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LESSONS IN GEOMETRY.-No. XXX. ACP was shown to be equal to the angle B AC. Therefore the

whole exterior angle acd is equal (Åx. 2) to the two interior LECTURES ON EUCLID.

and opposite angles c A B and A B C. To each of these equals, (Continued from page 44.)

add the angle A CB; the two angles A CD and ACB are there BOOK I.

fore equal (A.r. 2) to the three angles CAB, A B Q and A CE. PROPOSITION XXXII. THEOREM.

But the two angles acd and ACB

are equal (1. 13) to two

right angles. Therefore also the three angles C'A B, A B C and
If a side of any triangle be produced, the exterior angle is equal to Ac B are equal (Ax. 1) to two right angles.

Wherefore, if a
the two interior and opposite angles ; and the three interior side of any triangle be produced, etc. Q. E. D.
angles of every triangle are together equal to two right angles,

Corollary 1.- All the interior angles of any rectilineal figure,
Let a Bo be a triangle, fig. 32, No. 1, and let one of its sides together with four right angles, are equal to twice as many
BC be produced to . Then the exterior angle a cd is equal right angles as the figure has sides.
to the two interior and opposite

Fig. 32. No. 1, Let ABCD E, fig. 32, No. 2, be any rectilineal figure. All
angles CA B and ABC. And the three

the interior angles ABC, B C D, etc., together with four right interior angles ABC, DCA, and CAB

angles, are equal to twice as many right

Fig. 32, No. 2. are equal to iwo right angles.

angles as the figure has sides. Through the point c draw the

Divide the rectilineal figure ABCDE into straight line,(1.317 CE, parallel to the

as many triangles as the figure has sides, side BA.

by drawing straight lines from a point
Because ce is parallel to B A, and

F, within the figure to each of its angles.
ac meets them, the angle ACE is equal (I. 29) to the
ternate angle BAC.
Again, tecause ce is parallel to AB triangle are equal (1.32) to two right

Because the three interior angles of a
d BD falls upon them, the exterior angle BCD is equal angles, and there are as many triangles
29) to the interior and opposite angle a b c. But the angle | in the figure as it has sides, all the angles

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of these triangles are equal to twice as many right angles as the figure has sides. But all the angles of these triangles are

EXERCISE II. TO PROPOSITION XXXII. equal to ihe interior angles of the figure, viz. A B C, B C D, CIC, To triscct a giren finite straight line; that is, to divide it into three together with the angles at the point F, which are equal (1 15.

equal parts. Cor. 2) to four right angles. Therefore all the angles of these

Let A B, fig. 9, be the given straight line. It is required to triangles are equal (At. 1) to the interior angles of the figure together with four right angles. But it has been proved that trisect it ; that is, to divide it into three equal parts. all the angles of these triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. Let fig. 32, No.3, represent any rectilineal figure.

Because the interior angle A B C, Fig. 32. No. 3. and its adjacent exterior angle A B D are (I. 13) together equal to two right angles. Therefore all the interior angles together with all the exterior angles of the figure, are equal to twice as many D B right angles as the figure has sides. But it has been proved by the

Through the point A draw the straight line a c, making any foregoing corollary, that all the

angle with AR. Produce a c indefinitely to D. From od cut interior angles together with four right angles are equal to off c E and e f each, of them equal to Ac. Join FB. Through twice as many right angles as the figure has sides. Therefore all the points c and E, draw cg and eh parallel to P B and meetthe interior angles together with all the exterior angles are equal ing A B in G and 1 (Prop 31). Then A B is divided into three (45. 1) to all the interior angles and four right angles. Take equal parts in the points o and H. Through the points c and. from these equals all the interior angles; therefore all the , draw ci and e k, parallel (Prop. 31) to A B, and meeting PB exterior angles of the figure are equal (Ax. 3) to four right in the points 1 and K; and let co meet eu in L. Join G l. angles.

Because AG, CL and EK are parallel (Const.), the angles FEK, Corollary 1.-If two angles of a triangle be given, the third E CL, and caG are equal (Prop. 29). And because co, EL, is given; for it is the difference between their sum and two

and FK are parallel (Const.), and a B falls upon them, the angles right angles.

