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Now the angle DAE in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to AE, is also perpendicular to AD: therefore AE is the tangent of CDA the half sum; and DF, the tangent of DAB, the half difference of the angles to the same radius AD, by the definition of a tangent. But the tangents AE, DF being parallel, it will be as BE : BD:: AE : DF; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles to the tangent of half their difference.

It is to be observed, that in the third term of the proportion the cotangent of half the given angle may be used instead of the tangent of the half sum of the unknown angles. Example. In the plane triangle ABC (fig. 400.),

Let AB=345 ft.

AC=174.07 ft.
LA= 37° 20'.

Fig. 19.
Now, the side AB being 345


180° 00 The side AC 174.07

Take LA

37 20 Their sum is 519.07

Sum of C and B 142

Their difference 170.93

Half sum of do. 71 20
As the sum of the sides AB, AC=519.07

Log. 2715226
To difference of sides AB, AC=170-93

So tung. hall sum Ls C and B 710 20

To tang. half diff. Ls C and B = 44 16'

These added, give LC =115 36
And subtracted give 2 B = 27 4'

By the former theorem:
As sine LC 115° 36', or 64° 24'

Log. 9-955126
To its opposite side AB 345

So sine L A 37° 20'

9.782796 To its opposite side BC 232

2.365488 1052. THEOREM III. When the three sides of a triangle are given.

Let fall a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles, the proportion will be

As the base or sum of the segnients
Is to the sum of the other two sides,
So is the difference of those sides

To the difference of the segments of the base. Then take half the difference of these segments, and add it to the half sum, or the half base, for the greater segment; and for the lesser segment subtract it.

Thus, in each of the two right-angled triangles there will be known two sides and the angle opposite to one of them, whence, by the first theorem, the other angles will be found.

For the rectangle under the sum and difference of the two sides is equal to the rectangle under the sum and difference of the two segments. Therefore, forming the sides of these rectangles into a proportion, their sums and differences will be found proportional. Example. In the plane triangle ABC (fig. 401.),

Let AB=345 ft.

AC=232 ft.

Letting fall the perpendicular CP,
Base AB: AC + BC:: AC-BC: AP-BP;

Fig. 401.
That is, 345 : 406•07:: 57.93 : 68.18 = AP-DP;
Its half is

The half base is 172.50
The sum of these is 206:59= AP

And their difference 138·41 = BP
Tlien, in the triangle APC right-angled at P,
As the side AC

Log. 2.365488
To sine opp. LP

= 232


10.000000 So is side AP

= 206.59

2.315109 To sine opp. LACP 62° 56'

9.949621 Which subtracted from 90 0 Leaves LA

27 04




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Again, in the triangle BPC, right-angled at P,

As the side BC = 17407 Log. 2.440724
To sine opp. ZP
= 90° C

So is side BP
= 138:41

To sine opp. Z BCP 520 40

Which taken from 90 00

Leaves the L B 37 20
Also the angle ACP

= 62 56
Added to the angle BCP = 52 40

Gives the whole angle ACB 115 36
So that the three angles are as follow, viz. LA 27° 4'; L B 37° 20'; 2C 115° 36.

1053. Theorem IV. If the triangle be right-angled, any unknown part may be found by the following proportion :

As radius
Is to either leg of the triangle,
So is tangent of its adjacent angle
To the other leg ;
And so is secant of the same angle

To the hypothenuse.
For AB being the given leg in the right-angled triangle ABC, from the
centre A with any assumed radius AD describe an arc DE, and draw
DF perpendicular to AB, or parallel to BC. Now, from the definitions,
DF is the tangent and AF the secant of the arc DE, or of the angle A,
which is measured by that arc to the radius AD. Then, because of the
parallels BC, DF, we have AD : AB::DF: BC, and :: AF: AC, which
is the same as the theorem expresses in words.

