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SOLUTION OF TRIANGLES.

Fig. 401.

Let

B

(11) Right-angled Triangles. ABC be any right-angled triangle. Denote the sides opposite the angles by the corresponding small letters. Then any one side and one acute angle, or any two sides being given, the other parts can be obtained by one of the following equations:

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THEOREM I.-In every plane triangle, the sines of the angles are to each other as the opposite sides.

THEOREM IL-In every plane triangle, the sum of two sides is to their difference as the tangent of half the sum of the angles opposite those sides is to the tangent of half their difference.

THEOREM III.—In every plane triangle, the cosine of any angle is equal to a fraction whose numerator is the sum of the squares of the sides adjacent to the angle, minus the square of the side opposite to the angle, and whose denominator is twice the product of the sides adjacent to the angle.

All the cases for solution which can occur, may be reduced to four.

CASE 1.-Given a side and two angles. The third angle is obtained by subtract ing the sum of the two given angles from 180°. Then either unknown side can be btained by Theorem I.

sin. B

Calling the given side a, we have b=a.

and ca

sin. A

sin. C
sin. A

The third angle is obtained by The remaining side is then ob

CASE 2.-Given two sides and an angle opposite one of them. The angle opposite the other given side is found by Theorem I. subtracting the sum of the other two from 180°. tained by Theorem I.

a

Calling the given sides a and b, and the given angle A, we have sin. B=sin. A.Since an angle and its supplement have the same sine, the result is ambiguous; for the angle B may have either of the two supplementary values indicated by the sine, if b> a, and A is an acute angle.

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CASE 3.-Given two sides and their included angle. Applying Theorem II. (obtaining the sum of the angles opposite the given sides by subtracting the given included angle from 180°), we obtain the difference of the unknown angles. Adding this to their sum we obtain the greater angle, and subtracting it from their sum we get the less. Then Theorem I. will give the remaining side.

Calling the given sides a and b, and the included angle C, we have A+B=180°-C. Then

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In the first equation cot. C may be used in the place of tan.

(A + B).

of the three sides

CASE 4.-Given the three sides. Let s represent half the sum ==(a+b+c). Then any angle, as A, may be obtained from either of the following formulas, founded on Theorem III.:

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The first formula should be used when A < 90°, and the second when A > 90°. The third should not be used when A is nearly 180°; nor the fourth when A is nearly 90°; nor the fifth when A is very small. The third is the most convenient when all the angles are required.

APPENDIX B.

DEMONSTRATIONS OF PROBLEMS, ETC.

MANY of the problems, &c., contained in the preceding pages, require Demonstra tions. These will be given here, and will be designated by the same numbers as those of the Articles to which they refer.

As many of these Demonstrations involve the beautiful Theory of Transversals, &c., which has not yet found its way into our Geometries, a condensed summary of its principal Theorems will first be given.

TRANSVERSALS.

THEOREM I-If a straight line be drawn so as to cut any two sides of a triangle, and the third side prolonged, thus dividing them into six parts (the prolonged side and its prolongation being two of the parts), then will the product of any three of those parts, whose extremities are not contiguous, equal the product of the other three parts.

That is, in Fig. 403, ABC being the triangle, and DF the Transversal, BEXADXCF=EAXDCXBF.

Fig. 403.
A

To prove this, from B draw BG, parallel to CA. From the similar triangles BEG and AED, we have BG: BE:: AD: AE. From the similar triangles BFG and CFD, we have CD: CF :: BG: BF. Multiplying these proportions together, we have BGXCD: BEXCF:: ADXBG: AEX BF. Multiplying extremes and means, and suppressing the common factor BG, we have BEX ADXCF EAXDCXBF.

=

These six parts are sometimes said to be in involution.

If the Transversal passes entirely out

side of the triangle, and cuts the prolongations of all three sides, as in Fig. 404, the theorem still holds good. The same demonstration applies without any change.*

THEOREM II.-Conversely: If three points be taken on two sides of a triangle, and on the third side prolonged, or on the prolongations of the three sides, dividing them into six parts, such that the product of three non-consecutive parts equals the prod

F

B

Fig. 404.

A

2

D

C

uct of the other three parts; then will these three points lie in the same straight line, This Theorem is proved by a Reductio ad absurdum.

