NEW QUERIES, REBUSSES, &c. to be ANSWERED in the next Year's Diary.

W

Quare I. by Mr. G. Clarke Butler.

Hether is Friendship or Hatred the most durable?

Quare II. by Mr. J. Johnson.

Whether is the Drunkard, the Whoremonger, or the Gamefter, the mo pernicious to Society?

Quære III. by Mr. J. Randles.

Whether do Fools bring moft Profit to Wife-men, or Wife-men to Fools?

Quare IV. by T. B.

Whether, Mr. Pope, in L. 34. of Book iv. of his Dunciad,

("And running round the Circle finds its square.")

appears to have had a right Idea of what Geometers mean, by fquaring the Circle ?

A Rebus by Mr. Berington.

Of a Colour two Thirds, and what Roger gives Kitty,
Will point out a Thing that may puzzle the Witty.

Another by Mr. F. Sandling of Hindringham, in Norfolk.
To the Name of an Ox, in one Language defin'd,
Let his Price, in another, be properly join'd;

By which you'll discover the Name of a Town,
Where a Monarch in Fight loft his Life with his Crown.

A Paradox by Tempus Geographicus (of Ashford in Kent.)

I was born in a principal maritime City in England, on the 6th of Feb. 30 Minutes after eleven o'clock at Night, notwithstanding which, it fo happens that the Anniversary of my Birth is on the 18th Day of that Month,. 11 Minutes after four o'Clock in the Morning.

Another by Mr. Ja. Giles, wherein Mr. Gibbon's Paradox is answered.

Your great Exploits (Friend Gibbons) done

In th' Time between the Clock and Sun,

Might be perform'd, I own:

But what do you think one Day I did?

I eat a Calf, a Sheep, and Pig,

And picked ev'ry Bone.

Their Hides I tann'd, and made me Shoes,
The Wool I fpun, and knit me Hofe,

Idle I hate to be;

I flept away near half the Day;

And yet had Time for Sport and Play:

Pray what d'ye think of me?

A Paradoxical Problem by Mr. W. Ingram.

Ingenious Artifts, pray disclose,
How I nine Trees may fo difpofe,
That juft ten Rowes may form'd be,

And ev'ry Row contain juft three.

This Queftion (in other Words) was alfo propofed by our ingenious Contributor Mr. Ja. Taylor: which is bere taken Notice of to prevent a MifunderStanding, and not with a View to infinuate any thing to the Disadvantage of either of the Gentlemen to whom we are obliged.

ANSWERS to the MATHEMATICAL QUESTIONS in the laft Year's Diary.

I. QUEST. 421. anfw. by Mr. W. Bacon of Ipfwich.

Let the Share of A bex; then that of B will be 1

3x
4

and that of

C=x+70 (by the Queft.) whence 2x+ 370=1500 (by the Queft.]

The fame answered by Mr. R. Flitcon.

Let 4x, 3x, and 4x+70 denote the Shares of A, B, and C respectively; then will 11x+70=1500 (per Quest.) Whence x=130; and the three Shares 520, 390, and 590 Pounds.

Thus the Problem is alfo anfwered by Mr. W. Bamfield, Mr. Edw. Barrass, Mr. Fra. Bell, Mr. Ja. Beresford, Mr. J. Chapman, Mr. G. Clarke, Mr. E. Cock, Mr. T. Corbet, Mr. Ja. Fellowes. Mr. Ja. Giles, Mr. J. Hampfon, Mr. E. Hare, Mr. W. Harrison, Mr. G. Hickes, Mr. W. Honnor, Mr. T. Howe, Mr. R, Hudson, Mr. T. Jeffery, Mr. J. John fon, Mr. T. Knight, Mr. E. Langworth, Mr. J. Lewin, Mr. W. Litfon, Mr. R. Marfe, Mr. W. Mathewfon, Mr. Herbert Nokes, Mr. J. Rennard, Mr. R. Richardfon, Mr. Jof. Rofe, Mr. Alex. Rowe, Mr. T. Sandling, Mifs T. S-1, Mr. Jof. Scott, Mr. T. Sims, Sinbad, Mr. G. Stapley, Mr. H. Stephens, Mr. J. Stothart, Mr. R. Walton, and many others, as will appear by the Catalogue.

