true with either triangle; for the angle CB'D = CBA, and the sine of CB'D is the same as the sine of AB'C. In practice we can determine which of these triangles is proposed, by the side AB being greater or less than AC; or, by the angle at the vertex C being large, as ACB, or small, as ACB'.

In the solitary case in which AC, CB, and the angle A, are given, and CB less than AC, we can determine both of the A's AB and ACB'; and then we surely have the right one.

PROPOSITION V.

If from any angle of a triangle, a perpendicular be let fall. on the opposite side, or base, the tangents of the segments of the angle are to each other as the segments of the base.

Let ABC be the triangle. Let fall the perpendicular CD, on the side AB.

Take any radius, as Cn, and de

scribe the arc which measures the A

D

angle C. From n, draw qnp parallel to AB. Then it is obvious that np is the tangent of the angle DCB, and ng пр is the tangent of the angle ACD.

Now, by reason of the parallels AB and qp, we have,

If a perpendicular be let fall from any angle of a triangle to its opposite side or base, this base is to the sum of the other two sides, as the difference of the sides is to the difference of the segments of the base.

(See figure to Proposition 5.)

Let AB be the base, and from C, as a center, with the shorter side as radius, describe the circle, cutting AB in G, and AC in F; produce AC to E.

It is obvious that AE is the sum of the sides AC and CB, and AF is their difference.

Also, AD is one segment of the base made by the perpendicular, and BD DG is the other; therefore, the difference of the segments is AG.

=

As A is a point without a circle, by Cor. Th. 18, B. III, we have

AE × AF = AB × AG

Hence,

AB: AE = AF

AG.

PROPOSITION VI1.

The sum of any two sides of a triangle is to their difference, as the tangent of one half the sum of the angles opposite to these sides, is to the tangent of one half their difference.

Let ABC be any plane triangle. Then, by Proposition 4, we have,

BC: AC sin.A sin.B.

Hence,

=

:

BC+AC:BC-AC-sin.A+sin. B: sin.A—sin. B (Th. 9, B. II).

II),

Comparing the two latter proportions, (Th. 6, B.

Given, the three sides of any plane triangle, to find some relation which they must bear to the sines and cosines of the respective angles.

Let ABC be the triangle, and let the perpendicular fall either upon, or without

the base, as shown c

in the figures.

AA

a

By recurring to Th. 41, B. I, we shall find

α

B

Equating these two values of CD, and reducing, we

In this expression we observe, that the part c, whose square is found in the numerator with the minus sign, is the side opposite to the angle; and that the denominator is twice the rectangle of the sides adjacent to the angle. From these observations we at once draw the following expressions for the cosine A, and cosine B:

As these expressions are not convenient for logarithmic computation, we modify them as follows:

In the preceding expression, (n), if we consider radius unity, and add 1 to both members, we shall have

Considering b + c as one quantity, and observing that (b+c)2 — a2 is the difference of two squares, we have (b+c)2—a2=(b+c+a)(b+c—a); but (b+c—a)=b+c+a—2a.

In every triangle, the sum of the three angles is equal to 180°; and if one of the angles is small, the other two must be comparatively large; if two of them are small, the third one must be large. The greater angle is always opposite the greater side; hence, by merely inspecting the given sides, any person can decide at once which is the greater angle; and of the three pre

ceding equations, that one should be taken which applies to the greater angle, whether that be the particular angle required or not; because the equations bring out the cosines to the angles; and the cosines to very small arcs vary so slowly, that it may be impossible to decide, with sufficient numerical accuracy, to what particular arc the cosine belongs. For instance, the cosine 9.999999, carried to the table, applies to several arcs; and, of course, we should not know which one to take; but this difficulty does not exist when the angle is large; therefore, compute the largest angle first, and then compute the other angles by Proposition 4.

But we can deduce an expression for the sine of any of the angles, as well as the cosine. It is done as follows:

EQUATIONS FOR THE SINES OF THE ANGLES. Resuming equation (m), and considering radius unity, we have

Equating the second members of (1) and (2),