EXAMPLES. 153. 1. Given B = 33° 50', a = = 108°; find A, b, and c. By the rule of Art. 149, the formulæ from Art. 144 are, cos A = cos a sin B, tan b = sin a tan B, tan c= Hence, log cos A = log cos a + log sin B, tan a cos B + Since cos A and tan c are negative, the supplements of the angles obtained from the tables must be taken (Art. 152). Note. When the supplement of the angle obtained from the tables is to be taken, it is convenient to write 180° minus the element in the first member, as shown below in the cases of A and c. By the rule of Art. 150, the check formula for this case is cos A = tan b tan c or log cos A = log tan b — log tanc. The values of log tan b and log tan c may be taken from the first part of the work, and their difference should be equal to the result previously found for log cos A. sin a sin c sin A, tan b = tan c cos A, cot B = cos c tan A. The side a is determined from its sine; but the ambiguity is removed by the principles of Art. 138; for a and A must be in the same quadrant. Therefore a is > 90°, and the supplement of the angle obtained from the table must be taken. By Art. 150, the check formula is Note 1. The check formula should always be expressed in terms of the functions used in determining the required parts; thus, in the case above, the check formula is transformed so as to involve cot B instead of tan B. Note 2. We observe here a difference of .0001 in the two values of log sin a. This does not necessarily indicate an error in the work, for such a small difference might easily be due to the fact that the logarithms are only approximately correct to the fourth decimal place. cos ccot A cot B, or cos c tan A tan B = 1. That is, log cos c + log tan A+ log tan B = log 1 = 0 4. Given A= 105° 59′, a = 128° 33'; find b, B, and c. In this example, each of the required parts is determined from its sine; and as the ambiguity cannot be removed by Art. 138, both the acute angle obtained from the tables and its supplement must be retained in each case. It does not follow, however, that these values can be combined promiscuously; for by Art. 138, since a is > 90°, with the value of b less than 90° must be taken the value of c greater than 90°, and the value of B less than 90°; while with the value of b greater than 90° must be taken the value of c less than 90°, and the value of B greater than 90°. Thus the only solutions of the example are: 1. b = 21° 3.9', c = 125° 33.3', B = 26° 13.5'. 2. b = 158° 56.1', c = 54° 26.7', B= 153° 46.5'. Note. The figure shows geometrically why there are two solutions in this case. For if AB and AC are produced to A', forming the lune ABA'C, the triangle A'BC has the side a and the angle A' equal, respectively, to the side a and the angle A of the triangle ABC, and both triangles are right-angled at C. It is evident that the sides A'B and A'C and the angle A'BC are the supplements of the sides c and b and the angle ABC, respectively. SOLUTION OF QUADRANTAL TRIANGLES. 154. A spherical triangle is called quadrantal when it has one side equal to a quadrant. By Art. 136, (ƒ), the polar triangle of a quadrantal triangle is a right triangle. Therefore to solve a quadrantal triangle we have only to solve its polar triangle, and take the supplements of the parts obtained by the calculation. 1. Given c = 90°, a = 67° 38', b = 48° 50'; find A, B, and C. |