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. And this arc AF is the measure of the angle ACF in
the figure of the crystal.
[Illustration]
In the same figure, if the plane CGHF cuts the crystal so that it
divides the obtuse angles ACB, MHV, in the middle, it is stated, in
Article 10, that the angle CFH is 70 degrees 57 minutes. 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.
It was stated, in Article 26, that the straight line CS, which in the
preceding figure is CH, being the axis of the crystal, that is to say
being equally inclined to the three sides CA, CB, CF, the angle GCH is
45 degrees 20 minutes. This is also easily calculated by the same
spherical triangle. 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. Now in the
triangle QAS, which is right-angled, one knows also the angle A, which
is 52 degrees 30 minutes, and the side AQ 50 degrees 56 minutes;
whence the side SQ is found to be 45 degrees 20 minutes.
In Article 27 it was required to show that PMS being an ellipse the
centre of which is C, and which touches the straight line MD at M so
that the angle MCL which CM makes with CL, perpendicular on DM, is 6
degrees 40 minutes, and its semi-minor axis CS making with CG (which
is parallel to MD) an angle GCS of 45 degrees 20 minutes, it was
required to show, I say, that, CM being 100,000 parts, PC the
semi-major diameter of this ellipse is 105,032 parts, and CS, the
semi-minor diameter, 93,410.
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