Note. Radius is equal to the sine of 90°, or the tangent of 45°, and is
expressed by 1 in a table of natural sines, or by 10 in logarithmic sines. Fig. 402.
Example 1. In the right-angled triangle ABC,

Let the leg AB =162

=53° 7' 48"
As radius

: tang. 45°

Log. 10-000000
To leg AB

LA =53° 7' 48'

To leg BC

So secant LA
= 53° 7' 48"

To hypothenuse AC = 270

2.431363 Note. There is another mode for right-angled triangles, which is as follows:

ABC being such a triangle, make a leg AB radius; or, in other words, from the centre A with distance AB describe an arc BF. It is evident that the other leg BC will represent the tangent and the hypothenuse AC the secant of the arc BF or of the angle A.

In like manner, if BC be taken for radius, the other leg AB represents the tangent, and the hypothenuse AC the secant of the arc BG or angle C.

If the hypothenuse be made radius, then each leg will represent the sine of its opposite angle; namely, the leg AB the sine of the arc AE or angle C, and the leg BC the sine of the arc CD or

Fig. 403. angle A.

Then the general rule for all such cases is, that the sides of the triangle bear to each other the same proportion as the parts which they represent. This method is called making every side radius.

1054. If two sides of a right-angled triangle are given to find the third side, that may be found by the property of the squares of the

sides (Geom. Prop. 32. ; viz. That the square of the hypothenuse or longest side is equal to both the squares of the two other sides together). Thus, if the longest side be sought, it is equal to the square root of the sum of the squares of the two shorter sides; and to find one of the shorter sides, subtract one square from the other, and extract the square root of the remainder.

1055. The application of the foregoing theorems in the cases of measuring heights and distances will be obvious. It is, however, to be observed, that where we have to find the length of inaccessible lines, we must employ a line or base which can be measured, and, by means of angles, which will be furnished by the use of instruments, calculate the lengths of the other lines.


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1056. The conic sections, in geometry, are those lines formed by the intersections of a plane with the surface of a cone, and which assume different forms and acquire different properties, according to the several directions of such plane in respect of the axis of the conc. Their species are five in number. 1057. Derinitions.-1. A plane passing through the vertex of a cone meeting the plane

of the base or of the base produced is
called the directing plane. The plane

VRX (fig. 404.) is the directing plane, 2. The iine in which the directing plane

meets the plane of the base or the plane
of the base produced is called the di-

recirix. The line RX is the directrix. 3. If a cone be cut by a plane parallel to the

directing plane, the section is called a
conic section, as AMB or AHI (fig.

4. If the plane of a conic section be cut by RA

another plane at right angles passing
along the axis of the cone, the common
section of the two planes is called the

Fig. 401.

Fig. 405. line of the axis. 5. The point or points in which the line of the axis is cut by the conic surface is or are

called the vertex or vertices of the conic section. Thus the
points A and B (figs. 404. and 405.) are both vertices, as is the

point A or vertex (fig. 406.).
6. If the line of the axis be cut in two points by the conic surface,

or by the surfaces of the two opposite cones, the portion of
the line thus intercepted is called the primary axis. The line
AB (figs. 404, and 405.) and AH (fig. 406.) is called the

primary axis.
7. If a straight line be drawn in a conic section perpendicular to

the line of the axis so as to meet the curve, such straight line

is called an ordinate, as PM in the above figures. 8. The abscissa of an ordinate is that portion of the line of axis contained between the vertex and an ordinate to that line of

Fig. 106, axis. Thus in figs. 404, 405, and 406. the parts AP, BP of the line of axis are

the abscissas AP, BP. 9. If the primary axis be bisected, the bisecting point is called the centre of the conic

section. 10. If the directrix fall without the base of the cone, the section made by the cutting

plane is called an ellipse. Thus, in fig. 404., the section AMB is an ellipse. It is evident that, since the plane of section will cut every straight line drawn from the vertex of the cone to any point in the circumference of the base, every straight line drawn within the figure will be limited by the conic surface. Hence the axis, the

ordinates, and abscissas will be terminated by the curve. 11. If the directrix fall within the base of the cone, the section made by the cutting plane

is called an hyperbola. Hence it is evident, that since the directing plane passes alike through both cones, the plane of section will cut each of them, and therefore two sections will be formed. And as every straight line on the surface of the cone and on the same side of the directing planc cannot meet the cutting plane,

neither figure can be enclosed. 12. If the directrix touch the curve forming the base of the cone, the section made by

the cutting plane is a parabola.


