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tained angle may be used; or the co-tangent of half the contained angle, which are all equal to each other.

H

PROPOSITION V. (G) In any plane triangle, double the base or longest side, is to the sum of the other two sides, as their difference is to the distance of a perpendicular from the middle of the base ; this distance, added to half the base, gives the greater segment, and being subtracted leaves the less.

Let ABC be the triangle, with c as a centre, and the radius cb=the least side, describe a circle, produce ac to h; then because cr=CB=CH; AA = AC + cb the sum of the sides, and AF = AC – BC the

А

E difference between the sides.

Because cd is perpendicular to GB,GD=BD (Euclid 3 of III.) therefore ag is the difference between the segments of the base; but ED=AG (C. 35.) for it is the excess of the greater segment ad above half their sum AE; AB being the base, or sum of the segments AD and DB, and as its half.

Then (Euclid 36 of III. corol.) AH X AF = AB X AG, therefore AB : AH::AF : AG, that is AB : AC+CB::AC-CB: 2ED, or, 2AB : AC+CB:: AC-CB: ED the distance of the perpendicular cd from the middle e of the base; and that Ed added to half the base ae gives at the greater segment, &c. is obvious.

Q. E. D.

PROPOSITION VI. (H) From half the sum of the three sides of any triangle subtract the side opposite to the angle sought, and note the half sum and remainder. Then, the rectangle of the sides containing that angle, is to the square of the radius, as the rectangle of the half sum and remainder is to the square of the cosine of half the angle taken.

Let abc be the triangle; bisect the angle cab by the indefinite line ah, and draw Ec perpendicular to ad. Draw DG parallel and equal to CB, join GB and produce it to H.

* Since the sum of the three angles of every triangle is equal to two right angles, it is plain that half the sum of any two angles is half the supplement of the third angle. And, that the tangent of half the supplement of any angle is equal to the co-tangent of half that angle may be thus shown :

Let HBP (Fig. 1. Plate 1.) be any arc bisected in B, and ruf, its supplement, bisected in À; then ak, the co-tangent of the half arc er, is evidently the tangent of the balf supplement Ar.

Now the triangles ade and apc have the two sides pc and DE equal to each other (by construction), and the side ad common, also the angles about D are right angles, therefore ac= AE, (Euclid 4 of I.) and consequently eB is equal to the difference between the sides ac and ab.

Because the side Ec of the triangle cez is bisected in D, and DF is parallel to CB (by construction), EF=FB, therefore FB= half the difference between the sides ac and ab, and af=half their sum. (C. 35.)

By construction cb is equal and parallel to no, therefore (Euclid 33 of I.) GB is equal and parallel to cd or de, for the same reason DBGE is a parallelogram, and FDIFG, for the diagonals of a parallelogram bisect each other. And, since GB is parallel to ED, Gh is also parallel to it, and the angle at D being a right angle, the angle at h is also a right angle. With F as a centre, and radius FD=FG=CB describe a circle, which will pass through the point H. (Euclid 31 of 1II.)

Since FG=CB, FI=CB, but AF=ý (AC+AB), to each of these equals add fi or the half of CB, then will ai =} (AC + AB + BC), but ki =DG = CB, therefore ak

A SAI - CB=(AC+AB + BC)— BC

C

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radius: AE::sine aed=cosine Dae : AD
radius: AB::sine ABH=cosine HAB=DAE : AH

(Z. 34.) Hence, square radius : AE X AB:: square cosine DAE : AD XAH. But as X AD = AI X AK (Euclid 36 of III.), and AE = AC; therefore

square radius : AC X AB:: square cosine DAE : AIX AK; or AC X AB : square radius:: AI XAK : square cosine DAE; that is, ac X AB : square radius:: 1 (ac+ AB + BC)+(1 Ac+ AB+BC) - BC : square cosine DAE=} CAE or CAB.

Q. E. D.

PROPOSITION VII.

(I). From half the sum of the three sides of any plane triangle subtract the side opposite to the angle sought, and note the remainder: Also from the said half sum subtract each of the other sides of the triangle. Then,

The rectangle contained by half the sum of the sides and the first remainder, is to the square of the radius; as the rectangle contained by the other two remainders is to the square of the tangent of half the angle sought.

The same construction being made as in the preceding proposition, A1=5 (4C + AB + BC), and Ak=) (AC+ AB + BC) --- BC.