ACG, CE L and Erk are equal (Prop. 29). And the sides A c,
Corollary 2.-If two angles of one triangle be equal to two

CE and e f of the triangles ACG, ECL and EFK are equal engies of another triangle, the third angle of the one is equal (Const.); therefore, (Prop. 26) the sides A G,CL and Exare to the third angle of the other.

equal. But because cL is parallel to gy, and G L meets them, Corollary 3.-Every angle of an equilateral triangle is equal the angles CL Q and LG H are equal (Prop. 29); and because ca to one-third of two right angles, cr two-thirds of a right angle. is parallel to Ly, and al meets them, the angles ca 1 and Hence, a right angle can be trisected.

IL G are equal (Prop. 29); aud G L is common to the two triCorollary 4.-If one angle of a triangle be a right angle, the angles GCL and chl; therefore c is equal (Prop. 26) to Q H. sum of the other two is a right angle.

But c L was shown to be equal to Ag. Therefore Gh is equal Corollary 5.-If one angle of a triangle be equal to the sum (Ax. 1) to a g. In like manner, by joining L K and u 1, it may of the other two, it is a right angle.

be shown that Ek is equal to Li, and that Li is equal to y B. Corollary 6.--If one angle of a triangle be greater than the But e K was shown to be equal to a 0. Therefore in is equal sum of the other two, it is obtuse; and if less, acute.

But &x is also equal to Ag. Therefore, the three Corollary 7.-In every isosceles right-angled triangle, each straight lines A G, GH, and u are equal to one another. of the acute angles is equal to half a right angle.

Wherefore the straight line A B has been crisected in the points Corolary 8. --All the interior angles of every quadrilateral and 1.

Q. E. F.
figure are together equal to four right angles. This is only a Note.-By this problem a straight line may be divided into
particular case of Euclid's Cor. 1: but it is very necessary to any number of equal parts.
be remembered,

EXERCISE III. TO PROPOSITION XXXII. *
EXERCISE I. TO PROPOSITION XXXII.

Any angle of a triangle is right, acute, or obtuse, according as the

straight line drawn from its tertex bisecting the opposite side is If a straight line be drawn from one of the angles of a triangle, equal to, greater than, or less than half that side. making the exterior angle equal to the tuo interior and opposite

In fig. h, let A B be a triangle, and let A d be drawn from a, angles, it is in the same straiyht line with the adjacent side.

bisecting ecin D. Let op, fig. f, be a straight line drawn from the angle BCA

No. 2. of the triangle A B C, making the exterior angle a co equal to the two interior angles C B A and BAC; then cd is in the same straight line

Fig.f.

А Because the angle acd is equal (Hyp.) to the two angles CBA and HAC, add to each of these equals the angle BCA. Then the two angles ACD and BCA, are equal to the

B three angles BCA, CAB and ABC. But these three angles (Prop. 32) are equal to two right angles. Therefore the angles A c B and ACD are also equal to two right has the merit of generalising it, so as to make it applicable to ediniaiba angles. Therefore the straight line o p (Prop. 14) is in the same

of a straight line into any number of equal parts; a problem which Dr. straight line with Bc. Q. E. D.

Thomson postpones till he comes to Prop. xxxiv.
VARIN (East Dereham); E. J. BREMNER (Carlisle); J. II. EASTWOOD

(Middleton); and others.
* This exercise was solved by Q. PRINGLE (Glasgow), who added other

as abore by Q. (Glasgow); J. to kelis I. Bocock (Great Waliotes, Buenosotros para pc, Lab Phong thu); t. J. Bremser (Carlisle); D. H. Dridield): "T! it. EASTWOOD

moderni solution. equally correct; it was also solled by . u. (Writt BOCOCK (Great Warley); C. L. HADPIPLD (Molton-le Motorr); 11.SO. PUCH FIELD (Builon-le doors); J. u. EASTWood (Middleton); and others.

Middleton); and others.

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to four night an.3.68. * * IV, TO PROPO4ITION XXX11. Me shwallow down from the verter of any angle of a triangle A1 Ryhmmy the woeste side, to equal to, greater than, or be than Asof 14.1 mule, irronding as the anyle ta righe, acute, or obtuse.

其 tut, A, let a no bo triangle, and let ad be drawn from a Nurol in II,

1 live the angle # A No, I be a right angle. Then AD In rywall it, if not, thon A D in either greater or lors

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ANSWERS TO CORRESPONDENTS. ** per te mbi awansular pentagon (or fire.

