د a 62 b 64 a sin+sinc cosC+ sinC cosC+ sinCcosC3+&c.; and, in substituting the powers of this expression for those of b the tangent in the series of Note 9., we obtain B=-sinC+ S a 64 a -sinC cosC+33 (4cosC2-1) sinC+(2 cos C2 - 1) sin C b 62 63 64 Cos C + &c.; or sinC + sin2C+ sin3C+ sin4C + In certain extreme cases, approximations can likewise be employed with advantage. Thus, suppose the angles A and B to be exceedingly small; then, by the last paragraph of page 247, their versed sines are very nearly equal to half the squares of the sines. Wherefore, sinC, or sin(A+B)=(art. 1. Note 3.), sinA (I-sinB2)+sinB (1-sin A2) nearly, and consequently, by art. 5., c=(a+b) (1-sinA sinB); or, the arcs being nearly equal to their sines, substitute c for a + b in the second or differential term, and c=a+b-cAB. Again, put C = - 0, or 1 = A + B, and (a+b)(sin AsinB)=sin AsinB (a+b)2=102 15. Proposition twenty-fifth, which is employed with great advantage in maritime surveying, admits likewise of a convenient analytical solution. Let the given distances AB, BC and AC be denoted by a, b and c, and the observed angles ADB and CDB by m and n; then (art. 5. Note 3.) BD = sinBAD-sinBCD = (art. 18. Note 7.) tan(BAD-BCD) sinBAD+sinBCD tan (BAD+BCD) But the angles ABC and ADC of the quadrilateral figure DABC being evidently given, the sum of the remaining angles BAD and BCD is given, and each of them is consequently found. Hence the triangles ABD and CBD are immediately determined. This most useful problem was first proposed by Mr Townley, and solved in its various cases by Mr John Collins, in the Philosophical Transactions for the year 1671. The second solution given in the text is borrowed from Legendre. 16. The reduction of oblique angles to their projection on a horizontal plane, is commonly solved by the help of spherical trigonometry. It admits, however, of a simple and elegant general solution, derived from the arithmetic of sines. Let a and b denote the two vertical angles, or the acclivities of the diverging lines, A the oblique angle which these contain, and A' the reduced or horizontal angle. Since the magnitude of an angle depends not on the length of its sides, assume each of them equal to the radius or unit, and it is evident that the base of the isosceles triangle thus limited will be the chord of the oblique angle A, the perpendiculars from its extremities to the horizontal plane, the sines, and the horizontal traces or projections, the cosines, of the vertical angles a and b. The base of the isosceles triangle forms the hypotenuse of a right-angled vertical triangle, of which the perpendicular is the difference between the vertical lines. Consequently the square of the reduced base is equal to the excess of the square of the chord of A above the square of the difference of the sines of a and b, or (cor. 6. def. Trig.) 2-2cosA-(sin a-sin b)2= (II. 16. El.) 2-2cos A-sin a2-sin b2 + 2sin a sin b (2. cor. def. Trig.) cos a2 + cos b2+2sin a sin b-2cosA. Wherefore (Prop. 11. Trig.) in the triangle now traced on the horizontal plane, 2cos a cos b cos A'=2cosA-2sin a sin b; and multiplying by sec a sec b, there results (cor. 4. def. Trig.), 1. Cos A'seca secb cos A-tan a tan b. This expression appears concise and commodious, but it may be still variously transformed. For vers A'= 1 - cos A' = 1 + tana tanb-sec a sec b cosA = sec a sec b (cos a cos b+ sin a sin b-cosA)= (Prop. 2. Trig.) seca sec b(cos(a-b)-cosA): whence 2. Vers A'=sec a sec b(vers A-vers(ab).) Again, because (2. cor. 1. and 3. cor. 5. Trig.)vers A'=2sin A and vers A-vers(ab)=2sin A+(ab). tain, by substitution, = sec a 2 sin A-(a-b) 2 , we ob 3. Sin A sec b(sin A+ (a - b). sin A-(a + b)). 