Here the hypotenuse being made the middle part the other two will, obviously, be the opposite parts, being separated from the hypotenuse by the intervening angles A, C. Hence by the rule rad. sin. comp. c = cos. a X cos.b; that is, Cos. a cos. b rad. cos. ¿cos, a cos. b ... cos. c = rad. and as cos. a, cos. b, are both positive, cos. c is positive, and, therefore, 2. In the spherical triangle ABC, right-angled at C, are given b=46° 18′ 23′′, A = 34° 27′ 39", to find the other oblique angle B. Making B the middle part, the other two will be the opposite parts. Consequently, by the rule, rad. sin. comp. B cos. b x cos. comp. A; that is, cos. b sin. A ; rad. cos. B cos. b sin. A .. cos. B= and as cos. b, sin. A, are both positive, B is acute, 3. In the spherical triangle, right-angled at C, are given the two perpendicular sides, viz. a=116° 30′ 43′′, b=29° 41′ 32′′, to find the angle A. Making 6 the middle part, the others will be the adjacent parts, and, therefore, by the rule rad. sin. btan. a x tan: comp. A, that is, rad. sin. b tan. a cot. A .. cot. A = rad. sin. b and as sin. b is positive, and tan. a negative, cot. A will be negative, and, therefore, A will be obtuse, or the supplement of the angle given by the tables, 4. In a spherical triangle, right-angled at C, are given b=29° 12′ 50", and B = 37° 26′ 21′′, to find the side a. Taking a for the middle part, the other two will be adjacent parts; hence, by the rule, rad. X sin. atan. b x tan. comp. B that is, tan. b cot. B rad. sin. a tan. b cot. B .. sin. a = rad. In this case there are two solutions, viz. a and the supplement of a, both of which have the same sine. As sin. a is necessarily positive, b and B must necessarily be always of the same species, that is, either both acute or both obtuse, so that, as observed at p. 85, the sides including the right-angle are always of the same species as the opp. angles, a circumstance which must be attended to in framing examples, C It appears, therefore, that there exists two rightangled triangles, having an oblique angle, and the opposite side in one equal to an oblique angle and the opposite side in the other, but the remaining oblique angle in the one the suppleinent of the remaining oblique angle in the other. These triangles are situated, with respect to each other, on the sphere, as the triangles ABC, AB'C, in the annexed diagram, in which, with the exception of the common side AC, and the equal angles B, B', the parts of the one triangle are supplements of the corresponding parts of the other. 5. Given the angle A=23° 28′, the side b=49° 17′, to find the hypotenuse c. c=51° 42′ 37′′. 6. Given the hypotenuse c = 66° 32′, the side a=37° 48′, to find the angle B. B 70° 19' 18". 7. Given the perpendicular sides a = 59° 38′ 27′′, b=48° 24′ 16′′, to find all the other parts. c=70° 23′ 42′′, A=66° 20′ 40′′, B=52° 32′ 55′′. 8. Given b=121° 26' 25", and the opposite angle B=111° 14′ 37′′, to find all the other parts. Solution of Quadrantal Triangles. (54.) The rules for right-angled triangles will serve also for the solution of quadrantal triangles, or those in which one side is a quadrant. For by changing such a triangle for its supplemental triangle we shall then have to consider a right-angled triangle, of which the hypotenuse will be the supplement of the angle opposite the quadrantal side, the two perpendicular sides supplements of the other two angles of the proposed triangle, and the two oblique angles of the new triangle supplements of the oblique sides of the primitive triangle. That is, the sides of the primitive or quadrantal triangle being a, b, and c = 90°, and its angles A, B, C, the sides of the supplemental triangle will be 180° - A, 180°-B, and 180°C, this latter being the hypotenuse; and the opposite angles will be 180°. ·a, 180° — b, and 90°. But the parts of a quadrantal triangle may be determined without the aid of the sup - B A C. E plemental triangle. Thus let AD be the quadrantal side in the triangle ABD. Produce DB, if necessary, till DC becomes a quadrant, and draw the arc AC, which will, obviously, measure the angle D, since D will be the pole of the arc AC, and C will be a right-angle: also the angle CAB will be the complement of the angle BAD in the proposed triangle, and the angle ABC will either be identical with ABD in the proposed, or supplemental to it, accordingly as DC exceeds, or falls short of, a quadrant; hence all the parts of the proposed triangle are easily determined from those of the right-angled triangle ABC. If the angle DAB is less than 90°, or than the angle DAC, the side DB must, obviously, be acute; but if DAB is greater than 90°, DB will be obtuse, and conversely. Hence the angles adjacent to the quadrantal side are of the same species as the sides opposite to them. The same may be inferred from the polar triangle. It must be remarked that the solution will be ambiguous whenever the determination of the right-angled triangle becomes ambiguous, whether we employ the polar triangle or the triangle ABC in the above diagram. This ambiguity occurs only when the given parts in the right-angled triangle are one of the perpendicular sides and the angle opposite to it. (See ex. 4, p. 88.) EXAMPLES. In the triangle DAB, DA=90°, A=54° 43', and D = 42° 12′, required the other parts. As the angle DAB is less than 90°, that is, less than the angle DAC, DB is less than a quadrant, and, therefore, the right-angled triangle ABC is situated as in the figure, BC being the prolongation of DB. Of the parts of this right-angled triangle we have given A= 90° B Let A be the middle part, then b and c will be adjacent parts, |