โล

II. QUEST. 422. anfw. by Mr. J. Hampfen of Leigh.

D

Let A be the Place of Cannon, B that of the Target, and C that of the Spectator; and in CA take CDCB: Then the Sound of the Cannon will be at D, on the Ball's striking the Target at B; and fo AD will be to AB in the given Proportion of 2 to 3 (by the Question.) Putting, therefore, AB3a, AD=2 a, and BC (= CD)=x, we have ACx+2a; and from thence (by A Eu. 47. 1.) x x+4ax+4aa±x x+9aa: From which x

Yards.

In the very fame Manner, it is answered by Mr. W. Allen, Mr. J. Bank, Mr. Ja. Beresford, Mr. W. Bevil, Birchoverenfis, Mr J. Chapman, Mr. T. Corbett, Mr. J. Herrington, Mr. G. Hicks, Mr. J. Honey, Mr. W. Honnor, Mr. T. Howe, Mr. R. Hudfon, Mr. J. Milbourn, Mr. J. Rennard, Mr. T. Sims, Mr. H. Stephens, Mr. W. Stoker, Mr. W. Terrill, Mr. J. Wilson, and feveral others.

Ε

B

A

III. QUESTION 423, anfwered by Mr. Will. King fton of Bath. Let A, B, C and D be the four given Points: Bifect AB in E, and CD in F; then affuming P for the required Point, and putting AE: - BE = DF CF6, EP=x, = a, and FP, we fhall (by the 12 of the 24 of Simpson's Geometry) have AP2+ BP2= 2a2+2x2, and DP2 + CP2 = 26 2+2x2: Hence AP+BP +DP2+CP2 = 2a2+ 262 +2x2 +2x2, a Minimum: And confequently x2+22 (EP2+FP) a Min. alfo:

P

which will evidently be when EP+FP is a

Min. that is, when EP and FP maké a right Line; and EP will then be =FP; because (by Eu. 9. 2.) the Sum of the Squares of the two Parts of a Line equally divided is always less than the Sum of the Squares of the Parts, when the Line is divided unequally.

The fame answered by Mr. W. Lucas.

Let A, B, C and D represent the four given Points, and P any Point taken at Pleasure ; let PE and PF be drawn to bifect AB and DC in E and F ; then (by a Theorem in Geometry) AP+BP2 = 2 AE2+2 EP2, and DP+ CP2: 2 DF+2FP; therefore AP+BP + DP2 + CP2 =2 × AE2+DF2+EP'+FP2), a Minimum (per Queft.) And confequently EP2+FP a Minimum alfo: But, drawing PO to bifect EF, we EP2+FP2= 2 EO2+2 PO2: Therefore PO2 a Minimum, and fo PO bengo, O must be the Point required.

get

2

2

In this Manner, and from the very fame Property, the Problem is elegantly folved by Mr. Rob. Butler, Mr. Patrick O'Cavanah, and Mr. W, Davies.

IV. QUEST.

IV. QUESTION 424, answered by Mr. Thomas Baxtondans the Propofer.

F

E

KA

A

R

B

=

Let the Angles BAC and BAR be the given Elevations of the Top and Bafe of the Tower above the Plane of the Horizon AB, at the first Station A; Let D be the fecond Station, and DE its perpendicular Distance below the Plane ABE of the firft. In BC produced, fuppofe CF to be taken DE; let FK be parallel to CA, and let EF, EB, FK and AD be drawn. Then, becaufe DE is equal and parallel to FC, EF will also be equal and parallel to DC, and confequently the Angle EFB (= DCB) the Complement of the given Elevation at the fecond Station D. Now, from AD (100) and the Angle ADE (5° 30'), we have given AE=99,5396; and DE (FC)=9,5846: And it will be as Rad. Tang. BCA (60°) (:: BC: BA) :: FC: KA= 16,601 : From which, together with AE and the contained Angle KAE (135°), the Angle AKÉ will be found = 38° 59′. Moreover the Angles FBE and FBK being both right Ones, it will be Tang. BFE (51° 30'): Tang. BFK (60°) ::(BE:BK) :: Sin. BKE (38° 59′): Sin. BEK = 60o 5': Whence ABE is alfo given: =8c° 56, AEB 54° 4: From which, and AE (=99,5396) will be found AB 81,61; and from thence BC= 47,12; BR 17,34, and RC=29,77➡ Tower's true Height.

