29.) All the preceding formulæ have been deduced from O, and are therefore true, whatever changes may be made in a, A, &c. provided the formula 0, remain true when changed in the same manner. 1 Now, form the following product from Og, cos. A + cos. B cos. C, which gives, reducing to a common denominator, and writing 1 instead of sin. 2a in the numerator, (cos.a-cos.b cos.c) (I−cos 2a)+(cos.b-cos.c cos. a) (cos. c-cos.a cos.b) sin. 2a sin. b sin, c Which, developed, will be seen to be cos. a v2 sin. 2a sin. b sin. c or cos, a sin. B sin. C, from 010 |