R. 8. T. (Leeds): Pronounce Latin In general like English. Long a ls when a bus wale të A* miest, the angles formed at the pronounced as in fate. Final is in the dative and ablative plural Is pro. me-spey are put &**peei la ww right angles.

nounced as the English word ice; but many adopt the shori sound, as in

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D. D. N. (slough!: Pronounce o, . ro, as hse, he, toe, wv 88 seen, and

It is unnecessary to learn the English-Greek exercises. The
exercises are well done-particularly the English-Greek. Instead of "The
ding he is a general," you should hare said " The king himself," etc.

1. T. 8, H. (Oxford): We fear that his scheme for an air pamp, however inzentous, would not answer: we strongly doubt the applicability of glows pasions.-OMEGA: We must try and insert the biography he wants in the P.E.-J. NO BAIN Fester-lane): The Cotton Account is not debited for the amount of cash received, but for the amount due by the merchant for the axoa purchased on credit or oubet wise ; and the same account is credited by the amount du for the catsoa sid. The reason why the amounta due arv eutered, less the diseauna, is because the lalter are small and cusvaart, and not worth opera separate account for; besides, the same, oss the diseunts, are the assive which the cotton was purchased sodul and are, it** 2, the mo per sems to be entered in the Ledger

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diminished in volume, and the index at B was forced towards

the bulb at a. ON PHYSICS, OR NATURAL PHILOSOPHY.

The point where it became stationary was

marked, and this determined the volume of the air at 00 CenNo. XXXV.

tigrade, since the capacity of the bulb was known. The ice

was then removed and replaced by water or oil, and the box was (Continued from page 110.)

heated over a furnace. The air in the bulb expanded, and the

index advanced from A towards B. The point where it became EXPANSION AND DENSITY OF THE GASES. stationary was then marked, and at the same time the tempe

rature indicated by the two thermoineters D and e was noted, Laws of Expension.-M. Gay-Lussac first proposed the two following laws on the expansion of gas, which have been, till

so that the volume of air and its temperature were both ascer

tained. very recently, considered as correct, and universally received. Ist, That all gases have the same co-efficient of expansion,

Now if in the first instance we suppose that the atmospheric that is, the same increase in volume in passing from *0° to 10 pressure did not vary during the experiment, and neglect the

expansion of the glass, which is very small, the total expansion Centigrade, and that its numerical value is 0.00375 or

1

of the air in the apparatus will be found by subtracting the 2663' volume which it had at 0° Centigrade from the volume which

it had at the end of the experiment. Then, by dividing the 2nd. That this co-efficient is independent of the pressure, the expansion corresponding to 10 Centigrade will be deter:

remainder by the number of degrees in the final temperature, that is, of the original density of the gas. These laws are founded on the supposition that all gases units contained in the volume at 0° Centigrade, we obtain the

mined; and again dividing this quotient by the number of expand equally under an equal increase of temperature, and expansion corresponding to one degree and one unit of volume, that the expansion is in direct proportiin to the temperature that is, the co-efficient of expansion. In the following probThe method adopted by Gay-Lussac in the determination of these laws was the following the expansion of air

and other lems, the corrections for the variations in the pressure of the gases was measured by an air-thermometer formed of a

atmosphere and the expansion of the glass are taken into

account. spherical bulb a, and a capillary stem A B, fig. 183, The part of the stem between B and c was divided into

Problems on the Expansion of the Gases.-1st. Given v the parts of equal capacity; and the number of these parts which volume of a gas at 09 Centigrade, and the co-efficient of expan. the bulb A contained, was ascertained by weighing the appa- sion a, what will be its volume v' at the temperature to Centi. ratus when full of mercury at 0. Ceritigrade, and then heating grade, the pressure remaining constant. Here, the same reait gently in order to force out of it a little of the mercury soning being employed as in the case of linear expansion By weighing it again; the weight of the mercury forced given in a former lesson, we shall, at once, have v' =vtavt; nut was determined. By cooling the remainder down whence v'=v(1 + at) (1.) to O® Centigrade, a vacuüm was produced which showed 2nd. Given v' the volume of a gas at the temperature to, and the volume of the corresponding weight of the mercury the co-efficient of expansion a, what will be its volume v at de which was forced out. From this, the volume of the mera I Centigrade, the pressure remaining.constant! Here, from

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stant.

cury remaining in the apparatus was deduced, and conse: formula (1.), we have by division, v=
quently
, the volume of the bulb, by proportion, as already

(2). shown in the case of the piesometer, p. 105, vol. iv. The bulb

1+ at and the stem were now filled with dry air, in the following Centigrade and the co-efficient of expansion a, to find its

3rd. Given the volume v of a gas at the temperaturen in life , bulo in order to dry it ; a tube er

