Fig. 12.

30. I proceed now to treat of the application of trigonometrical tables to the resolution of triangles. It must be recollected, that, by means of these tables, when an angle is known, the value of its sine, that of its cosine, tangent, and cotangent are also known, and that, reciprocally, when the value of one of these lines is given, that of the arc is to be regarded as given.

Let CDE (fig. 12) be a triangle, right-angled at D; from one of the acute angles C, we describe, with a radius equal to that of the tables, the arc AM, and let fall the perpendicular PM upon AC; we then raise the tangent AN in order to form the two triangles of the tables, namely, CPM, which will be that of the sine and cosine, and CAN, that of the tangent and secant. These will be each similar to the triangle proposed; and by comparing them successively with this, we obtain the following proportions; CM: PM:: CE: DE

CM: CP:: CE : CD or
CA: AN:: CD: DE

R: sin C::CE: DE
R: cos C::CE: CD

R: tang C:: CD: DE.

The angle E being the complement of the angle C, we have cos Csin E; and the first two propositions admit of being united in one, and may be enunciated thus; in any right angledtriangle, radius is to the sine of one of the acute angles, as the hypothenuse is to the side opposite to this angle.

The third shows, that radius is to the tangent of one of the acute angles, as the side of the right angle adjacent to this acute angle is to the side opposite.

Radius being always given, it is sufficient to know two of the three other terms of each of the proportions, which I have just stated, in order to find the remaining one. Thus by the first proportion when two of these three things, namely, the hypothenuse, a side, and an acute angle are known, the third is readily determined.

I say simply an acute angle, although the proportion requires, that this angle should be opposite to the side given, or to that required, because one of the acute angles enables us to find the other immediately; therefore, if that which is known, or that which is sought, do not fulfil the condition, we may employ its complement.

By the second proportion when two of these these three things are known, namely, the two sides of a right angle and an acute angle, the third is readily determined.

It follows from this, 1. that knowing a side and an angle of a right-angled triangle, we can calculate the two other sides; 2. that any two sides whatever being known, we can calculate the acute angles.

These two cases do not comprehend that in which any two sides being given to find the third; but this is immediately resolved by the known property of a right-angled triangle, which

If we have given the hypothenuse CE and one of the sides of the right angle DE, for example, we have

CD=

2

Recollecting that CE — DE = (CE + DE) (CE — DE) (Alg. 34), if we take the logarithms of the two members of the equation CD = √(CE + DE) (CE

DE), we shall have 1CD = [(CE + DE) +1(CE -- DE)].

When we construct formulas to be used in numerical calculations, we should endeavour to prepare them in such a manner that logarithms may be conveniently applied to them, that is, so that it will be necessary to pass from logarithms to numbers and from numbers to logarithms, as little as possible. By applying logarithms to the determination of CD, by means of the first expression above given, we shall perceive very clearly the object of this remark.

I will conclude this exposition of the principles, that are employed in the resolution of right-angled triangles, by observing, that the two cases last treated may be resolved also by the two propositions given at the commencement of this article. For 1. if, having CD and DE, we would determine CE, we can calculate one of the acute angles, C, for example, by the proportion R: tang C: CD: DE; having found this angle, we calculate the hypothenuse CE by the proportion R: sin C:: CE: DE, in which the three terms R, sin C, and DE are known. 2. When the hypothenuse CE and one of the other sides, CD, for example, are known, we calculate the acute angle opposite to the side sought, by the proportion, R: sin E or cos C:: CE: CD; then the side DE is found by the proportion R sin C:: CE: DE.

:

31. What has been said upon right-angled triangles may be put into a convenient form, by using the letters A, B, C, to denote the angles, A being the right angle, and a, b, c, to denote the sides respectively opposite to these angles, as is shown by figure Fig. 13. 13; we have then by the first principle

R: sin C: a: c, R: sin B::a: b,

Exterminating a from these two equations, which is done by dividing the two members of the first each by the corresponding member of the second, we find

an equation, which represents the second principle enunciated in the preceding article.

Lastly, if we square each member of the first two equations, and add the results, member to member, observing, at the same time, that

sin Casin B2 sin C2 + cos C2 = R2 (10),

we have

are sufficient, together with the relation subsisting between the angles B and C, for the resolution of all cases of right-angled triangles.

32. The principle upon which the resolution of right-angled triangles is founded, leads also to that of triangles of whatever kind. By letting fall from the angle B of the triangle ABC Fig. 14. (fig. 14) a perpendicular BD, we form two triangles ABD, BDC, right-angled at D; we have in the first

RX BD sin A × AB, R x BD = sin C X BC, whence

sin AX AB sin C x BC, or sin A: sin C:: BC: AB.

=

When the perpendicular falls without the triangle, the angle C is not common to the two triangles ABC, BCD; but the angles BCD, BCA, being together equal to two right angles, have the same sine (22).

The proportion just given admits of a general application, and may be enunciated thus; In any triangle whatever the sines of the angles are to each other as the sides opposite to these angles.

33. The same proposition may be demonstrated in the following manner, which may appear more conformable to the idea I have given of trigonometry in art. 1 and 2.

Having inscribed the triangle ABC (fig. 15) in a circle, if Fig. 15. from the centre O of this circle, and with a radius O a, equal to that of the tables, we describe a circle a b c, and then join by the lines ab, bc, and a c, the points where the radii AO, BO, CO, meet the circle of the tables, we form a triangle abc similar to the triangle proposed, the sides of which are deduced from the tables.

The similarity of the two triangles ABC, abc, (Geom. 209), becomes evident, when we consider that the right lines a O, b 0, and c O, being equal, as radii of the same circle, as well as the straight lines AO, BO, and CO, the triangles AOB, BOC, and AOC, have their sides AO and BO, BO and CO, AO and CO cut proportionally in the points a and b, b and c, a and C, and consequently the right lines AB and ab, BC and bc, AC and a c, are respectively parallel; we have then

or

AB: BC: AC::ab: bc: ac,

: : 1 ab : 1 b c : 1 ac.

This being the case, the angles of the triangle abc, having their summit in the circumference, are measured by half of the arc subtending the side opposite to them, and each of these arcs has evidently for its sine half of the same side (14); whence