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4 AB, BC, CA, A, B, C the 2 AC x CB: ACq+CBq the three sides. three angles. - A Bq:: R: CʊS, C. If

ACq+CBq be greater
than ABq. FIG. 16.
2 ACX CB: ABq-ACq
-CBq:: R: CoS, C. If
ABq be greater than ACq
+CBq. FIG. 17. (4.)
Otherwise,

Let AB+BC+AC=2P.|
Px P-AB: P-AC x
P-BC:: Rq: Tq, 1⁄2 C,
and hence Cis known. (5.)
Otherwise,

Let AD be perpendicular to BC. 1. If ABq be less thanACq+CBq. FIG. 16. BC: BA+AC:: BA - AC: BD - DC, and BC the sum of BD, DC is given; therefore each of them is given. (7.)

2. If ABq be greater than ACq+CBq. FIG. 17. BC: BA+AC:: BA-AC: BD+DC; and BC the difference of BD, DC, is given, therefore each of them is given. (7.)

And CA CD :: R : CoS, C. (1.) and C being found, A and B are found by case 2 or 3.

g. 28:

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CONSTRUCTIONS

OF THE

TRIGONOMETRICAL CANON.

A Trigonometrical Canon is a Table, which, beginning
from one second or one minute, orderly expresses the
lengths that every sine, tangent, and secant have, in
respect of the radius, which is supposed unity; and
is conceived to be divided into 10000000 or more
decimal parts.
And so the sine, tangent, or secant
of an arc, may be had by the help of this table; and,
contrariwise, a sine, tangent, or secant being given,
we may find the arc it expresses.
Take notice that
in the following tract, R signifies the radius, S a sine,
Cos. a cosine, T a tangent, and Cot. a cotangent;
also ACq signifies the square of the right line AC;
and the marks or characters, +,−,=, :,::, and ✔✅,
are severally used to signify addition, subtraction,
equality, proportionality, and the extraction of the
square root. Again, when a line is drawn over the
sum or difference of two quantities, then that sum or
difference is to be considered as one quantity.

Constructions of the Trigonometrical Canon.

PROP. I. THEOR.

The two sides of any right angled triangle being given, Fig. 28: the other side is also given.

For (by 47. 1.) ACq=ABq+BCq, and ACq-BCq =ABq, and interchangeably AСq—ABq = BCq. Whence, by the extraction of the square root, there is given AC=✔ ABq+BCq; and AB=√ ACq—BCq; and BC ACq—ABq.

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