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Mr. Sheridan, in answer, said fome- House, Mr. Fox, who sat next to Mr. thing which indicated that proceedings Burke, pulled him down by his fleeve. at law were carrying on against them. Sir Edward Astley spoke to the fame

Mr. W. Pitt objected to an irregular purpose as Mr. Martin, when conversation upon a subject not pro Mr. Fox rose, and said, that if the perly before the House, and more efpe honourable baronet and honourable gencially in the absence of certain persons, tleman who fpoke before him confwho had it in their power to give the dered a moment, they surely would House authentic information with re- think, that calling what his honourable gard to the proceedings respecting Mr. friend had done a daring infult to the Powell and Mr. Bembridge. He de public, was an expression not more harsh fended the conduct of the last admini- than unjustiñable. With regard to the ftration, in ordering the discharge of honourable gentleman (Mr. Martin) he the cashier and accountant from their had on some occasion touched on huoffices, and owned he thought their re- manity; surely, if that honourable genftitution to their places suddenly, and tleman would reflect ever so little, he without any explanation respecting their would see that it was the fixed princiconduct, an extraordinary proceeding. ple of human justice, to prefume every

Mr. Sheridan rose again, and said, person innocent till some criminality the delay of the legal proceedings againit was proved againit him. Mr. Powell, Meff. Powell and Bembridge was by he said, had ever the character of a no means chargeable as matter of cen man of the stricteft honour and intesure against the present ministry. The grity, and he saw no reason, therefore, Attorney-General had commenced the for accepting the accusation, eren (if business, and there had been time accusation there was) against such a enough elapsed for him to have gone character for proof, nor of condemning on with it.

him unheard, any more than of conMr. Kenyon upon this rose, and ex- demning any other person, accused of plained his conduct respecting Mesirs. any other offenc:, before he had been Powell and Bembridge. He said he tried. With regard to the degree of could only judge from the case that had responfibility beionging to the offices been laid before him; as far as that of cashier and accountant, he declared went, it appeared to him that they were he was wholly ignorant; his honouraenormous offenderse

ble friend was responsible to the pubMr. Burke took fire at the words lic, not only for his own conduct, but enormous offenders, and argued upon the for that of every clerk under him, and, injustice of applying a term to men un- therefore, it was not to be presuned, convicted of any offence whatever. that his honourable friend wouid have Mr. Burke said, he held himself re restored two persons to their offices unsponsible for those he employed in of- der him, of whose unimpeachable confice, and appealed to the tenour of his duet he was not in his own mind perlife, in proof of the improbability of fectly convinced. But that his having his countenancing men guilty of enor- done fo was a daring insult to the pulmous offences.

lic, was furely not only a very harsh afMr. Martin said he knew nothing of sertion, but an assertion by no means the gentlemen who had been discharged, true. It was posible for the late payand restored to their places in the Pay, master to have seen the same conduct Office; but as they had been restored in a reprehensible point of view, and without one syllable being faid of the for his honourable friend to have seen matter, he could not but consider the it in a different point of view. His fact as a daring insult to the public -- honourable friend could never have

Mr. Burke rose in great heat, and been so weak as to fuppose, that the cried out “ I say it is not a daring in- act of restoring the cathier and accompsult to the public”-when the noise tant would pais unnoticed, or that it becoming general, and the cry of bear, would not call forth observations, and bear, coming from fome parts of the proroke enquiry. Undoubtedly it was

ob:iuus.

obvious, that notice would be taken to his general character, contrary to of it in that House, and he had no what he believed, Mr. Powell should manner of doubt but that his honour not turn out a man of honour and how able friend would be able to shew that nefty, the House must fee, that in so he had not, by any imprudent and hafty large, so complicated a transaction as measure, done a thing so culpable as the executorfhip of his father's affairs, fome gentlemen chose to suppose it. there muft have been great opportunity Mr. Fox declared, that, for his own for wronging his family, and though part, he knew nothing of the two gen 40,00ol. or 60,oool. was a trifle with tlemen being restored, till his honour- regard to the public, yet when it came able friend told him of it, as he was to be the case of an individual, the conentering the closet at St. James's. fideration was a very large one. With regard to an enquiry, it was a After this, a desultory conversation matter which concerned him more took place, which the Speaker ended, nearly than any other person whatever. by reminding the House that it was Mr. Powell was the acting executor of disorderly, and that no motion was his father, and said, that if contrary before them on that business.

ri

MATHEMATICS. ANSWERS TO MATHEMATICAL QUESTIONS.

Ons. j. QUESTION (I. July) answered by the Proposer. A

SSUME x2 + 1 X *3+ mx+r, = x5 + m +1.x3 + rx2 + innx +'nr = 35+ 12 + sr?

*3+rx2+x+1, the given expression. Then, by comparing the ho. mologous terms, we shall have t=nr, or n=-, and mn=s, orm=:-,

-=-; and, 12 + Spo2 confequently, 1+= as it ought to be. Hence the given equat.becomes x+ Xx3+*+r=o. Consequently two of the values of x are vand the other three may be had from the resolution of the cubic equation x3 +

*+r=.

