Let ABC be the proposed triangle ; AD the perpendicular, let fall from the vertex A on the opposite side BC: there may be two

:

cases.

First. If the perpendicular falls within

B D the triangle ABC, the right-angled triangles ABD, ACD, will give,

R:sin B :: AB: AD.

R:sin C::AC : AD.
In these two propositions, the extremes are equal ; hence,

sin C:sin B :: AB: AC.
Secondly. If the perpendicular falls A
without the triangle. ABC, the right-
angled triangles ABD, ACD, will still
give the proportions,
R: sin ABD :: AB : AD,

с R: sin C :: AC:AD;

D B from which we derive

sin C:sin ABD :: AB: AC. But the angle ABD is the supplement of ABC, or B; hence sin ABD=sin B; hence we still have

sin C:sin B :: AB: AC.

THEOREM IV.

In

every rectilineal triangle, the cosine of either of the angles is equal to radius multiplied by the sum of the squares of the sides adjacent to the angle, minus the square of the side opposite, divided by twice the rectangle of the adjacent sides.

Let ABC be a triangle: then will
AB2+ BC2_AC?

А.
cos B=R

2AB x BC. First. If the perpendicular falls within the triangle, we shall have AC?—AB? + BC?—2BC BD (Book IV. Prop. XII.); B

D AB + BCP_AC? hence BD=

But in the right-angled triangle

2BC ABD, we have

R:cos B :: AB: BD;

RX BD hence, cos B=

AB

or by substituting the value of BD, AB + BCP_AC2

+] cos B=Rx

2AB x BC Secondly. If the perpendicular falls

A
without the triangle, we shall have
ACAB+BC+2BC x BD; hence

AC2_AB2-BC2
BD-

2BC
But in the right-angled triangle BAD,
RX BD

D B с we still have cos ABD

; and the angle ABD being

AB supplemental to ABC, or B, we have

RX BD cos B=-COS ABD

AB hence by substituting the value of BD, we shall again have

AB2+BC_AC? cos B=Rx

2AB X BC

Scholium. Let A, B, C, be the three angles of any triangle ; a, b, c, the sides respectively opposite them: by the theorem,

a' +_bo we shall have cos B=RX

And the same principle,

2ac when applied to each of the other two angles, will, in like manbo+cao

a'+b% ner give cos A=Rx

and cos C=RX 2bc

2ab Either of these formulas may readily be reduced to one in which the computation can be made by logarithms.

Recurring to the formula RP-R cos A=2 sino {A (Art. XXIII.), or 2sin ?A=R?—RcosA, and substituting for cosa, we shall have

32 +-?
2sin” A=R?__Rox

2bc
Rox 2bc-Ro(bo+c -a a_12_0+2bc
2bc

2bc
= Rox
a?—(6—c) –R? x (a+b_c) (a+c—),

Hence
260

2bc
sin LA=Rv ((a+bc) (a+c—b)

4bc For the sake of brevity, put 1 (a+b+c)=p, or a+b+c=2p; we have a+bc=2p-2c, a+c-b=2p-2b; hence

17 .

bc

Х

sin dA=Rv (6.) (p).

p-b—c)

THEOREM V.

In

every rectilineal triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides, to the tangent of half their difference.

:

For, AB :BC::sin C: šin A (Theo.

B rem III.). Hence, AB+BC: AB-BC :: sin C+sin A : sin C-sin A. But

C+A sin+sin A: sin C-sin A :: tang

2 C-A?

Α.

с tang (Art. XXIV.); hence, 2

C+A C-A AB+BC : AB-BC :: tang

tang

which is 2

2 the property we had to demonstrate.

With the aid of these five theorems we can solve all the cases of rectilineal trigonometry.

Scholium. The required part should always be found from the given parts ; so that if an error is made in any part of the work, it may not affect the correctness of that which follows.

SOLUTION OF RECTILINEAL TRIANGLES BY MEANS OF

LOGARITHMS.

It has already been remarked, that in order to abridge the calculations which are necessary to find the unknown parts of a triangle, we use the logarithms of the parts instead of the parts themselves. Since the addition of logarithms answers to the multiplica

of their corresponding numbers, and their subtraction to the division of their numbers; it follows, that the logarithm of the fourth term of a proportion will be equal to the sum of the logarithms of the second and third terins, diminished by the logarithm of the first term.

Instead, however, of subtracting the logarithm of the first term from the sum of the logarithms of the second and third terms, it is more convenient to use the arithmetical complement of the first term.

The arithmetical complement of a logarithm is the number which remains after subtracting the logarithm from 10. Thus 10–9.274687=0.725313: hence, 0.725313 is the arithmetical complement of 9.274687.

=

It is now to be shown that, the difference between two logarithms is truly found, by adding to the first logarithm the arithmetical complement of the logarithm to be subtracted, and diminishing their sum by 10. Let a = the first logarithm.

b=the logarithm to be subtracted.

c=10—=the arithmetical complement of b. Now, the difference between the two logarithms will be expressed by a—b. But from the equation c=10—6, we have C-10=-6: hence if we substitute for _b its value, we shall have

a-b=a+c-10, which agrees with the enunciation.

When we wish the arithmetical compleinent of a logarithm, we may write it directly from the tables, by subtracting the left hand figure from 9, then proceeding to the right, subtract each figure from 9, till we reach the last significant figure, which must be taken from 10: this will be the same as taking the logarithm from 10.

а

Ex. From 3.274107 take 2.104729.

Common method.

By ar.-comp. 3.274107

3.274107 2.104729

ar.-comp. 7.895271 Diff. 1.169378

sum 1.169378 after rejecting the 10.

We therefore have, for all the proportions of trigonometry, the following

RULE.

Add together the arithmetical complement of the logarithm of the

the first term, the logarithm of the second term, and the logarithm of the third term, and their sum after rejecting 10, will be the logarithm of the fourth term. And if any expression occurs in which the arithmetical complement is twice used, 20 must be rejected from the sum.

SOLUTION OF RIGHT ANGLED TRIANGLFS.

a

angle.c.

C

Let A be the right angle of the proposed

C right angled triangle, B and C the other two angles; let a be the hypothenuse, b the side

6 opposite the angle B, c the side opposite the Here we must consider that the B

А. two angles C and B are complements of each other; and that consequently, according to the different cases, we are entitled to assume sin C=cos B, sin B=cos C, and likewise tang B= cot C, tang C=cot B. This being fixed, the unknown parts of a right angled triangle may be found by the first two theorems; or if two of the sides are given, by means of the property, that the square of the hypothenuse is equal to the sum of the squares of the other two sides.

EXAMPLES.

::

Ex. 1. In the right angled triangle BCA, there are given the hypothenuse a=250, and the side b=240; required the other parts.

R: sin B : :a : b (Theorem I.). or,

a : 6 R : sin B. When logarithms are used, it is most convenient to write the proportion thus, As hyp. a

250 ar.-comp. log. 7.602060 To side 6 240

2.380211 So is R

10.000000 To sin B 73° 44' 23" (after rejecting 10) 9.982271

But the angle C=90°-B=90°—73° 44' 23"=16° 15' 37". or, C might be found by the proportion,

250

ar.-comp. log. 7.602060 To side b 240

2.380211 So is R

10.000000 Toros C 16° 15' 37"

9.982271

As hyp. a

To find the side c, we say,
As R

log.
To tang. C 16° 15' 37"
So is side b 240
To side c 70.0003