Since the sine of an arc or angle is equal to the sine of the supplemental arc or angle, the values of these sines are doubtful; and by producing the arcs, or sides ẞ and y, until they intersect each other on the surface of the sphere, the angle formed at their intersection will = A, and will also be opposite to a; so that there will be two right-angled triangles having the same data. This is the only ambiguous case in the solution of right-angled triangles. CASE VI. Given the two angles A and B. 90. These rules are to be applied in subserviency to the considerations already made in (art. 127. pt. I.). Thus, if a and y be given to determine B, since if B be small, or nearly 180°, it cannot be determined by Napier's rules with sufficient accuracy; the difficulty is to be avoided in the following manner : B tan is necessarily positive, and the value of B will be 2 found from this equation with sufficient accuracy. 91. Similarly, if A and B be given to find y, which is small, or nearly, it must be found from the equation 93. It must be remembered in the solution of right-angled triangles, that a + ẞ is greater or less than , according as A + B is greater or less than 180°; and that A + B necessarily exceeds one right angle. II. On the solution of quadrantal triangles. 94. Let C, be the angle of the polar triangle correspond that is, the polar triangle has a right angle; also, any two parts of the primary triangle being given, the two corresponding parts of the polar triangle are also given; hence the polar triangle being solved by Napier's rules, the sides and angles of the quadrantal triangle become known from the equations in (art. 88.). But Napier's rules may be immediately applied to the quadrantal triangle by making the two angles adjacent to the quadrantal side, the complements of the two other sides, and the complement of the angle opposite to the quadrantal arc, the circular parts: thus, by making a, the middle part, sin a1 = tan ẞ1. tan (90° B1), in the same way it may be shown, that Napier's rules are true for the other eight cases. III. On the solution of oblique-angled triangles. 95. From (art. 59. 61. and 76.) it appears that if any three parts of an oblique-angled triangle be given, a fourth part may be found; hence, since there are six parts of the triangle in question, namely, the three sides, and three angles, it follows that there will be in all fifteen cases; because six things combined four and four together produce fifteen combinations; but these fifteen cases are evidently contained in the following six, in which are specified the parts given to find the remaining, that is: (1.) The three sides. (2.) The three angles. (3.) Two sides and the included angle. (5.) Two sides and the angle opposite to one of them. |