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On the queer alphabetical table which the two pieces of paper made, he now wrote quickly the alphabet again, horizontally across the top, starting with H, and vertically down the side, starting with R, thus: H I J K L M N O P Q R S T U V W X Y Z A B C D E F G R a- b- c- d- e- f- g- h- i- j- k- l- m- n- o- p- q- r- s- t- u- v- w- x- y- z- S aa ba ca da ea fa ga ha ia ja ka la ma na oa pa qa ra sa ta ua va wa xa ya za T ab bb cb db eb fb gb hb ib jb kb lb mb nb ob pb qb rb sb tb ub vb wb xb yb zb U ac bc cc dc ec fc gc hc ic jc kc lc mc nc oc pc qc rc sc tc uc vc wc xc yc zc V ad bd cd dd ed fd gd hd id jd kd ld md nd od pd qd rd sd td ud vd wd xd yd zd W ae be ce de ee fe ge he ie je ke le me ne oe pe qe re se te ue ve we xe ye ze X af bf cf df ef ff gf hf if jf kf lf mf nf of pf qf rf sf tf uf vf wf xf yf zf Y ag bg cg dg eg fg gg hg ig jg kg lg mg ng og pg qg rg sg tg ug vg wg xg yg zg Z ah bh ch dh eh fh gh hh ih jh kh lh mh nh oh ph qh rh sh th uh vy wh xh yh zh & ai bi ci di ei fi gi hi ii ji ki li mi ni oi pi qi ri si ti ui vi wi xi yi zi A aj bj cj dj ej fj gj hj ij jj kj lj mj nj oj pj qj rj sj tj uj vj wj xj yj zj B ak bk ck dk ek fk gk hk ik jk kk lk mk nk ok pk qk rk sk tk uk vk wk xk yk zk C al bl cl dl el fl gl hl il jl kl ll ml nl ol pl ql rl sl tl ul vl wl xl yl zl D am bm cm dm em fm gm hm im jm km lm mm nm om pm qm rm sm tm um vm wm xm ym zm E an bn cn dn en fn gn hn in jn kn ln mn nn on pn qn rn sn tn un vn wn xn yn zn F ao bo co do eo fo go ho io jo ko lo mo no oo po qo ro so to uo vo wo xo yo zo G ap bp cp dp ep fp gp hp ip jp kp lp mp np op pp qp rp sp tp up vp wp xp yp zp H aq bq cq dq eq fq gq hq iq jq kq lq mq nq oq pq qq rq sq tq uq vq wq xq yq zq I ar br cr dr er fr gr hr ir jr kr lr mr nr or pr qr rr sr tr ur vr wr xr yr zr J as bs cs ds es fs gs hs is js ks ls ms ns os ps qs rs ss ts us vs ws xs ys zs K at bt ct dt et ft gt ht it jt kt lt mt nt ot pt qt rt st tt ut vt wt xt yt zt L au bu cu du eu fu gu hu iu ju ku lu mu nu ou pu qu ru su tu uu vu wu xu yu zu M av bv cv dv ev fv gv hv iv jv kv lv mv nv ov pv qv rv sv tv uv vv wv xv yv zv N aw bw cw dw ew fw gw hw iw jw kw lw mw nw ow pw qw rw sw tw uw vw ww xw yw zw O ax bx cx dx ex fx gx hx ix jx kx lx mx nx ox px qx rx sx tx ux vx wx xx yx zx P ay by cy dy ey fy gy hy iy jy ky ly my ny oy py qy ry sy ty uy vy wy xy yy zy Q az bz cz dz ez fz gz hz iz jz kz lz mz nz oz pz qz rz sz tz uz vz wz xz yz zz
This again is easily shown in the same spherical triangle ABF, in which it appears that the arc FQ is as many degrees as the angle GCF in the crystal, the supplement of which is the angle CFH. Now the arc FQ is found to be 109 degrees 3 minutes. Then its supplement, 70 degrees 57 minutes, is the angle CFH.
Or rather, having taken AS equal to 2/3 of AT, I observe that 3/2 of GD ought to be equal to 3/2 of SB, plus BQ; and, deducting both of them from FD or FQ, that FD less 3/2 of GD ought to be equal to FB less 3/2 of SB. And this last difference is a given length: and all that is required is to draw the straight line FD from the given point F to meet VG so that it may be thus.
To find from this the obtuse angle BCA, I imagined a sphere having its centre at C, and on its surface a spherical triangle, formed by the intersection of three planes which enclose the solid angle C. In this equilateral triangle, which is ABF in this other figure, I see that each of the angles should be 105 degrees, namely equal to the angle OCN; and that each of the sides should be of as many degrees as the angle ACB, or ACF, or BCF. Having then drawn the arc FQ perpendicular to the side AB, which it divides equally at Q, the triangle FQA has a right angle at Q, the angle A 105 degrees, and F half as much, namely 52 degrees 30 minutes; whence the hypotenuse AF is found to be 101 degrees 52 minutes.
For by drawing the other arc AD which cuts BF equally, and intersects FQ at S, this point will be the centre of the triangle. And it is easy to see that the arc SQ is the measure of the angle GCH in the figure which represents the crystal.
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