That is the area of the spherical triangle exceeds the area : of the plane triangle by a2 + b2 + c2 part of the latter. 24,2 Cor. 2. If we omit terms of the second degree in Er2 = A. Hence, if the sides of a spherical triangle be very small compared with the radius of the sphere, its area is approximately equal to the area of the plane triangle having sides of the same length. 228. Roy's Rule. The area of a spherical triangle on the Earth's surface being known, to establish a formula for computing the spherical excess in seconds. Let A be the area of the triangle in square feet, and n the number of seconds in the spherical excess. Then we have Now, the length of a degree on the Earth's surface is found by actual measurement to be 365155 feet. Substituting this value of r in (1), and reducing, we get 9.3267737 (2) log n = log A This formula is called General Roy's rule, as it was used. by him in the Trigonometric Survey of the British Isles. He gave it in the following form: From the logarithm of the area of the triangle, taken as a plane triangle, in square feet, subtract the constant logarithm 9.3267737; and the remainder is the logarithm of the excess above 180°, in seconds, nearly. Ex. If the observed angles of a spherical triangle are 42° 2' 32", 67° 55′ 39′′, 70° 1′ 48", and the side opposite the angle A is 27404.2 feet, required the number of seconds in the sum of the errors made in observing the three angles. Here the apparent spherical excess is The area of the triangle is calculated from the expression a2 sin B sin C 2 sin A (Art. 101) and by Roy's Rule the computed spherical excess is found to be .23". Now since the computed spherical excess may be supposed to be the real spherical excess, the sum of the observed angles ought to have been 180° +.23". Hence it appears that the sum of the errors of the observations is .23"-(-1")=1".23, which the observer must add to the three observed angles, in such proportions as his judgment may direct. One way is to increase each of the observed angles by one-third of 1".23, and take the angles thus corrected for the true angles. ·Given the 229. Reduction of an Angle to the Horizon. angles of elevation or depression of two objects, which are at a small angular distance from the horizon, and the angle which the objects subtend, to find the horizontal angle between them. Let a, b be the two objects, the angular distance between which is measured by an observer at O; let OZ be the direction at right angles to the observer's horizon. Describe a sphere round O as a centre, and let vertical planes through Oɑ, Ob, meet the horizon at OA, OB, respectively; then the horizontal angle AOB, or AB, is required. α A Let ab 0, AB=0+x, Aah, B This gives the exact value of AB; by approximation we obtain, where x is essentially small, EXAMPLES. 1. Prove that the angles subtended by the sides of a spherical triangle at the pole of its circumcircle are respectively double the corresponding angles of its chordal triangle. 2. If A1, B1, C1; A2, B2, C2; A, B, C.; be the angles of the chordal triangles of the colunars, prove that cos A1 = =cosa sin S, cos A,= sina sin (S-C), cos B2 = cos {b sin S, = cos B1 = sinb sin (S-C), cos C, sin csin (S-B), cos C2 = sinc sin(S—A), cos A, sina sin (S-B), cos B-sin b sin (S-A), cos C=cos csin S. = 3. Prove Legendre's Theorem from either of the formulæ for sin A, cos A, tan A, respectively, in terms of the sides. 4. If C = A + B, prove cos C : =-tana tan b. 230. Small Variations in the Parts of a Spherical Triangle. It is sometimes important in Geodesy and Astronomy to determine the error introduced into one of the computed parts of a triangle from any small error in the given parts. If two parts of a spherical triangle remain constant, to determine the relation between the small variations of any other two parts. Suppose C and c to remain constant. (1) Required the relation between the small variations. of a side and the opposite angle (a, A). We suppose a and A to receive very small increments da and dA; then we require the ratio of da and dA when both are extremely small. Thus メ sin (A + dA) sin c = sin C sin (a + da), or (sin A cos dA+ cos A sin dA) sin c = sin C (sin a cos da + cos a sin da) . (2) Because the arcs dA and da are extremely small, their sines are equal to the arcs themselves and their cosines equal 1: therefore (2) may be written (2) Required the relation between the small variations of the other sides (a, b). We have cos c = cos a cos b + sin a sin b cos C. (1) .. cosc= cos(a+da)cos(b+db)+sin(a+da)sin(b+db)cos C, =(cos asin ada) (cos b- sin bdb) or +(sin a + cos ada) (sin b + cos bdb) cos C. (2) Subtracting (2) from (1) and neglecting the product cos a sin b cos C) da +(cos a sin b sin a cos b cos C) db, (3) Required the relation between the small variations of the other angles (A, B). |