* This Theorem may be extended to polygons.

THEOREM III-If from the summits of a triangle, lines be drawn, to a point situated either within or without the triangle, and prolonged to meet the sides of the triangle, or their prolongations, thus dividing them into six parts; then will the product of any three non-consecutive parts be equal to the product of the other three parts. That is, in Fig. 405, or Fig. 406,

AE X BF X CD= EB X FC X DA. For, the triangle ABF being cut by the transversal EC, gives the relation (Theorem I.),

AE X BC X FPEB X FC X PA.

The triangle ACF, being cut by the transversal DB, gives

DC X FB X PA=AD X CB X FP.

Multiplying these equations together,

and suppressing the common factors

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Fig. 406.

A

PA, CB, and FP, we have AE X BF X CD = EB X FC X DA.

THEOREM IV.-Conversely: If three points are situated on the three sides of a triangle, or on their prolongations (either one, or three, of these points being on the sides), so that they divide these lines in such a way that the product of any three non-con secutive parts equals the product of the other three parts, then will lines drawn from these points to the opposite angles meet in the same point.

This Theorem can be demonstrated by a Reductio ad absurdum.

COROLLARIES OF THE PRECEDING THEOREMS.

COR. 1.-The MEDIANS of a triangle (i. e., the lines drawn from its summits to the middles of the opposite sides) meet in the same point.

For, supposing, in Fig. 405, the points D, E, and F to be the middles of the sides, the products of the non-consecutive parts will be equal, i. e., AE X BF X CD= DAXEB×FC; since AE= EB, BF=FC, CD=DA. Then Theorem IV. applies. COR. 2.-The BISSECTRICES of a triangle (i. e., the lines bisecting its angles) meet in the same point.

For, in Fig. 405, supposing the lines AF, BD, CE to be Bissectrices, we have (Legendre IV. 17):

BF: FC:: AB: AC,

CD: DA:: BC: BA, whence
AE: EB:: CA: CB,

BF X AC FC X AB,
CD X BA=DA X BC,
AE X CB

EB X CA.

Multiplying these equations together, and omitting the common factors, we have BF XCD X AE = FC X DA X EB. Then Theorem IV. applies

COR. 3.-The ALTITUDES of a triangle (i. e., the lines arawn from its summits perpendicular to the opposite sides) meet in the same point.

For, in Fig. 405, supposing the lines AF, BD, and CE, to be Altitudes, we have three pairs of similar triangles, BCD and FCA, CAE and DAB, ABF and EBC, by comparing which we obtain relations from which it is easy to deduce BFXCDXAE =EBXFCXDA; and then Theorem IV. again applies.

COR. 4.-If, in Fig. 405, or Fig. 406, the point F be taken in the middle of BC, then will the line ED be parallel to BC.

For, since BFFC, the equation of Theorem III. reduces to AEXCD=EB×DA; whence AE: EB :: AD: DC consequently ED is parallel to BC.

COR. 5.-Conversely: If EI be parallel to BC, then is BF=FC.

For, since AE: EB :: AD: DC, we have AE × DC=EB × AD; whence, in the equation of Theorem III., we must have BF = FC.

COR. 6.-From the preceding Corollary, we derive the following: If two sides of a triangle are divided proportionally, starting from the same summit, as A, and lines are drawn from the extremities of the third side to the points of division, the intersections of the corresponding lines will all lie in the same straight line joining the summit A, and the middle of the base.

COR. 7.-A particular case of the preceding corollary is this:

Fig. 407.
A

B

F

In any trapezoid, the straight line which joins the intersection of the diagonals and the point of meeting of the non-parallel sides produced, passes through the middle of the two parallel bases.

COR. 8.-If the three lines drawn through the corresponding summits of two triangles cut each other in the same point, then the three points in which the corresponding sides, produced if necessary, will meet, are situated in the same straight line.

This corollary may be otherwise enunciated, thus:

If two triangles have their summits situated, two and two, on three lines which meet in the same point, then, &c.

This is proved by obtaining by Theorem I. three equations, which, being multiplied together, and the six common factors cancelled, give an equation to which Theorem II. applies.

Triangles thus situated are called homologic; the common point of meeting of the lines passing through their summits is called the centre of homology; and the line on which the sides meet, the axis of homology.

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