which shows that (1) holds when a and b are each greater which shows that (2) also holds in this case. Similarly it may be shown that formulas (3) to (10) hold when a and b are each greater than 90°. It will be noticed that the 129. Sufficiency of formulas. ten formulas of Arts. 127 and 128 contain all possible combinations of the five parts of a spherical right triangle, taken three at a time; hence they are sufficient to solve any spherical right triangle directly from two given parts. 130. Comparison of formulas of plane and spherical right triangles. By rearranging the formulas of the previous article, the analogy between the formulas of the plane and the spherical right triangles is made apparent. IN PLANE RIGHT TRIANGLES * IN SPHERICAL RIGHT TRIANGLES *The above comparison is taken from Chauvenet's "Plane and Spherical Trigonometry." 131. Napier's Rules. The ten formulas used in the solution of spherical right triangles can all be expressed by means of two rules, known as Napier's rules of circular parts. со а со с со β b a, α Napier's circular parts are the sides a and b, the complements of the angles opposite or 90°, 90° - ß, and the complement of the hypotenuse or 90° — c. They are usually written b, co α, co B, CO C. It will be noticed that the right angle is not one of the circular parts. со с со а сов b α Let the five circular parts be placed in the sectors of a circle in the order in which they occur in the triangle. Whenever any three parts are considered, it is always possible to select one of them in such a manner that the other two parts will either be adjacent to this part, or opposite this part. The part having the other two parts adjacent to it or opposite it is called the middle part. Thus let co a, b, and a be the parts under consideration. Then b is the middle part and co a and a are adjacent parts. If co c, co ẞ, and b are the parts under consideration, b is the middle part and co c and co ẞ are opposite parts. Napier's rules may now be stated as follows: The sine of the middle part is equal to the product of the cosines of the opposite parts. The sine of the middle part is equal to the product of the tangents of the adjacent parts.* *To associate cosine with opposite and tangent with adjacent, it may be noticed that the words cosine and opposite have the same 132. Theorem. In a spherical right triangle, a side and the angle opposite terminate in the same quadrant. From the equation cos α = cos a sin ß it is seen that cos a and cos a must always have the same sign, since sin ẞ is always positive. Hence a and a terminate in the same quadrant. 133. Theorem. Of the three parts a, b, c, if any two terminate in the same quadrant, the third terminates in the first quadrant; if any two terminate in different quadrants, the third terminates in the second quadrant. This follows directly from the equation cos ccos a cos b by noticing that if any two of the quantities cos a, cos b, and cos c have like signs, the third is positive; if any two have unlike signs, the third is negative. 134. Two parts determine a triangle. In order to solve a spherical right triangle two parts, in addition to the right angle, must be given. Each of the required parts should be obtained directly from the given parts. using the formulas for spherical right triangles, or Napier's Rules. If, in the solution of a problem, the sine of any required part is found to be negative, no triangle is possible, since no part of a spherical triangle can be greater than 180°. Likewise if the logarithmic sine or cosine of any required part is found to be greater than zero, the triangle is impossible, since no sine or cosine is numerically greater than unity. 135. The quadrant of any required part. Since the parts of a spherical triangle may have any value between 0° and 180°, it is always necessary to determine whether the required parts are greater or less than 90°. This can be done by the theorems of Arts. 132 and 133. Thus, given we have b=50°, c=110°, a > 90°, a > 90°, and ẞ< 90°. The quadrant in which any required part terminates may also be determined from the formula used in calculating that part, by observing the signs of the functions involved. But when the unknown part is determined from its sine, the part terminates in both quadrants, giving two solutions, unless limited by the theorems of Arts. 132 and 133. Thus, given b=50°, c = 110°, by writing the signs of each function above the function, we have Then a > 90° and a > 90°, since their cosines are negative, and B 90° by Art. 132. 136. Check formula. The formula containing the three computed parts may always be used as a check formula. Thus, having given b = 50°, c=110°, the check formula is 137. Solution of a right triangle. Given b=77° 35' 16" and a = 112° 19' 42" side and the angle opposite are given, there are two solutions. Thus if a and a are given, the only formulas by which b, B, and c can be determined are Since the unknown parts are obtained from their sines, each may have two values, giving two solutions, the theorems of Arts. 132 and 133 not restricting the values of the parts to one solution. Having found the two values for each part, the theorems |