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dicular to AB: now DE is the sine of the angle A, aud CF is the sine of B, to the same radius AD or BC. But by similar triangles,
AD : DE :: AC : OF.
BC: sin A :: AC : sin B, or
BC : AC :: sin A : sin B. By comparing the sides AB, AC, in a similar manner, we should find,
AB : AC :: sinc: sin B.
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 angles, to the tangent of half their difference.
22. Let ACB be a triangle: then will AB+AC : AB-AC :: tan }(C+B) : tan }(C-B).
With A as a centre, and a E radius AC, the less of the two given sides, let the semicircumference IFCE be described, meeting AB in I, and BA produced, in E. Then, BE will be the sum of the
C-----...--'FGH sides, and BI their difference. Draw CI and AF.
Since CAE is an exterior angle of the triangle ACB, it is equal to the sum of the interior 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 }(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, thu extremity of the diameter IE,
to }(C+half the site at the ce
since the right angle ICE must be E inscribed in a semicircle.
But CE is the tangent of CIE =(C+B); and IH is the tan
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 j(C+B) : tan }(C-B).
In any plane triangle, if a line is 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 two other sides, as the difference of those sides to the difference of the segments.
23. Let BAC be a triangle, and AD perpendicular to the base; then
BC : CA + AB :: CA – AB : CD- DB. For, AB-BD®+ AD (Bk. IV., Prop. XI.); and AQ’=DO + AD by subtraction, AC – AB* = CD’ – BD.
But since the difference of B D the squares of two lines is equivalent to the rectangle contained by their sum and difference (Bk. IV., Prop. X.), we have,
AT" - ABC (AC+ AB).(AC-AB)
OD” – DB.(CD+DB).(CD- DB) therefore, (CD +. DB).(CD) – DB)=(AC+ AB). (AC-AB) hence, CD + DB : AC+ AB :: AC- AB : CD-DB.
In unij right-angled plane triangle, radius is to the tangent of either of the acute angles, as the side adjacent to the side opposite.
24. Let CAB be the proposed triangle, and denote the radius by R: then R : tan C :: AC : AB.
AG For, with any radius as CD de- _ scribe the arc DHI, and draw the tangent DG. From the similar triangles CDG and CAB, we have,
OD : DG :: CA : AB; hence,
R : tan C :: CA : AB. By describing an are with B as a centre, we could show in the same manner that, .
R : tan B :: AB : AC.
· TIIEOREM V.
In every riglit-angled plane triangle, radius is to the cosine of either of the acute angles, as the hypothenuse to the side adjacent.
25. Let ABC be a triangle, right-angled at B: then
R : cOS A :: AC : AB.
For, from the point A as a centre, with a radius AD=R, describe the arc DF, which will measure the angle A EF B 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,
K : cos A :: AC : AB. 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 (if the six parts which make up a plane triangle, three must be given, and at least one of these a side, before the others can be determined.
If the three angles only 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.
Assurning, with this restriction, any three parts of a triangle as giyen, one of the four following cases will always be presented.
I. When two angles and a side are given.
When two angles and a side are given. 26. Add the given angles together, and subtract their sum from 180 degrees. The remaining parts of the triangle can then be found by Theorem I.
1. In a plane triangle, ABC, there are given the angle A = 58° 07', the angle B=22° 37', and the side AB=408 yards. Required the 'oth er parts.
GEOMETRICALLY. 27. 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 with the protractor (see page 270), and when so measured, will be found equal to
99° 16'. The sides AC and BC may be measured by referring them to the scale of equal parts (see page 268). We shall find AC=158.9 and BC=351 yards.
TRIGONOMETRICALLY BY LOGARITHMS.
To the angle ..A=58° 07'
Their sum, = 80° 44'
leaves C . . . . 99° 16', of which, as it exceeds 90°, we use the supplement 80° 44'.
- To find the side BC.
REMARK. The logarithm of the fourth term of a proportion is obtained by adding the logarithm of the second terin 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 (Int. Art. 13): hence, the arithinetical complement of the logarithm of the first term added to the logarithms of the second and third terms, minus ten, will give the logarithm of the fourth term.
2. In a triangle ABC, there are given A=38° 25', B= 57° 42', and AB= 400 : required the remaining parts.
Ans. C= 83° 53', BC=249.974, AC= 340.01.