angles. The first Case is solved by a single analogy, as in rightangled triangles; the third is solved in a manner almost equally simple, by means of Napier's analogies. As for the second, it requires two analogies; and also it sometimes admits of two solutions, while the first and third never admit of more than one. XC. To distinguish whether, in the second Case, for the single given values of A, a, b, there are two triangles which satisfy the question or only one, let us first suppose the angle A≤100°, and let the two sides AC, AB be produced till they meet again in A'. If we take the arc AC 100°, and draw CD perpendicular to AB, the sides AD, CD of the right-angled triangle ACD will be each less than 100°; the line CD will be the shortest distance from the point C to the arc AB; and taking DB'=DB, the oblique lines CB', CB will be equal, and the longer the more they diverge from the perpendicular. Put AC=b, CB-a; it appears then that a triangle which has A 100°, b 100°, and ab, must necessarily admit of two solutions ACB, ACB': but if, A and b being still supposed less than 100°, we have ab, in that case the point B would pass beyond the point A, and there would be only one solution represented by ABC. Next suppose AC 100°: if C'D' is drawn perpendicular to ABA', we shall as before have C'D'A'C'; and the arc C'B"" drawn between D' and A' will be greater than C'D' and less than C'A': hence making AC=b, C'B" C'B""a the supposition A 100° and b 100° will evidently give two solutions, if a+b200°, and only one if a+b>200°, because the point B" would then pass beyond A'. By examining upon the same principles the case where the angle A is greater than 100°, the circumstances which determine, if in Case second, the question admits of two solutions or only of one, may be established as follows. A≤ 100°, b100° (ab la <b one solution. two solutions. A≤100°, b100° ( a + b200° one solution. a + b 200° two solutions. a+b200° two solutions. 1a + b<200° one solution. A100°, b100° There will only be one solution if A= 100°, whether a=b, or a+b=200°. There will be two if b=100°. XCI. These same results may be applied to the fifth Case by means of the polar triangle; and the following circumstances may be deduced from it, to shew whether for given values of A, B, a, there are two triangles which satisfy the question or only one. AZB one solution. two solutions. a 100°, B100° a100°, B100° a100°, B100° (AZB two solutions. There will only be one solution, if any of the following equalities have place, a 100°, A = B, A+B=200°. There will be two, if B-100°. XCII. In every case, to avoid all useless or false solutions, we must consider First, that every angle or every side must be less than 200°; Secondly, that the greater angles lie opposite the greater sides; so that if we have AB, we must likewise have ab, and vice versa. Examples of the Solution of Spherical Triangles. XCIII. Example 1. Let 0, M, N be three points situated in a plane inclined to the horizon: if from these three points, the perpendiculars OD, M m, Nn are drawn to the horizontal plane DEF, the objects situated in O, M, N will be represented on the horizontal plane by their projections D, m, n; and the angle MON by m Dn. being granted, suppose the angle MON, and the inclinations of its two sides OM, ON to the This C B F n E vertical OD were given; and that we had to find the angle of projection m Dn. From the point O as a centre and with a radius = 1, describe a spherical surface, meeting the sides OM, ON and the vertical OD, in the points A, B, C; we shall have a spherical triangle ABC whose three sides are known; we shall therefore be able to determine C equal to m Dn, by the formula Case first. Suppose, for example, the angle MON=AB=64° 44′, 60′′, the angle DOM-AC-98° 12, and the angle DON=BC= 105° 42'; by the formula referred to we shall have sin2 C=R2. sin 28° 57′ 30′′ sin 35° 87′ 30′′ A value which may be computed thus: Sum+2 L. R......39. 3650518 19.9982348 L. sin 98° 12′....9.9998106 19.9982348 Hence the angle 64° 44′ 60", measured on a plane inclined to the horizon, is reduced to 64° 9′ 41′′ when it is projected on the plane of the horizon. This problem is useful in the art of taking plans, when the surface to be operated on presents any sensible inequalities, and it is at the same time required to determine the principal positions with great accuracy. XCIV. Example 2. Knowing the latitudes of two points on the globe, and their difference of longitude, to find the shortest distance between them. Conceive a spherical triangle ACB to be formed by the North Pole C and the two places A and B, whose distance we are required to find. In this triangle we shall know the angle at the Pole ACB, since it is the difference in longitude of the two points A and B; we shall like B wise know the two including sides AC, CB, since they are the complements of the latitudes of the points A and B. The third side AB may therefore be determined by the formulas of Case third. Let A and B, for example, be the Observatories of Paris and Pekin: the north latitude of the one of these places is 54° 26/ 36", that of the other is 44° 33′ 73", and their difference in longitude is 126° 80′ 56′′. Thus we shall have a = 45° 73' 64" b 55 66 27 C=126 80 56 According to these data, for determining c we shall have the The angle answering in the tables to this logarithmic tangent is 28° 94' 23." We must consider, however, that cos C is negative, and that tang, being consequently negative, we must take 28° 94′ 23′′. Having settled this, and observing that cos (—) =cos 4, we shall finish the calculation thus: Hence the required distance c82° 16′ 05′′. This same distance may also be expressed in myriametres by 821.605; for a myriametre is the length of an arc measuring ten minutes, a metre being that of an arc measuring one-tenth of a second. XCV. Example 3. To give an example of Case fifth, let us undertake to solve the spherical triangle, wherein are known the two angles A=78° 50′, B = 54° 0', and the side opposite one of them a= 99° 20′ 17′′. By means of this, we find from the Table in Art. 91., that there can be only one solution, since we have at the same time a 100°, B 100° and A⇒ B. This solution is computed as follows: First. The side b will be found by the formula sin b= sin B sin a Which gives b=58° 50′ 14′′, or its supplement 141° 49′ 86"; but since the angle B is less than A, the side b must also be less than a; hence the first value is the only proper one. L. sin .......... .9.9999220 LR...0.0547193 .9.5455236 L. sin B............9.8204063 L. tang B L. tang a- LR..1.9016731 1 L. cot A......... L. tang 9. .11.7220794 L. sin (c-4).... 9.6001649 @ 98° 79' 28" . 8 C Φ 26° 770.5. Here we have again the choice of taking for c- the value. 26° 7' 70.5, or its supplement 173 92′ 29". 5; but by adopting the second value, we should have c≈ 100°; therefore we must keep by the first, which gives c = 124° 81' 99" . 3. Thirdly. In fine, to calculate the angle C directly, we shall take the formulas cot = cos a tang B cos A sin↓ cos B R sin (C = |