Now, by adding the two first of these equations together, we have mn + Do x OC (DG x OC) Om x CF + Dm xOF: whence DG is known. Moreover, by taking the latter from the former, we get mn - Do x OC (BE × OC) Om x CF-DmX OF whence BE is known. In like manner, by adding the third and fourth equations together, we have On + OC (OE x OC) = Om x OF + Dmx CF; and, by subtracting the latter from the former, we have On-mex OC (OG × OC) = Om X OF CF; whence E and OG are also known. Hence, if the sines of two arches be denoted by S and s, their co-sines by C and c, and radius by R, then will Cc + Ss Hence the sine of the double of either arch (when they C2 S2 and its co-sine = R whence it appears, that the sine of the double of any arch is equal to twice the rectangle of the sine and co-sine of the single arch, divided by radius; and that its co-sine is equal to the difference of the squares of the sine and co-sine of the single arch also divided by radius Moreover, because Dmx CF = OC x my OCX EG OCX OE-OG); and Om X OF OC x On (OC × 20n = 10Cx OE + OG), it follows, that the rectangle of the sines of any two arches AC, CD (BC) is equal to a rectangle under half the radius, and the difference of the co-sines, of the sum and difference of those arches; and that the rectangle of their co-sines is equal to a rectangle under half the radius, and the sum of the co-sines, of the sum and difference of the same arches. The tangents of two arches being given, to find the tangents of the sum and difference of those arches. Let AN and AM (fig. 20. and 21 be the two arches, and AB and AC their tangents; also, NE be the tangent of their sum, in the first case, and the tangent of their difference, in the second, and let CF, perpendicular to the radius DN, be drawn: then, because of the equiangular triangles BAD and BFC, we shall have (BD x CF = DA x BC BDX BF = BAX BC} by 4. and 16. 6. Take each of the last equal quantities from BD2, and there will remain BD2 BD BF (BD x DF) = BD2 - BA x BC now BD x DF (BD2 BAX BC): BD x CF = (DAX BC):: DF: CF:: DN (DA): NE:: DA2: DA x NE; whence, alternately, BD2 BA x BC: DA2 (:: DA x BC: DAX NE): BC: NE. But the first term of this proportion, because BD2 DA2+ BA2, will also be expressed by DA2 + BA2- BA x BC, or by DA2 +BABA x AB AC; or, lastly, by DA2 ‡ BA × AC therefore, the three first terms of the proportion being known, the fourth, NE, will likewise be known. 2. E. I COROLLARY. Hence, if radius be supposed unity, and the tangents of two arches be denoted by T and t, it follows that the tangent of their sum will be = T + t 1-tT' and the tangent of But it will be proper to take notice here, once for all, that, if in these, or any other theorems, the tangent, secant, cosine, co-tangent, &c. of an arch greater than 90 degrees be concerned; then, instead thereof, the tangent, secant, cosine, &c. of an arch, as much below 90 degrees, is to be taken, with a negative sine, according to the observation in page 7. Thus, for instance, let BA (fig. 22.) be an arch greater than 90°, and let the tangent of the sum of AB and AC be required; supposing T to represent the tangent of AD (the supplement of AB) and t the tangent of AC: then, by writing-T' instead of T, in the first of the foregoing theorems, As the sum of the tangents of any two angles BAC, BAD (fig. 23.) is to their difference, so is the sine of the sum of those angles to the sine of their differences = Let BC and BD be the two proposed tangents, to the radius AB; take Bd BD, join Ad, and draw DE and dF perpendicular to AC. It is manifest, because Bd= BD, that Ad AD, and dAB = DAB, and, consequently, that CAd is the difference of the two angles BAC and BAD. = Now, by reason of the similar triangles CDE and CɗF, it will be, CD (CB+ BD): Cd (CB-BD):: DE : dF; but DE and dF are sines of DAE and dAF to the equal radii AD and Ad: whence the truth of the proposition is manifest. COROLLARY. Hence it also appears, that the base (CD) of a plane triangle is to (Cd) the difference of its two segments, made by letting fall a perpendicular, as the sine of the angle (CAD) at the vertex, to the sine of the difference of the angles at the base. PROP. V. In any plane triangle ABC (fig. 24.), it will be, as the sum of the two sides plus the base, is to the sum of the two sides minus the base, so is the co-tangent of half either angle at the base to the tangent of half the other angle at the base. In AC produced take CD = BC, and let BD be drawn: then (by Theor. 5. p. 9.) it will be, AD + AB : AD — AB : ABD D/ABD CBD 2 2 = + AB: AC + BC-AB: co-tang. A tang. ABC. 2. E. D. PROP. VI. In any plane triangle ABC (fig. 25.), it will be, as the base plus the difference of the two sides is to the base minus' the same difference, so is the tangent of half the greater angle at the base to the tangent of half the lesser. In the lesser side CA, produced, take CD = CB, so that AD may be the difference of the two sides; and let BD be drawn: then it is manifest that the angle CBD will be equal to D: but (by Theor. 5. p. 9.) AB + AD: AB D + DBA CAB AD tangent D-DBA 2. 2 2 = (by 32. 1.) tangent 2 |