To find BC.

Having found B, make sin B: sin A:: AC; BC. But BC may also be found without seeking for the angles B and C; for BC AB2-2 cos AAB.X AC +AC2, Prop. 6.

This method of finding BC is extremely useful in many geometrical investigations, but it is not very well adapted for computation by logarithms, because the quantity under the radical sign cannot be separated into simple multipliers. Therefore, when AB and AC are expressed by large numbers, the other solution, by finding the angles, and then computing BC, is preferable.

CASE IV.

The three sides AB, BC, AC, being given, to find the angles A, B, C.

SOLUTION.

Take F such that BC: BA+AC:: BA-AC: F, then F is either the sum or the difference of BD, DC, the segments of the base, (5). If F be greater than BC, F is the sum, and BC the difference of BD, DC; but, if F be less than BC, BC is the sum, and F the difference of BD and DC. In either case, the sum of BD and DC, and their difference being given, BD and DC are found. (Lem. 2.)

Then, (1.) CA: CD:: R: cos C; and BA: BD:: R cos B: wherefore C and B are given, and consequently A.

SOLUTION II.

Let Dbethe difference of the sides AB,AC. Then (Cor.7.) 2 √ AB.AC: √ (BC + D) (BC—D):: R: sin BAC.

SOLUTION III.

Let S be the sum of the sides BA and AC. Then (1.Cor.8.) 2 √ AB.AC : √ (S + BC) (S—BC):: R: cosBAC.

SOLUTION IV.

S and D retaining the significations above, (2 Cor. 8.) √(S+BC)(S—BC): √(BC+1)(BC-D)::R:tan BAC.

It is also convenient to reduce this last solution into another form, by putting P for half the perimeter of the triangle. Then S+BC=2P, and S-BC=2P—2BC; and therefore (S+BC) (S-BC) = 2P (2P-2BC) = 4P (P-BC). In like manner, BC+D=2P-2AC, and BC—D = 2P—AB; wherefore (BC+D) (BC-D) (2P-2AC) (2P-2AB)=4(P-AC) (P-AB). Hence √ P(P—BC) : √ (P—AC) (P—AB):: R: tan BAC.

It may be observed of these four solutions, that the first has the advantage of being easily remembered, but

that the others are rather more expeditious in calculation, The second solution is preferable to the third, when the angle sought is less than a right angle; on the other hand, the third is preferable to the second, when the angle sought is greater than a right angle; and in extreme cases, that is, when the angle sought is very acute or very obtuse, this distinction is very material to be considered. The reason is, that the sines of angles, which are nearly =90°, or the cosines of angles, which are nearly = 0, vary very little for a considerable variation in the corre sponding angles, as may be seen from looking into the tables of sines and cosines. The consequence of this is, that when the sine or cosine of such an angle is given, (that is, a sine or cosine nearly equal to the radius,) the angle itself cannot be very accurately found. If, for ins stance, the natural sine .9998500 is given, it will be immediately perceived from the tables, that the arch corre sponding is between 89° and 89°, 1'; but it cannot be found true to seconds, because the sines of 89° and of 89°, 1', differ only by 50 (in the two last places), whereas the arches themselves differ by 60 seconds. Two arches, therefore, that differ by 1", or even by more than 1", have the same sine in the tables, if they fall in the last degree of the quadrant.

The fourth solution, which finds the angle from its tangent, is not liable to this objection; nevertheless, when an arch approaches near to 90, the variations of the tangents become very great, and too irregular to allow the proportional parts to be found with exactness, so that when the angle sought is extremely obtuse, and its half of consequence very near to 90, the third solution is the best.

It may always be known, whether the angle sought is greater or less than a right angle by the square of the side opposite to it being greater or less than the squares of the other two sides.

It may be useful to have all the solutions of the oblique angled triangle reduced to a form purely arithmetical, not requiring the inspection of a diagram, and brought together in one table,

Let A, B, C, be the angles of the triangle, and a, b, c, the sides respectively opposite to them.

SECTION III.

CONSTRUCTION OF TRIGONOMETRICAL TABLES.

In all the calculations performed by the preceding rules, tables of sines and tangents are necessarily employed, the construction of which remains to be explained.

These tables usually contain the sines, &c. to every minute of the quadrant from 1' to 90°, and the first thing required to be done is to compute the sine of 1', or of the least arch in the tables.

1. If ADB be a circle, of which the centre is C, DB any arch of that circle, and the arch DBE double of DB; and if the chords DE, DB be drawn, and also the perpendiculars to them from C, viz. CF, CG, it has been

demonstrated, (8. 1. Sup.) that CG is a mean propor tional between AH, half the radius, and AF the line made up of the radius and the perpendicular CF. Now CF is the cosine of the arch BD, and CG the cosine of the half of BD; whence the cosine of the half of any arch BD, of a circle of which the radius=1, is a mean proportional between and I+ cos BD. Or for the greater