2 2 Of these formula, the first, I presume, is new, and appears distinguished by its simplicity and elegance. The last one however, is, on the whole, the best adapted for logarithmic calculation. When the vertical angles are small, the problem will admit of a very convenient approximation. For, assuming the arcs a, b as equal to their tangents, it follows, by substitution, that cos A'=cosA/(1+a2)/(1+b2)-ab=cosA((1+a*x1+ =cosA(1+a2+b2-)ab, nearly. Whence, by Note 13, the decrement of the cosine of that oblique angle is (II. 18. El.) a2+b2=(+)+(); 2 2 2 a+b 2 (+)*(1-cosA)-(-) (1+cosA) 2 2 2 Consequently the increment of the oblique angle itself is, by Such is the theorem which the celebrated Legendre has given, for reducing an oblique angle to its projection on the horizontal plane. It is very neat, and extremely useful in practice. But to connect it with our division of the quadrant, requires some adaptation. Let a and b express the vertical angles in minutes; then will 2 a 2 b 2) cotA) 1 3438 a+b -)*tan Adenote, likewise in minutes, the quantity of reduction to be applied to the oblique angle. 17. In computing very extensive surveys, it becomes necessary to allow for the minute derangements occasioned by the convexity of the surface of our globe. The sides of the triangles which connect the successive stations, though reduced to the same horizontal plane, may be considered as formed by arcs of great circles, and their solution hence belongs to Spherical Trigonometry. But, avoiding such laborious calculations, for which indeed our Tables are not fitted, it seems far better to estimate merely the deviation of those incurved triangles from triangles with rectilineal sides. For effecting that correction two ingenious methods have lately been proposed on the Continent. The first is that employed by M. Delambre, who substitutes the chords for their arcs, and thus converts the small spherical, into a plane, triangle. This conversion requires two distinct steps. 1. Each spherical angle, or that formed by tangents at the surface of the globe, is changed into its corresponding plane angle contained by the chords. Let a and s express the sides or arcs in miles; and the angles of elevation, or those made by the tangents and the respective chords, will 21600 21600 be (III. 29. El.) denoted by 24856a and 248568 in minutes, 1350′ or 3107 & and 1350' β. Insert these, therefore, in place of a and b in the formula of the preceding note, and the quantity of reduction of the angle A, contained by the small arcs a and ẞ, 1 will be ((a + b) tan A - (a - b)2 cot-A) 1214 in seconds. 2. Each arc is converted into its chord: But, by the Scholium to Proposition VI. of the Trigonometry, an arc a is to its chord, 2 as 1 to 1-6D; wherefore the diminution of that are in pass ing into its chord, amounts to the whole. 375,600,000 part of the These reductions bestow great accuracy, and are sufficiently commodious in practice. But the second method of correcting the effects of the earth's convexity, and which was given by M. Legendre, is distinguished by its conciseness and peculiar elegance. That profound geometer viewed the spherical triangle as having its curved sides stretched out on a plane, and sought to determine the variation which its angles would thence undergo. Analysis led him, through a complicated process, to the discovery of a theorem of singular beauty. But the following investigation, grounded on other principles, appears to be much simpler. Let A and B denote any two angles in the small spherical triangle, and a and ẞ express in miles the opposite sides, or those or its extension upon a plane. Since (Prop. 9. Trig.) α:β::sinA: sinB, there must exist some minute arc 1, such that sina: sinß:: sin(A + 0) : sin(B + 6.). But (art. 1. Note 3.) sin(A + 0) = sinA + 0 cosA, and (Schol. Prop. VI. Trig.) + e cosB. Now β:α: :sinB : sin A, and therefore, (V. 9. El.) β 1:10: sinA sinB + cos A sin B : sinA sinB + O sin A cosB. But the first and second terms being very nearly equal, and likewise the third and fourth, it is obvious that the analogy will not be disturbed, if each of those pairs be increa sed equally. Hence 1:1+6 :: sinA sinB : sinA sinB + (sin A cos B cos A sinB); and since (Prop. I. Trig.) sinA cos B cos A sin B = sin (A-B), therefore (V. 10. El.) 1: αβ :: sinA sinB: 0 sin(A-B). Consequently, since a and s are proportional to sinA and sinB, 0 (sinA-B) = sinA sinB |