AMB; those of the arc AMC are the only ones from which AC can be seen under an angle equal to AMC: hence the point M, where those arcs intersect, is likewise the only point from which AB and AC can at once be seen under the angles AMB, AMC. We are now to calculate the position of the point M trigonometrically, from this construction. Suppose the given quantities, AB=2500 yards, AC=7000 yds., BC 9000 yds., AMB = 30° 80′, AMC=121° 40′. In the triangle ABC, whose three sides are known, we shall find the angle BAC (Art. 57.) by the formula sin2 A= 6750 × 2250 R2 ; from which we obtain 2 log sin & A= log_sin | 2500 × 7000 19.9384483, log sin A9.9692241, A=76° 31′. 5, and, finally, A=152° 63. Draw the diameter AD, and join DB; in the triangle BAD, which is right-angled at B, we shall have the side BA= 2500, and the opposite angle BDA= BMA = 30° 80; whence results the hypotenuse AD = BA X R sin BDA =5374.6 yds. By drawing the diameter AE, in like manner, and joining CE, we shall have ACE a right-angled triangle, in which are known the side AC=7000, and the adjacent angle CAE = AMC-100° 21° 40'; whence, we shall conclude, that AE = RX AC cos CAE =7415 yds. Now, if MD and ME are drawn, the two angles AMD, AME being right, the line DME will be straight. It remains then to resolve the triangle DAE in which the line AM, whose magnitude and position we are required to determine, is perpendicular to DE. Now, in this triangle, we have the given sides AD=5374.6, AE=7415, and the included angle DAE = BAC+CAE-DAB 104° 83. Hence we shall obtain the angle ADE = 50° 93′; and, finally, by the right-angled DAM, we shall have AM 4190.83 yds. This distance and the angle BAM = 112° 27', completely determine the position of the point M.* = In order to compute these examples by means of the common tables, we have only to change the angles from the centesimal into the nonagesimal scale; in other respects, the calculation will be exactly the same. The easiest method of effecting this change, in other words, of diminishing the given angles by one tenth of their amount, is to subtract from each, expressed in degrees and decimals of a degree, the same expression having the decimal point a place farther to the left. Thus the angle BAM=112° 27′= 112°.27 (centes.)=112°.27—11°. 227=101°. 043=101°2′ 34′′ 8. (nonages.) ED. PRINCIPLES FOR THE SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. LXII. In every right-angled spherical triangle, radius is to the sine of the hypotenuse, as the sine of one of the oblique angles is to the sine of the opposite side. Let ABC be the proposed spherical triangle; A its right angle; B and C the other two angles, which we shall call oblique, although one or both of them may be right: we shall have the proportion R: sin BC:: sin B: sin AC. From O, the centre of the F B A sphere, draw the radii OA, OB, OC; then take OF equal to radius in the tables, and from the point F draw FD perpendicular to OA: the line FD will be perpendicular to the plane OAB, because the angle A being right by hypothesis, the two planes OAB, OAC are thus perpendicular to each other. From the point D, draw DE perpendicular to OB; and join EF: the line EF will also be perpendicular to OB, and thus the angle DEF will measure the inclination of the two planes OBA, OBC, and be equal to the angle B of the triangle ABC. This being proved, in the triangle DEF, right-angled at D, we have R : sin DEF:: EF: DF; now the angle DEF = B; and since OF=R, we have EF = sin EOF sin BC, DF = sin AC. Hence R : sin B:: sin BC: sin AC, or R: sin BC: : sin B: sin AC. If we designate by a the hypotenuse or side opposite the right angle A, by b the side opposite the angle B, by c the side opposite the angle C, we shall thus have R: sin a sin B: sin b:: sin C: sin c; a formula, which of itself furnishes two equations among the parts of the right-angled spherical triangles. X LXIII. In every right-angled spherical triangle, radius is to the cosine of an oblique angle, as the tangent of the hypotenuse is to the tangent of the side adjacent to that angle. Let ABC again be the proposed triangle right-angled at A; we are to shew that R:cos B:: tang BC: tang AB. B E F D A For, performing the same construction as before, the right-angled triangle DEF gives the proportion R cos DEF:: EF: ED. But we have DEF = B, EF = sin BC, OE=cos BC; and in the triangle OED, right-angled at E, we have DE = R: cos B: tang BC : tang AB. Making, as above, BC=a and AB=c, we shall have R: cos B:: tanga: tang c, or cos B= R tang c tang c cot a tang a R The same principle, applied to the angle C, will give cos C = LXIV. In every right-angled spherical triangle, radius is to the cosine of a side containing the right angle, as the cosine of the other side is to the cosine of the hypotenuse. Let ABC (see the preceding figure) be the proposed triangle right-angled at A; we are to shew that R: cos AB:: cos AC cos BC. For, the same construction remaining, the triangle ODF, which is right-angled at D and has the hypotenuse OF= R, will give OD=cos DOF=cos AC: also the triangle ODE, rightOD cos DOE cos AC cos AB angled at E, will give OE= R R But in the right-angled triangle OEF, we have OE=cos BC; hence cos BC= cos AC cos BC or what amounts to the same, R R: cos AC:: cos AB : cos BC. This third principle is expressed by the equation R cos a = cos b cos c; it cannot form a second equation, like the two preceding principles, because a change of b for c in it would produce no alteration. LXV. By means of these three general principles, three others may be found, which are requisite for the solution of right-angled spherical triangles. They might be demonstrated directly, each by a particular construction; but it seems preferable to deduce them, by way of analysis, from the three which are already proved. We shall now do so. third principle, to COS C Ꭱ R Hence we have this fourth principle, sin B cos C:: R: cos C, from which also, by changing the letters, there results, R: tang B:: sin c: tang b; from which also, by changing the letters, there results R: tang C:: sin b: tang c; R2 tang b tang c Lastly, these two formulas give tang Btang C sin b sin c R3 = cos a tang B tang C, or cot B cot CR cos a; or tang B: cot C:: R: cos a. Hence This is the sixth and last principle: it cannot furnish another equation, because the change of B for C in it produces no alter ation. We subjoin a recapitulation of these six principles, whereof four give each two equations: sin a sin B, R sin c = I. R sin b II. R tang b = III. R cos a cos b cos c, sin a sin C. tang a cos B. IV. R cos B sin C cos b, R cos C sin B cos c. = V. R tang b = sin c tang B, R tang c = sin b tang C. From these are obtained ten equations including all the relations that can exist between three of the five elements B, C, a, b, c; so that two of these quantities with the right angle being given, the third will immediately be discovered in the form of its sine, cosine, tangent, or cotangent. LXVI. It is to be observed, that when any element is discovered in the form of its sine only, there will be two values for this element, and consequently two triangles that will satisfy the question; because, the same sine which corresponds to an angle or an arc, corresponds likewise to its supplement. This will not take place, when the unknown quantity is determined by means of its cosine, its tangent, or cotangent. In all these cases, the sign will enable us to decide whether the element in question is less or greater than 100°; the element will be less than 100°, if its cosine, tangent, or cotangent has the sign +; it will be greater if one of these quantities has the sign -. this point, likewise, some general principles might be established, which would merely be consequences of the six equations demonstrated above. On From the equation R cos a = cos b cos c, for example, it results, that either the three sides of a right-angled spherical triangle are all less than 100°; or that of those three sides, two are greater than 100%, while the third is less. No other combination can render the sign of cos b cos c like that of cos a, as the equation requires. In like manner, the equation R tang csin b tang C, in which sin b is always positive, proves that tang C has always |