73. Having given a side b and the adjacent angle A. tan c = tan b cos A' tan a = tan A sin b, cos B = cos b sin A. Here c, a, B are determined without ambiguity, and the triangle is always possible. 74. Having given the two sides a and b. Here we have from (1) and (4) of Art. 62, cos c = cos a cos b, cot A = cot a sin b, cot B = cot b sin a. Here c, A, B are determined without ambiguity, and the triangle is always possible. 75. Having given the hypotenuse c and a side a. Here we have from (1), (3) and (2) of Art. 62, Here b, B, A are determined without ambiguity, since A must be of the same affection as a. It will be seen from these formulæ that there are limitations of the data in order to insure a possible triangle; in fact, c must lie between ɑ and π — -a in order that the values found for cos b, cos B, and sin A may be numerically not greater than unity. If c and a are right angles, A is a right angle, and b and B are indeterminate. 76. Having given the two angles A and B. Here we have from (5) and (6) of Art. 62, cos c = cot A cot B, COS A cos B cos b = sin A' Here c, a, b are determined without ambiguity. There are limitations of the data in order to insure a possible triangle. First π suppose A less than then B must lie between П next suppose A greater than then B must lie between 2' π 2 (-A) and +(-A), that is, between A and 3п - A. 2 π 2 2 77. Having given a side a and the opposite angle A. Here we have from (2), (4) and (6) of Art. 62, Here there is an ambiguity, as the parts are determined from their sines. If sin a be less than sin A, there are two values admissible for c; corresponding to each of these there will be in general only one admissible value of b, since we must have cos c = cos a cos b, and only one admissible value of B, since we must have cos ccot A cot B. Thus if one triangle exists with the given parts, there will be in general two, and only two, triangles with the given parts. We say in general in the preceding sentences, because if a = A there will be only one triangle, unless a and A are each right angles, and then b and B become indeterminate. It is easy to see from a figure that the ambiguity must occur in general. A B For, suppose BAC to be a triangle which satisfies the given conditions; produce AB and AC to meet again at A'; then the triangle A'BC also satisfies the given conditions, for it has a right angle at C, BC the given side, and A' = A the given angle. If a = A, then the formulæ of solution shew that c, b, and B are right angles; in this case A is the pole of BC, and the triangle ABC is symmetrically equal to the triangle ABC (Art. 57). If a and A are both right angles, B is the pole of AC; B and b are then equal, but may have any value whatever. There are limitations of the data in order to insure a possible triangle. A and a must have the same affection by Art. 64; hence the formulæ of solution shew that a must be less than A if both are acute, and greater than A if both are obtuse. EXAMPLES. If ABC be a triangle in which the angle C is a right angle, prove the following relations contained in Examples 1 to 5. 4. Sin a tan § 4 – sin b tan ¦ B = sin (a – b). 5. Sin (c-a) = sin b cos a tan 1 B, 10 Sin (ca) = tan b cos c tan B. 6. If ABC be a spherical triangle, right-angled at C, and cos A = cos2 a, shew that if A be not a right angle b+c= 1⁄2 or 3 2, according as b and c are both less or both greater than π 2 7. If a, ẞ be the arcs drawn from the right angle respectively perpendicular to and bisecting the hypotenuse c, shew that sin 1⁄2 · √(1 + sin2 a) = sin ß. 8. In a triangle, if C be a right angle and D the middle point of AB, shew that 4 cos sin CD = sin2 a + sin2 b. 9. In a right-angled triangle, if 8 be the length of the arc drawn from C perpendicular to the hypotenuse AB, shew that cot d = √(cot2 a + coť2 b). 1 1 10. OAA, is a spherical triangle right-angled at A, and acuteangled at A; the arc 4,4, of a great circle is drawn perpendicular to OA, 4, 4, is drawn perpendicular to OA,, and so on; shew that A A vanishes when n becomes infinite; and find the value of cos A4, cos 4,4, cos 4,4,.....to infinity. 2 3 n n+1 2 2 11. ABC is a right-angled spherical triangle, A not being the right angle; shew that if A = a, then c and b are quadrants. 12. If & be the length of the arc drawn from C perpendicular to AB in any triangle, shew that = cosec c (cos2 a + cos3 b − 2 cos a cos b cos c)3. 13. ABC is a great circle of a sphere; AA', BB, CC', are arcs of great circles drawn at right angles to ABC and reckoned positive when they lie on the same side of it; shew that the condition of A', B', C' lying in a great circle is tan A A' sin BC + tan BB' sin CA + tan CC'′ sin AB = 0. 14. Perpendiculars are drawn from the angles A, B, C of any triangle meeting the opposite sides in D, E, F respectively; shew that tan BD tan CE tan AF = tan DC tan EA tan FB. 15. Ox, Oy are two great circles of a sphere at right angles to each other, P any point in AB another great circle. OC=p is the arc perpendicular to AB from 0, making the angle Cox = a with Ox. PM, PN are arcs perpendicular to Ox, Oy respectively; shew that if OM = x and ON =y, cos a tan x + sin a tan y = tan p. 16. The position of a point on a sphere, with reference to two great circles at right angles to each other as axes, is determined by the portions 0, 4 of these circles cut off by great circles through π the point, and through two points on the axes, each from their 2 point of intersection; shew that if the three points (0, 4), (0′, p'), (", ") lie on the same great circle tan (tan - tan 6') + tan p′ (tan 0” – tan 0) + tan o" (tan – tan 0') = 0. 17. If a point on a sphere be referred to two great circles at right angles to each other as axes, by means of the portions of these axes cut off by great circles drawn through the point and two points on the axes each 90° from their intersection, shew that the equation to a great circle is tan cota + tan & cot ß = 1. A π 18. In a triangle, if 4 =, B, and C, shew that |