out the triangle, of course the difference, instead of the sum of the angles and sides found in the right angled triangles, is to be taken.

Thus in the annexed triangle let the side and angle opposite given be c and c, and the other given angle B. Then first

sin c sin c:: sin B: sin b

:

by means of which proportion ₺ may be calculated and will be legitimately ambiguous. Then

B

Ъ

C

there will be left unknown A and a. From A let fall a perpendicular A D upon a which we have not drawn, lest it should confuse the diagram, but which the student can imagine; then in the right angled triangle B A D, we know two parts в and c, and also in the right angled triangle c a D we know two parts c and b, the latter having been found by the proportion above. To calculate the partial angles at a, calling that in the first right angled triangle above mentioned w, and that in the second ', we have by Napier's rules

Thus all the parts of the triangle are determined.

For an application of this case of solution, find the angle

s and the side zs in the example of Art. 81.

We have now demonstrated formulæ for the solution of every possible case of plane and spherical triangles, including the more simple formula which apply exclusively to the right-angled triangles. We have had occasion for this purpose to derive many formulæ expressing the general relations of the trigonometrical lines. A treatise has thus been formed complete in itself, and containing nothing that could well be omitted. For greater convenience in committing to memory, we give in the next Part, a recapitulation of the results.

* Some few of the general formulæ which have been derived in the course of the work are here omitted; all that are found in this recapitulation, may with advantage be committed to memory.

+ This is derived from the preceding.

2

(22) cos p + cos q == cos (p+q) cos(pq) (Art. 86)

R

2

R

sin p

sin

tan (p-q)

or,

sum of the sines: diff. of the sines :: tan 1⁄2 sum : tan 1⁄2 difference.

II.

Formulæ and rules for the solution of triangles.

1. Right angled plane triangles.

A being the right angle and a the hypothenuse.

CASE I.

The hypothenuse being either given or required, and one of the angles.

(1) . . R : sin B:: a: b or R : sin c :: a : c

(Art. 38.)

i. e. Rad the sine of either acute angle: : the hypothenuse : the side opposite the angle.

CASE II.

The hypothenuse being neither given nor required, and one of the acute angles being either given or required.

(2).

R: tan Bcb or R: tan cb: c

(Art. 41) i, e. Rad tan of either acute angle: : the side adjacent that

angle: the side opposite.

In the second of proportions (1) and (2) sin and tan of c may be changed into cos, and cot of B.

CASE III.

When two of the sides are given and the third required. a = √ b2 + c2 or b = √ a2 — c2 or c = √ a2 — bi 2. Oblique angled plane triangles.

CASE I.

Two of the three given parts being a side and its opposite

or the sines of the angles are as the opposite sides.

This rule when used alone if an angle be required gives two solutions since sin of an arcsin of its supplement. If other parts of the triangle besides the four in the proportion are already determined, they serve to indicate which of the two is to be chosen.

CASE II,

Two angles and the interjacent side being given to find the other parts,

Subtract the sum of the two given angles from 180°, the remainder is the third angle. Then say as sine of this angle: the given side opposite: the sine of either of the other angles: the side opposite.