1

following are put down without demonstrations, for the exercise of the student.

Expression for the area of a triangle, in terms of the sides.

221. Let the sides of the triangle ABC (Fig. 23.) be expressed by a, b, and c, the perpendicular CD by p, the segment AD by d, and the area by S.

Then a2b3+c-2cd, (Euc. 13. 2.)

Reducing the fraction, (Alg. 150.) and extracting the root

4c2

of both sides,

This gives the length of the perpendicular, in terms of the sides of the triangle. But the area is equal to the product of the base into half the perpendicular height. (Alg. 518.) that is,

S=cp=}V/4bc _(b+c* − a )

Here we have an expression for the area, in terms of the sides. But this may be reduced to a form much better adapted to arithmetical computation. It will be seen, that the quantities 4b2c2, and (b2+c2 -a2) are both squares; and that the whole expression under the radical sign is the difference of these squares. But the difference of two squares is equal to the product of the sum and difference of their roots. (Alg. 235.) Therefore 4b2c2-(b2+c2 - a2)2 may be resolved into the two factors,

2bc+(b2+c2 - a2) which is equal to (b+c)2-a2 2bc-(b2+c2-a2) which is equal to a2 -(b-c)3

Each of these also, as will be seen in the expressions on the right, is the difference of two squares; and may, on the same principle, be resolved into factors, so that,

S (b+c)2 —a2=(b+c+a)×(b+c-a)'

{

a2 — (b—c)2 = (a+b−c)×(a−b+c)

Substituting, then, these four factors, in the place of the quantity which has been resolved into them, we have,

S=1√(b+c+a)×(b+c−a)×(a+b−c)×(a−b+c)

* The expression for the perpendicular is the same, when one of the angles is obtuse, as in Fig. 24. Let AD=d.

Then a2=b2+c2+2cd. (Euc. 12. 2.) And d=.

Therefore d2 =

1

-b2 — c2+a2

2€

( — b2 — c2 +a2 ) 2 __ (b2+c2 − a2)2 (Alg. 219.)

And p

4c2

4c2

Here it will be observed, that all the three sides, a, b, and c, are in each of these factors.

Let h=(a+b+c) half the sum of the sides. Then

S=√hx (h− a)× (h—b)× (h−c)

222. For finding the area of a triangle, then, when the three sides are given, we have this general rule;

From half the sum of the sides, subtract each side severally; multiply together the half sum and the three remainders ; and extract the square root of the product.

SECTION VIII.

COMPUTATION OF THE CANON.

ART. 223. THE trigonometrical canon is a set of tables containing the sines, cosines, tangents, &c. to every degree and minute of the quadrant. In the computation of these tables, it is common to find, in the first place, the sine and cosine of one minute; and then, by successive additions and multiplications, the sines, cosines, &c. of the larger arcs. For this purpose, it will be proper to begin with an arc, whose sine or cosine is a known portion of the radius. The cosine of 60° is equal to half radius. (Art. 96. Cor.) A formula has been given, (Art. 210.) by which, when the cosine of an arc is known, the cosine of half that arc may be obtained.

By successive bisections of 60°, we have the arcs

If the radius be 1, and if a=60°, b=30°, c=15°, &c.; then

cos b=cos a=√ }+1×1=0.8660254

cos c=cos b+cos b=0.9659258
cosd=cos &c=+cos c=0.9914449
cos e cos id=✔+cos d=0.9978589

Proceeding in this manner, by repeated extractions of the square root, we shall find the cosine of

0° 0′ 52" 44" 3"" 45"""" to be 0.99999996732

And the sine (Art. 94.)=√1 — cos2 = 0.00025566346

This, however, does not give the sine of one minute exactly. The arc is a little less than a minute. But the ratio of very small arcs to each other, is so nearly equal to the ratio of their sines, that one may be taken for the other, without sensible error. Now the circumference of a circle is divided into 21600 parts, for the arc of 1'; and into 24576, for the arc of 0° 0′ 52" 44"" 3""" 45"""

Therefore,

21600 24576: 0.00025566346 0.0002908882,

which is the sine of 1 minute very nearly.*

And the cosine =√1-sin2 =0.9999999577.

224. Having computed the sine and cosine of one minute, we may proceed, in a contrary order, to find the sines and cosines of larger arcs.

Making radius=1, and adding the two first equations in art. 208, we have

sin(a+b)+sin(a-b)=2sin a cos b

Adding the third and fourth,
cos(a+b)+cos(a—b)=2cos a cos b
Transposing sin(a—b) and cos(a—b)
I. sin(a+b)=2sin a cos b-sin(a—b)
II. cos(a+b)=2cos a cos b-cos(a—b)

If we put b=1', and a=1', 2', 3', &c. successively, we shall have expressions for the sines and cosines of a series of arcs increasing regularly by one minute. Thus,

* See note H.