72

rt

and V"

1

This question was also answered by Mr. James Webb.

QUESTION (II. July) answered by NAUTICUS. Let ZPH represent an arch of the

Z meridian, where Z is the zenith, P the pole, and H the point of the horizon

P Р which is of the same name with the la. titude, Moreuver, let H A be the horizon, S the place of the given ftar, and

that of the required one; and Z * SA the vertical circle they are on when they pais each other in azimuth. Then, by the queftion, the difference of the flux

HI ions of the angles SZP and * ZP must be a maximum. Now, by Theo. 21, of Coies's traci De Estimatio Errorum, &c. in the spherica} triangle SZP, P : :: R X lin. 29 : fin. PS X cos. S; and by Sutiktimating in this proportion for fin. Is, lin. 'z, and cei. S their equals, ducd irom the runciples of spherical

trigonometry,

R

trigonometry, we shall have P: 2 :: R3: cos, PZ XR2-fin. PZxcof. Z x cotan. sz. In like manner we may derive P: 2 :: R3 : cof. PZxR2-in. PZx cof. 2x cotang. * 2. Hence, if radius be taken equal to unity, the fluxion of the angle PZS will be equal to P x cof. ZP-o x lin. ZP x col. Z x cotan. SZ; and the fluxion of PZ * will be equal to Pxcol. ZP-Pxfin. ZP Xcof. Zxcotan. *Z; and, consequently, their difference, or P x fin. ZP x cof. 2xcot. ZS-Px fin. ZP x cof. Z x cotan. Z * must be a maximum; or, because 2 x sin. PZ is constant, cof, ZX tur.Zs-cot. Z * will be a maximum. Now, as tlie coline of 7 can never exceed unity, and as the difference of the cotangents of ZS and Z * : that is, the difference of the tangents of A and AS will be infinite when the required staris in the zenith; it is manifelt that the cor. Zxcot. Zs - cor. 7 * will be a maximum when the declination of the required star is equal to the latitude of the place.

3. Question (III. July) answered by Mr. JAMES WEBB. Let CE be the given hyperbola, IQ the opposite hyperbola, and I the given point in it. From I, as a center, conceive a circle CEF to be described, which touches the hyperbola CE in the point C; this

;P point, it is evident, is that which is fouglit, and may be determined as follows : Let OP be the femi-conjugate axe of the iwohy

DIT perboias, =c, the semi-transverse EO=OR being =1: draw IC, and produce it to meet the axe, produced, in G, and draw IR, and CD perpendicular to EQ produced. Let OD be put = x, IR = a, and OR=b; then by the properties of the hyperbola, 12 :02::*:0*x = DG, 12:02: 1+ * XX-15 c? XXL-I=DC?; and by similar triangles c?x : 02*+*+6:30 V7?–1:x. Cona

6 fequently, multiplying means and extremes, and reducing the equation

It82
-XX

; and if x be now considered as the secant of an arc, the radius V x4 of which is unity, = will be the cosecant; and the excess of the corecant,

VX-1 above the secant drawn into a given quantity, is known. Hence x is readily found by the irethod poinied out at p. 470. Philofoph. Transac. for 1781.

Cor. Had the lenji-transverse EO, initead of being equal unity, been taken equal to any given quantity t; the final equation would then have beer

VX-2 1-72 12b

; and the only difference that would have arisen in finding the value of x, by the method pointed out above, would have been in taking xas the fecant of an arc, the radius of which is the given quantity t. Answered alto by Mr. James Eastwood, and the proposer Mr. W. Kay.

4. QUESTION (IV. July) answered by NauticOS, CONSTRUCTION.

B In the given angle H conftirute the parallogram HGAI, equal to twice the given area; and in HI, produced, take IK equal to the given difference of the inclu

C ding lides, allo KE 1o, that KEIE may be equal IAXAG. Through E and A draw EF meeting HG, produced, in F: then completing the parallograms, as in the figure, and joining BC; BAC will be the triangle required.

H

I

к Lond. Mac. 03. 1783.

PCMON

AC

ac

ас

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DEMONSTRATION., The angle BAC = the opposite vertical angle GAI (Euc. I. 15) = the given angle GHI (Luc. I. 34); and the triangle BAC = half the parallelogram ABCD (Euc. I. 34) = half the parallelogram HIAG (Euc. I. 43) the

given area, by construction. Lastly, fince IF:AG:: AI: KE, by contruction, and GF:GA: AI: IE hy fim. tuangles, ex æquo perturbato, IE:GF::JE: KE. Now, the aniecedents being here the fame, the consequents must be equal: that is GF (=AB)= KE; and AC (=IE) exceeds AB (=KE) by IK, the given difference. Q.E.D.

Mr. Geo. Sanderson and the proposer also gave elegant constructions to this question.

5. QUESTION (V. July) answered by Mr. Isaac DALBY:

ANALYSIS.

