t-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.

Let CP and CS be prolonged and meet the tangent DM at D and Z; and
from the point of contact M let MN and MO be drawn as perpendiculars
to CP and CS. Now because the angles SCP, GCL, are right angles, the
angle PCL will be equal to GCS which was 45 degrees 20 minutes. And
deducting the angle LCM, which is 6 degrees 40 minutes, from LCP,
which is 45 degrees 20 minutes, there remains MCP, 38 degrees 40
minutes. Considering then CM as a radius of 100,000 parts, MN, the
sine of 38 degrees 40 minutes, will be 62,479. And in the right-angled
triangle MND, MN will be to ND as the radius of the Tables is to the
tangent of 45 degrees 20 minutes (because the angle NMD is equal to
DCL, or GCS); that is to say as 100,000 to 101,170: whence results ND
63,210. But NC is 78,079 of the same parts, CM being 100,000, because
NC is the sine of the complement of the angle MCP, which was 38
degrees 40 minutes. Then the whole line DC is 141,289; and CP, which
is a mean proportional between DC and CN, since MD touches the
Ellipse, will be 105,032.

[Illustration]

Similarly, because the angle OMZ is equal to CDZ, or LCZ, which is 44
degrees 40 minutes, being the complement of GCS, it follows that, as
the radius of the Tables is to the tangent of 44 degrees 40 minutes,
so will OM 78,079 be to OZ 77,176. But OC is 62,479 of these same
parts of which CM is 100,000, because it is equal to MN, the sine of
the angle MCP, which is 38 degrees 40 minutes. Then the whole line CZ
is 139,655; and CS, which is a mean proportional between CZ and CO
will be 93,410.

At the same place it was stated that GC was found to be 98,779 parts.
To prove this, let PE be drawn in the same figure parallel to DM, and
meeting CM at E. In the right-angled triangle CLD the side CL is
99,324 (CM being 100,000), beca