Maitland Club > Works of Sir Thomas Urquhart

118 THE TRISSOTETRAS.
taught how to deale with its subordinat problems, Cathcedwd, Cathoetwt, and Ca-
thcethwth.
All the three works being thus specified apart, according to our accustomed method,
we will declare what way the last two are to be joyned into one ; for the better effec-
tuating whereof, their resolvers, Neg — To — Nu03°Nyr, and To — Neg — NeC^Nur,
must be interpreted ; the first being, As the sine complement of a first base to the
radius, so the sine complement of a first subtendent to the sine complement of the
perpendicular. And the second, As the radius to the sine complement of a second
base, so the sine complement of the perpendicular to the sine complement of a second
subtendent, which is the side required.
Now, seeing a multiplier must be dashed against a divider, being both quantified
alike, and that all unnecessary pestring of a work with superfluous ingredients is to be
avoided, we are to deale with the radius and perpendicular in this place, as formerly
we have done in the moods of Alamebna, Allamebne, Ammanepreb, and Enerablo,
where we did eject them forth of both the orders of proportionalls ; and when we have
done the like here, instead of Neg — To — NuC3"Nyr, and To — Neg — NeftS'Nur, we
may with the same efficacie say, Neg — Nu — NeCCrNur, that is, As the sine comple-
ment of one side is to the sine complement of a subtendent, so the sine complement of
another side to the sine complement of another subtendent ; or more determinatly,
As the sine complement of a first base to the sine complement of a first subtendent, so
the sine complement of a second base to the sine complement of a second subtendent.
This theorem eomprehendeth to the full both the last operations, and according to
the number of the cases of this mood, is particularized into three finall resolvers, the
first whereof for the first case, Dacforamb, is Naet — Nut — NcedC^f 3 Nwd*yr, that is,
As the sine complement of the first basal, or great base, to the sine complement of
the first subtendentall, or great subtendent, so the sine complement of the second
basidion, or little base, to the sine complement of the second subtendentine, or little
subtendent, which is the side required.
The second finall resolver, is for Damforac, the second case, and is set downe thus,
Nsed — Nud— Noetfr3"Nwt*yr, that is, As the sine complement of the first basidion
to the sine complement of the first subtendentine, so the sine complement of the second
basal to the sine complement of the second subtendentall, which is the side in this
ease required.
The third and last finall resolver is for Dakinatamb, and is expressed thus, Nasth —
Nuth — No?th{f3 = Nwth*yr, that is to say, As the sine complement of the first co-base
to the sine complement of the first co-subtendent, so the sine complement of the second
co-base to the sine complement of the second co-subtendent, which in the third case
is alwayes the side required.
The reason of all this is proved out of the fourth and last disergetick axiom, Niub-
prodnesver, whose directer Ennerra, sheweth by its determinater, the syllable Enn,