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1058. The primary axis of an ellipsis is called the major axis, as AB (fig. 407.); and a straight line DE drawn through its centro perpendicular to it, and terminated at each extremity by the curve, is called the minor axis.

1059. A straight line VQ drawn through the centre and terminated at each extremity by the curve is called a diameter. Hence the two axes are also diaineters.

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Fig. 103.

1060. The extremities of a diameter which terminate in ihe curve are called the vertices of that diameter. Thus the points V and Q are the vertices of the diameter VQ.

1061. A straight line drawn from any point of a diameter parallel to a tangent at either extremity of the diameter to meet the curves is called an ordinate to the two abscissas. Thus PM, being parallel to a tangent at V, is an ordinate to the two abscissas VP, PQ.

1062. If a diameter be drawn through the centre parallel to a tangent at the extremity of another diameter, these two diameters are called conjugate diameters. Thus VR and RS are conjugate diameters.

1063. A third proportional to any diameter and its conjugate is called the parameter or latus rectum.

1064. The points in the axis where the ordinate is equal to the semi-parameter are called the foci.

1065. THEOREM I. In the ellipsis the squares of the ordinates of an axis are to each other as the rectangles of their abscissas.

Let AVB (fig. 408.) be a plane passing through the axis of the cone, and AEB another section of the cone perpendicular to the plane of the former ; AB the axis of the elliptic section, and PM, HI ordinates perpendicular to it; then it will be

PM2 ; H12:: AP PB : All HB. For through the ordinates PM, HI draw the circular sections KML, MIN parallel to the base of the cone, having KL, MN for their diameters, to wbich PM, HI are ordinates as well as to the axis of the ellipse. Now, in the similar triangles APL, AHN,

AP: PL:: AH : HN, And in BPK, BHM,

BP: PK::BH : HM. Taking the rectangles of the corresponding terms,

AP X BP : PL x PK:: AH x BH : HNX HM.
By the property of the circle,

PL x PK=PMo and HN HM=HI2. Therefore,
APx BP: PM?:: AH ~ HB : HI”, or

PM : HI2:: APR BP : AH ~ HB.
Coroll. 1. If C be the centre of the figure, AP ® PB=CAP-CP2, and AH xHB=

Therefore PM?: HIP::CA?- CP2 : CAP-CH2 For AP=CA-CP, and PB= CA+CP: consequently AP ® PB=(CA-CP)(CA+CP)=CAP-CP?; and in the sainc manner it is evident that AH X HB=(CA+CH)(CA-CH)=CA?- CH2.

Coroll. 2. If the point P coincide with the middle point C of the semi-major axis, PM will become equal to CE, and CP will vanish ; we shall therefore have

PM : HI2::CA2-CP2 : CA2-CH3

Now CEP: HIP::CA? : CAP-CHạ, or CA? x HIP=CE (CA2-CH2). 1066. THEOREM II. In every ellipsis the square of the major aris is to the square of the minor aris as the rectangle of the abscissas is to the square of their ordinate.

Let AB (fig. 409.) be the major axis, DE the minor axis, C the centre, PM andi II ordinates to the axis AB; then will

CA2: CE2:: APX PB: PM2, For since by Theor. I., PM : HIP:: AP PB : AH X HB; and if A the point H be in the centre, then AH and HB become each equal to CA, and HI becomes equal to CE; therefore PM2 : CEP:: APR PB : CA2;

Fig. 400. And, alternately, CA2 : CE?:: AP X PB::PM Coro!), 1. Hence, if we divide the two first terms of the analogy by AC, it will be

::AP PB : PM? But by the definition of parameter, AB : DE::DE: pa. rameter, or CA: CE::2CE : parameter

Therefore is the parameter, which Ict us call P; then

AB: P:: AP PB : PM?. Coroll. 2. Hence CAP: CE :: CA- CP : PM'. For CA'-CP=(CA-CP) (CA + CP)=(AP x PB).