, álved with drying volume v" at the temperature t", the pressure remaining cons substances, such as chloride of calcium, was then fixed to the

Here, by forinula (1.) we have v" =v(1+at"); extremity of the stem by means of a cork. Into the stem AB, 1 and by formula (2.) v=

v' through the tube, was next introduced a fine platinum wire,

;; whence, by substitution for

1+at de mes at the same time held in an inclined position, so as to v, in the former expression, we have v"=V'(1 + ati

(3.) permit the mercury to flow out drop by drop, by slightly

1+ at

4th. Given the volume v of a gas at to Centigrade, and at bubble by bubble, after having been dried by passing through the

pressure a inches of the barometer, to find its volume v be chloride of calcium. Lastly, there was preserved in the at 00 Centigrade and at the barometric pressure of 7 inches, Stem a B a small portion of mercury to serve as an index, as

the co-efficient of expansion being a. Here, we have, by formade of tin. This box was at first filled with ice, the air was alde air-thermometer was then placed in a rectangular box mula (2.) the volume at 0° Centigrade equal to

1-at 113

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17 di

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pressure u. Now by the law of Mariotte, we have vh= dried; for the humidity which adhered to its sides is dissipated Vu

in steam, and the air which filled it each time that a vacuum ; whence, v=

(4.) 1+at'

was made, was dried in its passage through the U-shaped (1 tat)

tubes. When this is done, a space of time, say about half-an-
5th. Given the volume of a glass bulb at to Centigrade to hour, is allowed for the air to assume the temperature of the
find its volume v at 0° Centigrade. In solving this problem, steam; the dessiccating tubes are then removed, and the
we suppose that the bulb will expand for a given variation of extremity of the capillary tube is hermetically sealed, the
temperature, by the saine quantity that it would expand were height of the barometer at the same instant being noted. The
it a mass of glass of the same volume and at the same tempera- reservoir e is then cooled, and placed in the apparatus shown
ture. If we then represent the co-efficient of the cubic expan. at fig. 185,
sion of glass by d, we have according to formula (1.) v =vt

Fig. 183.
V
dvt=v(1 + dt); whence v=

6th. Given the weight r' of a certain volume of gas at to
Centigrade, what will be its weight p at 00 Centigrade, the
co-efficient of expansion being a ? Here, lei d' be the density

B of the gas at to, and d its density at 09 Centigrade. The

P' d' weights being proportional to the densities, we have

à. Now representing by 1 a certain volume of a given gas at 09 Centigrade, its volume at 1° will be 1 tat; but the densities are inversely proportional to the volumes, therefore we have al 1

; whence, and from the former expression, we
d ta

yuv
1
have

; and r =P'(1 + at).
itat

2C In the determination of the co-efficient of the expansion of gases, M. Regnault employed in succession, four different processes. In one set the pressure was constant, and the volume of the gas variable, as in the process of Gay-Lussac; in the other set, the volume re: ained the same, but the pressure was varied at pleasure. The following is the first process employed by M. Regnault, and in it the pressure remained constant; the same process was employed by M. Dulong and by M. It is then completely surrounded with ice, so that the air Rudberg. The experiments of M. Regnault are characterised by which it contains may be reduced to the temperature of 0° The greatest care to avoid the possibility of error in the results. Centigrade; and the extremity of the capillary tube is immer. His apparatus was composed of a cylindric bulb or reservoir B, fig. sed in a cup filled with mercury. When the reservoir B-is at 184, of a pretty large capacity, to which is cemented a bent capil- 0° Centigrade, the end 6 is broken off with a pair of small

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lary tube. In order to fill this bulb with perfectly dry air, it was | pincers, and the interior air being condensed, the mercury in arranged, as shown in the figure, in a tin vessel similar to that the cup enters the capillary tube in consequence of the presa ter; then by means

of sheets of caoutchouc, the capillary tube of which, added to the elastic force of the air remaining in the is connected in a series of U-shaped india: rubber tubes, filled apparatus, makes an equilibrium with the atmospheric pres: with dessiccating substances. These

tubes terminate in a sure. In order to measure the height of the column co, which tubes and in the reservoir

, whilst the latter is surrounded with the point o is level with the surface of mercury in the cup; and steam. The air is then allowed to enter slowly, and a new the difference of the height between the point g and the level vacuum. is made ; and this process is repeated a great number of the mercury at a is then measured with the cathetometer

. of times ; so that at last the air in the reservoir is completely Adding to this difference the length of the part go, which is

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