Suppose the thing done, and let ABCD be the filli. pond, and EFGHDCBAD the walk of equal breadth Jurrounding it; through A, D, C, B, draw Em, Hm, Gn, Fr, which, it is evident, will bitect the angles of all the trapeziums EG, AC, &c. whose fides are parallel, and whose angles fall in these lines, and confequently equally diftant all round. Make the trapeziun IKLNI so that its fides may be || to the tides of the given trapez. and produce EH, AD, and IM; take Hg=HG, Dc= DC, and MI=ML, draw gcl, which must be a right line becaule HDM and GCL are right lines, and (letting fall the perp.ch, Dt. Dz) Dw=D==ch; hence it tol. lows from Euc. I. 36, 37, 38, that the quadrilaterals DHgc = DHGC, McD = MLCD, therefore HMlg = HMLG; and, confequently, if gl was produced to meet Hm, the A formed thereby on the bale Hg would be equal in area to the A HZG.

In like manner, it there be taken gf=GF, cb=CB, Ik=LK, and to-FE, baBA, ki-K1, the quadrilateral lgtk=LGFk, and kfei=KFEI, and therefore the locus of the points E, A, I, &c. when transferred to the points e, a, i, &c. is a right line, and the quadrilateral. E Aae = the area of the walk by bypotb.) hence this easy

G

CONSTRUCTION.

R

Z

K

Take IKLMI || and equidiftant from ABCDA, and draw AI, DM, produced both ways at pleasure; produce AD, IM, and take Da=DC+EB+BA, alto Mi = ML + K + KI, that is Aa, lis the perimeters; FB through a, i draw ar to meet Al producer, then is the A ak A given; make the rectang. AO=ARA, and the reciang. aN=AaRAtarea of the walk, take aP a mean proportional between aA and as, erect the perp.

P PE to meet I A produced, and PE is the breadth of the walk required; and drawing Ee || Aa, the quadranga Ae is its area: For rectang. AO (APRA): rectang, AN :: aA :aS :: A ARa : A ERE (Euc. VI. coroł. 19.) therefore rectang. aN =AER=ARA +area of the walk (by construc.) therefore E Aae = area of the

S walk,

COROL. I. This method of solution Lolds good in any polygon, regular or irregular, or consisting of any number of fites.

COROL. II, When the given figure is a regular folygon, the points m, 2, n, R, fell in its center,

The

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The same answered by the Rev. Mr. HeLLINS, Teacher of the Mathematics and

Natural Philosophy.

ANALYSIS,
Let abcă represent the fill-pond, and ABCD
the outside bounds of the walk. Then, since the

D
walk is of the same breadth on every fide, it is
evident that if any two corresponding angular
points be joined, the line which joins them will
bileet those angles. Thus the line Aa bisects
the angle DAB, and the line Bb bilects the angle
ABC, &c. If now Aa and Bb be produced, till
they meet in e there will be given the triangle
aeb. In like manner the three other triangles
bfc, cgd, and dha become known. The pro-
biem, then, is reduced to this:

To four given triangles to add as many spaces,
by producing the sides of those triangles until i

B В they meet four right lines drawn parallel to their bases, which four spaces, taken together, thall be equal to a given space: to facilitate the construction of which I all premise the following

LE M M A.
If through two triangles of equal bases, and between the same parallels, a line
be drawn parallel to their bases, it will cut off equal spaces from thote triangles,
This is sufficiently evident from Euclid I. 38. & Ant.

COROLLARY.
If the fides of two triangles, having equal bases and altitudes, he produced to two
fines, drawn parallel to, and at equal distances below their bales, the spaces added
will be equal to each other.
CONSTRUCTION

PROBLEM.
Let the two indefinite lines
Is, in form a righe angle at l;
in Is take Imrabibe base of the
triangle aeb in fig. 1, and draw
mn so that the altitude of the
triangle inm may be equal to the
altitude of the triangle aeb. In
like manner, in the fame right
line, take mo, oq, qs, respectively
equal to the bales of the other
three triangles, making the al-
titudes of the triangles mpo, ora,

7

9
qts, forored on them, equal to
the altitudes of their correspond-

y

i ing ones in the given trapezium. Produce st until it cut in in v. Produce si, and make the triangle lxu, by Prob. IV. p. 218 of Simp. Geom. similar to the triangle lus, and equal to the given area of the walk. Join xv, and in vi, produced, take vy=vx, and ly will be the breadth of the walk.

DEMONSTRATION Produce vs till it cuts a line drawn through y, parallel to ls in %. Then (Euc.? VI. 19) the areas of the similar triangles yvz, lys, and lxu are as the Squares of the fides yo (=xv) lv, and lx ; and since yo? ( =*v*)=lv +x? ; it is evident the area of the triangle juz is equal to the fum of the areas of the triangles lu's and Ixu; and, confequenidy, that the quadrilaterai ylsz is equal to the triangle lxų. Produce now nm, po, and rq to meet the line uz in the points qu, j, k, respectively; then, by the foregoing Lemma, the quadrilateral ylm-w is equal to its corresponding Space AabB; and to ale the others, mwio, oikg, akzs to tbeir corresponding ones in the first figure, and their tum, or the quadrilateral yisk, is therefore equal to the area of the walk,

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SCHOLI

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Q. E. D.

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