Demonstration. First Case. When each of the legs differs from 90°, the equation (470), cos. A=cos. a sin. B, gives, by (520), cos. Asin. B ; sin. (90°-A) <sin. B. or, by (5), First. The only case in which it is necessary to prove that the sum of the angles is greater than 180°, or, that the sum of A and B is greater than 90°, is, when A and B are both acute. In this case, by (30') and (543), B> 90° - A; A+B > 90°. or or (541) (542) (543) Secondly. As the preceding equation expresses, that when the right angle is the greatest angle of the triangle it is less than the sum of the other two angles; (546) we have only to show farther, that, when either of the other angles, as B, is the greatest angle, and of course obtuse, it is less than the sum of the other two angles. We may suppose A to be acute. Then, as the difference between B and 90° is B tween 90° and 90° A is 90°. have, by (507) and (543), B - 90° < A, (544) (545) 90°, and as that be- (547) · (90° — A) or A; we B< 90° + A; from which we conclude that each angle of a right triangle is less than the sum of the other two. (548) (549) (550) Thirdly. The only case in which it is necessary to prove that the sum of all the angles is less than (551) 360°, or that the sum of A and B is less than 270°, is when A and B are both obtuse. But, if A is obtuse, (552) (553) (554) (555) 90° A is the negative of A-90°, which may by whence, by (507), or or A+B 270°. Second Case. When one of the legs is equal to 90°, its opposite angle is also 90°, by (522); and therefore whatever is the value of the third angle, it cannot but satisfy the conditions of the proposition (540). SECTION. II. Solution of Spherical Right Triangles. 18. To solve a spherical right triangle, two parts must be known in addition to the right angle. From the two known parts, the other three parts are to be determined, separately, by equations derived from Napier's Rules. The two given parts with the one to be determined are, in each case, to enter into the same equation. These three parts are either all adjacent to each other, in which case the middle one is (556) taken as the MIDDLE PART, and the other two are, by (473), ADJACENT PARTS; or one is separated from the other two, and then the part, which stands by itself, is the MIDDLE PART, and the other two are, by (473), OPPOSITE PARTS. 19. Problem. To solve a spherical right triangle, when the hypothenuse and one of the angles are known. Solution. Let ABC (fig. 2.) be the right triangle, right angled at C; and let the sides be denoted as in (427). Let h and A be given, to solve the triangle. A or and, by (7), First. To find the other angle B. The three parts which are to enter into the same equation are co. h, co. A, and co. B; and, by (556), as they are all adjacent to each other, co. h is the middle part, and co. A and co. B are adjacent parts. Hence, by (474), sin. (co. h) tang. (co. A) tang. (co. B), cos. hcotan. A cotan. B; = cotan. B = or cos. h cotan. A Fig 2 h sin. a = sin. h sin. A. 2* B 2 =cos. h tang. A. Secondly. To find the opposite leg a. The three parts are co. A, co. h, and a; of which, by (556), a is the middle part, and co. h and co. A are the opposite parts. Hence, by (475), sin. a = cos. (co. h) cos. (co. A), (557) (558) (559) Thirdly. To find the adjacent leg b. The three parts are co. A, co. h, and b; of which co. A is the middle part, and co. h and bare the adjacent parts. Hence, by (474), sin. (co. A) tang. (co. }) tang b (560) or cos. A = cotan. h tang. b; = 20. Scholium. The tables always give two angles, which are supplements of each other, corre(562) sponding to each sine, cosine, &c. But it is easy to choose the proper angle for the particular case, by referring to (495) and (517); or by having regard to the signs of the different terms of the equation, as determined by (496). 21. Scholium. When hand A are both equal to 90°, the values of cotan. B and tang. b (558) and (561), are indeterminate; since the numerators and denominators of the fractional values are, by (157) and (159), equal to zero; and in this case there are an infinite number of triangles which satisfy the given values of h and A. (563) The problem is impossible by (535) or (538), if the (564) given value of h differs from 90° while that of A is equal to 90°. EXAMPLES. 1. Given in the spherical right triangle (fig. 2.), h145° and A23° 28'; to solve the triangle. B, cotan. 9.55097 n; a sin. 9.35871; b tang. 9.80774 n. 2. Given, in the spherical right trangle (fig. 2.), h=32° 34' and A 44° 44', to solve the triangle. Ans. B 50° 8', = a = 22° 16', by (561), 22. Problem. To solve a spherical right triangle, when its hypothenuse and one of its legs are known. sin. asin. h sin A; Solution. Let ABC (fig. 2.) be the triangle; h the given hypothenuse, and a the given leg. and by (7), First. To find the opposite angle A; a is the middle part, and co. A and co. h are the opposite parts. Hence, by (475), sin. acos. (co. h) cos. (co. A); *The letter n placed after a logarithm indicates it to be the logarithm of a negative quantity, and it is plain that when the number of such logarithms to be added together is even, the sum is the logarithm of a positive quantity; but if odd, the sum is the logarithm of a negative quantity. (565) (566) |