Coroll. 3. Hence, also, AB:P::CA-CP : PM?.

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Fig. 411.

1067. THEOREM III. In every ellipsis, the square of the minor aris is to the square of the major axis as the difference of the squares of half the minor aris and the distance of an ordinate from the centre on the minor aris to the square of that ordinate.

Draw MQ (fig. 410.) parallel to AB, meeting CE in Q; then

Fig. 110.
For by Cor. 2. Theor. II., CAP: CAP-CP2:: CE2 : PM2;

Therefore, by division, CAP: CP::CE2 : CE!- PM.
Therefore, since CQ=PM and CP=QM; CAP: QMP::CE? : CEP-CQ:.

Coroll. 1. If a circle be described on each axis as a diameter, one being inscribed within the ellipse, and the other circumscribed about it, then an ordinate in the circle will be to the corresponding ordinate in the ellipsis as the axis belonging to this ordinate is to the axis belonging to the other; that is,

CA: CE::PG : PN,
and CE: CA::pg : pM;
and since CAP: CEP:: AP PB : PMS,
and because APR PB=PG%; CAP: CE2::PG2: PM,

or CA: CE::PG : PM. In the same manner it may be shown that CE: CA::pg : pm, or, alternately, CA: CE::PM: pg ; therefore, by equality, PG: PM::pM : pg, or PG: Cp::CP: ng: therefore CgG is a continued straight line.

Coroll. 2. Hence, also, as the ellipsis and circle are made up of the same number of corresponding ordinates, which are all in the same proportion as the two axes, it follows that the area of the whole circle and of the ellipsis, as also of any like parts of them, are in the same ratio, or as the square of the diameter to the rectangle of the two axes; that is, the area of the two circles and of the ellipsis are as the square of each axis and the rectangle of the two; and therefore the ellipsis is a mean proportional between the two circles.

Coroll. 3. Draw MQ parallel to GC, meeting ED in Q; then will QM=CG =CA; and let R be the point where QM cuts AB; then, because QMGC is a parallelogram, QM is cqual to CG=CE; and therefore, since QM is equal to CA, half the major axis and RM=CE, half the minor axis QR is the difference of the two semi-axes, and hence we have a method of describing the ellipsis. This is the principle of the trammel, so well known among werkmen.

If we conceive it to move in the line DE, and the point R in the line AB, while the point M is carried from A, towards E, B, D, until it return to A, the point M will in its progress describe the curve of an ellipsis.

1068. THEOREM IV. The square of the distance of the foci from the centre of an ellipsis is equal to the difference of the square of the semi-axes.

Let AB (fig. 412.) be the major axis, C the centre, F the focus, and FG the semi-para. meter; then will CE2=CAP-CF2. For draw CE perpendicular to AB, and join FE. By Cor. 2. Th. II., CA; CEP::CA2CFP : FG%, and the parameter FG is a third proportional to CA, CE; therefore CA2 : CEP::CE” : FG%, and as in the two ana. Jogies the first, second, and fourth terms are identical, the third terms are equal ; consequently CEO=CA2_CF2.

Fig. 112. Coroll, 1. Hence CF9=CA- CE2. Coroll. 2. The two semi-axes and the distance of the focus from the centre are the sides of a right-angled triangle CFE, and the distance FE from the focus to the extremity of the minor axis is equal to CA or CB, or to half the major axis.

Coroil. 3. The minor axis CE is a mean proportional between the two segments of the axis on each side of the focus For CE2=CA2-CF!=(CA+ CF) (CA-CF).

1069. THEOREM V. In an ellipsis, the sum of the lines drawn from the foci to any point in the curve is equal to the major aris.

Let the points F, f (fig. 413.) be the two foci, and M a point in the curve; join FM and f M, then will AB==2CA=FM +FM.

By Cor. 2. Th. II., CA2 : CEP::CA-CP: PA,
But by Th. IV., CEP=CA2_CF2;

Fig. 413.
And by taking the rectangle of the extremes and means, and dividing the equation by
CA?, the result is -


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