:

Given. Sought.

2 AC X CB : ACq+CBą - ABq :: R : CoS, C. If AC9+CBq be greater than ABq. Fig. 16.

2 AC XCB: AB-ACq

-CBq::R: Cos, c. 1 4 AB, BC, CA, A,B,C, the ABq be greater than ACq+ the three sides. three angles. CBq. Fig. 17. (4.)

Otherwise,
Let AB+BC + AC=2P.
PxP - AB:P - ACX
P-BC :: Rq: Tq, C
and hence C is known. (5.)

Otherwise,
Let AD be perpendicular
Ito BC.

1. If ABg be less than ACq+CBq. Fig. 16. BC : BA+AC: : BAAC : 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 ::

Co S, C. (1.) and C being found, A and B are found by case 2 or 3.

C

R:

485

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 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, Sa 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 v, 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 giveň, the other side is also given.

For (by 47. 1.) ACq=ABą + BCq, and Acq-BCq= Fig. 28. ABq, and interchangeably ACq- ABq=BCq. "Whence, by the extraction of the square root, there is given AC= VABq + BCq; and AB=V ACq-BCq; and BC= VACY-ABq.

PROP. II. PROB. The sine DE of the arc BD, and the radius CD, Fig. 29. being given, to find the cosine DF.

The radius CD, and the sine DE, being given in the right angled triangle CDE, there will be given by the last Prop.) ✓CDq-DEq= (CE=) DF.

PROP. III. PROB. Fig. 29. The sine DE of any arc DB being given, to find

DM or BM, the sine of half the arc.

DE being given, CE (by the last Prop.) will be given, and accordingly EB, which is the difference between the cosine and radius. Therefore DE, EB, being given, in the right angled triangle DBE, there will be given DB, whose half DM is the sine of the arc DL=1 the arc BD.

PROP. IV. PROB. Fig. 29. The siné BM of the arc BL being given, to find

the sine of double that arc.

The sine BM being given, there will be given (by Prop. 2.) the cosine CM. But the triangles CBM, DBE, are equiangular, because the angles at E and M are right angles, and the angle at B common: Wherefore (by 4. 6.) we have CB : CM::(BD, or) 2 BM: DE. Whence, since the three first terms of this analogy are given, the fourth also, which is the sine of the arc DB, will be known.

Cor. Hence CB : 2 CM :: BD: 2 DE; that is, the radius is to double the cosine of one half of the arc DB, as the subtense of the arc DB is to the subtense of double that árc. Also CB : 2 CM :: (2 BM : 2 DE ::) BM : DE::

CB: CM. Wherefore the sine of an arc, and the sine of its double being given, the cosine of the arc itself is given.

PROP. V. PROB. Fig. 30. The sines of two arcs, BD, FD, being given, to

find FI, the sine of the sum, as likewise EL, the sine of their difference.

Let the radius CD be drawn, and then CO is the cosine of the arc FD, which accordingly is given, and draw OP through O parallel to DK; also let OM, GE, be drawn parallel to CB. Then because the triangles CDK, COP,