In the right angled triangles ADE and AHB.

s radius : AD::tangent daE : DE radius: An:: tangent dae : BHS

(A. 34.) Hence, square radius : AD XAH:: square tangent dae : de

And AD XAH : square radius:, DE X BH : square tangent

}

BH,
DAE.

But AD X AH=AIXAK (Euclid 36 of III.) and because DEGB being opposite sides of a parallelogram, we have DE X BH= GB X BHKB X BI. (Euclid 35 of III.) Again, because, ki is bisected in F, and that FE and FB are equal; kE is equal to BI, and therefore kB is equal to Ei; but ei = AI – AE=AL - AC, consequently KBSAI - AC, and BI=AI - AB.

Hence AI X AK: square radius::(AI – AC) (AI – AB): square tangent DAE, viz. }(ac+ AB + BC) x (14C+AB + BC-BC): square radius:: (1 AC + AB + BC-AC) * (+ AC+ AB + BC-AB): square tangent DAE. Q. E. D.

PROPOSITION VIII. (K) Twice the rectangle of any two sides of a plane triangle, is to the sum of the squares of the same two sides diminished by the square of the third side; as radius, is to the cosine of the angle contained between those sides.

Let abc be any plane triangle and cd a perpendicular from the vertical angle upon the base or upon the base produced.

1. If the perpendicular falls within the triangle ac = AB? + BC2 2AB X BD. (Euclid 13 of II.)

1

2 2. If the perpendicular falls without the triangle AC? AB + BC2 + 2AB X BD. (EUCLID 13 of II.)

AB? + BC?- AC? From the first BD=

2AB

AC2_ AB- BC? From the second BDS

2AB But Bc : rad.::BD : cosine of the angle B (Z. 34.)

cos. LB X BC **BD

here LB is acute in the first triangle,

rad.
and obtuse in the second. By substitution,

cos. L. B X BC AB? + BC? - AC
rad.

2AB
COS. LB X BC AC2 - AB?
ci rad.

2BA

son

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2

BC2

or,

from either of these equations, by reduction,

rad. X (AB? + BC? - Aco) cos. Z B=

2AB X BC or, 2AB X BC : ABS + BC? — AC? :: rad. : cos. LB. Q. E. D. NOTE. If the sides of the triangle be large, the rule given above is inconvenient for arithmetical calculation.

SCHOLIUM.

(L) Various other propositions might be given, but the preceding are sufficient for solving every case of plane trigonometry. The analogies of plane trigonometry are only particular cases of spherical trigonometry, and may in general be deduced from thence by simple inferences only.

Vide Book III. Chap. V. Prop. xxviii. and the Scholium.

SOLUTIONS OF THE DIFFERENT CASES OF RIGHT ANGLED

PLANE TRIANGLES.

a

Rules for solving the different cases both of right angled plane triangles, and oblique angled plane triangles, have already been investigated. The solutions here given are adapted to the table of natural sines; and the method of notation is that used by Lagrange and Legendre, the three angles of the triangle being represented by A, B, C, and their opposite sides by a, b, c, as in the figure annexed; r=radius=sine of 90°

Case I. Given the angles and the hypothenuse, to find the base and perpendicular.

sine A.6

Cos C.6 SOLUTION. a=

tang A. 6 g.b

cosec A COS A.b sine t.b

cot A 6

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sec A

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Case II. Given the angles and the base, to find the hypothenuse and the perpendicular.

*

Logarithmical formulæ are easily supplied, thus in the first case (log. sine A+ log. ) – 10=log, a. The introduction of such formulæ would only incrcase the size of the book, without any real advantage to the student,

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Case III. Given the angles and the perpendicular, to find the base and the hypothenuse.

a sine c.a SOLUTION. C=

sine A

COS A

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cot A.

COS A

tang A

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Case IV. Given the hypothenuse and the base, to find the angles and the perpendicular.

6 Solution. Cos a=sine c=* *; sec a=coscc c=

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tang į a=r

6+ c Having found the angles, the perpendicular may be found by Case I. or by Case II. Or, a=v0? —c* = v(b+c).(6—c).

CASE V. Given the hypothenuse and the perpendicular, to find the angles and the base.

6 SOLUTION. Sine Acos c=

b

;

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a
;

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16-a bta

tang ; c=r

Having found the angles, the base may be found by Case I. Or, c=1b2-a?=v(6+a).(-a).

Case VI. Given the base and the perpendicular, to find the angles and the hypothenuse.

; cot Astang c="

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