An Algebraic Solution to the fame, by Mr. Rich. Mallock. "I have fome Reason to believe that this Queftion is wrong printed, viz. North for South, or North-Eaft for South-Eaft; which being corrected I solve it thus."

D

H

Let DE be the perpendicular Distance of the fecond Station D, below the horizontal Plane (ABE) of the first Station A; in CB (the Tower's Height) produced, take BH ED; join D, H, and upon AB let fall the perpendicular EF.

By Means of AD (= 100) and the given Angle DAE (5° 30') I find ED (BH)=9,58458, and AE =99,53955; from which laft AF (EF). 70,38 53, Now put BHb, AF (EF) c, and BC= x; and let m and n denote the Tangents of the Complements (BCA and HCD) of the Tower's Elevation at the two Stations A and D: Then (by Trigonometry) BA = mx, and HD = n x x + b = BE (becaufe BH being parallel, and equal to ED, BE will alfo be parallel and equal to HD.) Hence BF=mx-c; and confequently mx—c)2+ c2= nx+nb)2 mc+nnb

(Eu. 47. 1.): Which, ordered, gives x2. 2a2-262

mmnn

in Numbers, x2

mnnn

X 2 x =

193,11 x = 6877,8; whence

= 96,555 + 49,44; that is, the Tower's Height above AB

f

the required Distance A B is found 81,5 Yards.

In the very fame Manner the Problem is anfwered by Mr. W. Allen, Mr. R. Butler, Birchoverenfis, Mr. W. Marshall, Mr. J. Milbourn, Mr. S. Towndrew, and Mr. W. Stoker.

Through Want of Care in tranfcribing this Question for the Prefs (not from any Fault of the Printer) the Bearing of the Tower, and the firft Station from each other, was inverted. The Solutions, given above, in which the Tower is fuppofed to bear due North of the Obferver, are exactly conformable to the Propofer's Meaning.

I'

V. QUESTION 425 anfwered by Mr. O'Cavanah.

N the Cafe propofed, where the Sums of the Squares of the oppofite Sides of the Trapezium are equal (and where the Diagonals do, E therefore, cut each other at right Angles) the Problem may be thus conftructed.

Having made AB and AE perpendicular to each other, and equal each to any one of the given Sides (as 35) from the Centers B and E with

C

D

B

F

A

Radii equal to the two contiguous Sides (25 and 31) let two Arcs be defcribed, interfecting in C; draw A C, and alfo BD perpendicular, and equal thereto; fo fhall A B C D (when AD and CD are drawn) be the Trapezium fought.

2

2

2

For the Angles A B D and E AC are equal, being both Complements of CAB, to a right Angle (by Conftr.) and the Sides A B, BD, AE, AC, containing them, are alfo, refpectively, equal (by Confir.) Whence the remaining Sides A D and EC muft neceffarily be equal. Alfo A B - AD2 ·DF) BC-DC2, and therefore A B+DC2= =AD2+BC2, that is, in the present Cafe, 352 + DC2 = 312 + 25, and confequently DC 19, as it ought to be. The numerical Solution is, from hence, very eafy; whereby

2

2

=BF

12

A R

I

the Area comes out 4: I: 384.

Mr. W. Harrison, Mr. J. Honey, Mr. R. Mallock, Mr. J. Pearce, Mr. W. Stoker, and Mr. W. Terrill (whofe Solutions agree in almost every Step) give the following analytical Investigation of this Problem:

"Since the Sums of the Squares of the oppofite Sides of the Tra"pezium A B C D are equal (by the Queftion), the Diagonals will, "therefore, cut each other at Right-angles in F: fo that, putting "BC (25) = a, BA (= 35) b, DA (= 31)=c, CD 66 (19)=d, and AC (= BD)=x, we fhall have x: b+a::

C