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But AD being equal to BC, we have

BC : sin A:: AC: sin B, or
BC : AC

: sin B. By comparing the sides AB, AC, in a similar manner, we should find, AB: AC :: sin C: sin B.

:: sin A

THEOREM II.

D

A

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In any triangle, the sum of the two sides containing either angle, is to their difference, as the tangent of half the sum of the two other argles, to the tangent of kalf their difference.

58. Let ACB be a triangle : then will AB+AC: AB-AC: : tan }(C+B) : tan }(C-B).

With A as a centre, and a radius AC the less of the two given sides, let the semicircle IFCE be described, meeting AB in I, and BA produced, in E. Then, BE will be the sum of the sides, and BI

C

F GH

B their difference. Draw CI and AF.

Since CAE is an outward angle of the triangle ACB, it is equal to the sum of the inward angles C and B (Bk. I, Prop. XXV, Cor. 6). But the angle CIE being at the circumference, is half the angle CAE at the centre (Bk. III, Prop. XVIII); that is, half the sum of the angles C and B, or equal to i(C+B).

The angle AFC=ACB, is also equal to ABC+BAF; therefore, BAF=ACB-ABC.

But, ICF=;(BAF)=}(ACB-ABC), or }(C-B). With I and C as centres, and the common radius IC, let the arcs CD and IG be described, and draw the lines CE and IH perpendicular to IC. The perpendicular CE will pass through E, the extremity of the diameter IE, since the right angle ICE must be inscribed in a semicircle.

But CE is the tangent of CIE=}(C+B); and IH is the tangent of ICB=}(C-B), to the common radius CI.

But since the lines CE and IH are parallel, the triangles BHI and BCE are similar, and give the proportion,

BE : BI :: CE : IH, or by placing for BE and BI, CE and IH, their values, we have AB+AC: AB-AC :: tan ;(C+B). tan }(C-B).

THEOREM III.

In any plane triangle, if a line be drawn from the vertical angle perpendicular to the base, dividing it into two segments : then, the whole base, or sum of the segments, is to the sum of the other two sides, as the difference of those sides to the difference of the segments.

59. Let BAC be a triangle, and AD perpendicular to the base : then will

BC: CATAB :: CAAB: CD-DB For,

AB2=BD2+AD (Bk. IV, Prop. XI);

A and

AC-DCP+AD? by subtraction AC-AB=CDBD.

But since the difference of the squares B of two lines is equal to the rectangle contained by their sum and difference (Bk. IV, Prop X), we have,

AC-AB=(AC+AB). (AC-AB) and CD-DB=(CD+DB). (CD-DB) therefore, (CD+DB).(CD-DB)=(ACHAB). (AC--AB) hence, CD+DB : ACTAB :: AC-AB : CD-DB. !

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THEOREM IV.

In any right-angled plane triangle, radius is to the tangent of either of the acute angles, as the side adjacent to the side opposite. 60. Let CAB be the proposed triangle,

B and denote the radius by R: then will

R: tan C: AC: AB. For, with any radius as CD describe the arc DH, and draw the tangent DG.

D From the similar triangles CDG and CAB we shall have,

CD: DG :: CA : AB; hence,

R: tan C:: CA: AB. By describing an arc with B as a centre, we could show in the same manner that,

THEOREM V.

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D

B

In every right-angled plane triangle, radius is to the cosine of either of the acute angles, as the hypothenuse to the side adjacent.

61. Let ABC be a triangle, right angled at B then will

R: cos A: AC: AB. For, from the point A as a centre, and any radius as AD, describe the arc DF,

EF which will measure the angle A, and draw DE perpendicular to AB: then will AE be the cosine of A. The triangles ADE and ACB, being similar, we have

AD: AE :: AC: AB: that is,

R: cos A:: AC: AB. 62. REMARK. The relations between the sides and angles of plane triangles, demonstrated in these five theorems, are sufficient to solve all the cases of Plane Trigonometry. Of the six parts which make up a plane triangle, at least three must

, be given, and one of these a side; before the others can be determined.

If the three angles are given, it is plain, that an indefinite number of similar triangles may be constructed, the angles of which shall be respectively equal to the angles that are given, and therefore, the sides could not be determined.

Assuming, with this restriction, any three parts of a triangle as given, one of the three following cases will always be presented.

I. When two angles and a side are given.
II. When two sides and an opposite angle are given.
III. When two sides and the included angle are given.
IV. When the three sides are given.

a

CASE I.

When two angles and a side are given.

63. Add the given angles together and subtract their sum from 180 degrees. The remaining parts of the triangle can

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Draw an indefinite straight line AB, and from the scale of equal parts lay off AB equal to 408. Then at A lay off an angle equal to 58° 07', and at B an angle equal to 22° 37', and draw the lines AC and BC: then will ABC be the triangle required.

The angle C may be measured either with the protractor or the scale of chords (Arts. 23 and 24), and will be found equal to 99° 16'. The sides AC and BC may be measured by referring them to the scale of equal parts (Art. 22). We shall find AC=158.9 and BC=251 yards.

BY LOGARITHMS.

.

.

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To the angle A=58° 07'
Add the angle.

B=22° 37'
Their sum

=80° 44'
taken from,

180° 00' leaves C

99° 16' which, exceeding 90° we use its supplement

80° 44'.

To find the side BC. As sin C. 99° 16'

ar. comp.

0.005705 : sin A. 58° 07'

9.928972 :: AB. 408

2.610660 351.024 (after rejecting 10). . 2.545337 REMARK. The logarithm of the fourth term of a proportion is obtained by adding the logarithm of the second term to that of the third, and subtracting from their sum the logarithm of the first term. But to subtract the first term is the same as to add its arithmetical complement and reject 10 from the sum (Art. 17): hence, the arithmetical complement of the first term added to the logarithms of the second and third terms,

.

:

BC.

ar. comp.

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.

.

To find side AC.
As sin C
990 16'

0.005705
: sin B
22° 37'

9.584968 AB 408

2.610660 : AC 158.976

2.201333 2. In a triangle ABC, there are given A=38° 25', B=570 42', and AB=400: required the remaining parts.

Ans. C=83° 53', BC=249.974, AC=340.04.

CASE II.

When two sides and an opposite angle are given 64. In a plane triangle ABC, there are given AC=216, CB=117, the angle A=22° 37', to find the other parts.

d

INSTRUMENTALLY.

Draw an indefinite right line ABB': from any point as A, draw AC making BAC=22° 37', and make AC=216. With C as a centre, and a radius equal to 117, the other given side, describe the arc B'B; draw B'C and BC: then will either of the triangles ABC or AB'C, answer all the conditions of the question.

BY LOGARITHMS.

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To find the angle B. As BC 117

ar. comp. AC

216 :: sin A

22° 37' : sin B' 45° 13' 55", or ABC 134° 46' 05"

:

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7.931814 2.334454 9.584968

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9.851236

The ambiguity in this, and similar examples, arises in conse quence of the first proportion being true for either of the angles ABC, or AB'C, which are supplements of each other, and therefore have the same sine (Art. 43). As long as the two triangles exist, the ambiguity will continue. But if the side CB, opposite the given angle, is greater than AC, the arc BB' will

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