the square root as well as in multiplying and dividing. It will be advantageous to free the denominators of radicals; for example, let it be required to find the sine of 75°. But a table of sines is comparatively of little importance, as the logarithms of these numbers are generally preferable in practice. PROPOSITION XIV. To compute the logarithmic sines and tangents. Restricting (317) to the fourth power of x, we have sinx = x(1 − x2+10x1), ... log, sinx = log. x+log. (1 − x2 + + 1⁄2 σx1) = log. x+M[(— ‡x2+ıb ox1)1 — †(− x2)2] [(270)] = LOGARITHMIC SINES, &C. log. sinr=log. - No. to 28596331+2log. +log.[1+No. to (2log. x 1'477)]}. 189 (378) As an example, let it be required to find the logarithm of the Form (378) should not be employed when the arc exceeds 5o, and the last term, log. [1 + No. to (2log. x 1'447)], may be omitted if the arc be less than 3°. Imitating the process above, we find log. cost No. to {13367543+2log. x + log.[1+ log. cos5° =- - 0.0016558 = 1'9983442. The logarithmic tangent will be found from the relation (379) and log. sin5° 29402962, In order to avoid minus characteristics, 10 is usually added; thus in most tables we find log. sin5° = 8′9402960. Dividing the first of (320) by sina, there results sin(a+x) = cosx – cosa sin = cos (1+cota tanx), .. log. sin(a+x) = log. sina + log. cosx+M(cota tanz In order to avoid the accumulation of errors, the computations should be recommenced from new points of departure, for which purpose the above table of sines and cosines to every 3° may employed. be Fig. 57. ... 9 or (250), y' = (1)→+ (− ‡) (1)−•−1 • (− x2)2 + ( − 4) ( − + − 1) (1) +2 . (— 202)2 + *** 2 1 1 .2.3 1+3.2 2 28 +...; where no constant is to be added, since x and (383) makes known z, then (x + z) is determined by (384) and finally x. For an example, let √216 • sinx + √72 • cosx = 12; we find x = 15°. The second form will be preferable, when m is such a quantity as to be most readily computed by logarithms. What will the equations become when k=0? when l=0? The student may form an example for himself PROPOSITION XVIII. To resolve the quadratic equation, x2+2px = 9, when p and q are such as to require logarithmic tables. We have x+p = ±(p2 + q)* = ±(p2 + p2 tan2v) [putting q= p2 tan2v] = ±p(1+ tan2v)* = ±p secv, = Resolution of Triangles and Mensuration of Heights and Distances. PROPOSITION I. The length of a line and its inclination to a second line being given, to find its PROJECTION upon that line. Let a be the given line and l the line upon which its projection is to be made; drop the perpendiculars p, p, from the extremities of a upon l; then a, the portion of l intercepted (a,1) cos (a,1) Fig. 58. a between these perpendiculars, is called the projection of a upon 7. Through the extremity of a nearest l, draw a, parallel to a2 and terminating in P2, then a Also produce a to intersect l, |