that the required Perimeter may be diminished, and the faid Area and Diff. of Sides may remain the fame; which therefore cannot obtain; whence the Perimeter of any Trapezoid is the leaft poffible when the two Sides that join the Extremities of the parallel Sides are equal to each other. Q.E.D. Now put the given Area a, AL (1⁄2 AI) = b, and LB AB-FH

2

a
x; then LF= and (per47E.1.)

x

+x=Half the requir❜d Peri

+x2-√a2+

2

x2x

=0, or

2

+x2 -√ a2 + b2x2)=o, Confequently b2x2+

x2 √ a2+b2x2 -—a2 — b2x2 = 0, :. x2 √ a2 + b2x2 1 = a2 and b2x6a2xa—a, from whence the Value of x may be found.

The fame answer'd by Mr. William Bevil. Let ABHF represent a Trapezoid (fee the laft Fig.) Put 2x ABHF, 2a=AB-HF, b= the square Poles in one Acre, then x+a AB and x-a=HF, and

62

2x

x2

+2x is a Min. and in

√46a +4a2 = 462, and 6x*x*+a2x2 = 6*; Hence x= 12,5

x2

nearly.. AB 15.3211, DC 9.6786, and AFBH= 13.107191.

The fame answered by Mr Jofeph Orchard.

Let ABHF reprefent the Trapezoid (fee the laft _Fig.) Now the Perimeter of a Circle being less than that of any other Figure having the fame Area, it is plain the required Trapezoid must be infcribed in a Circle, and that ALGB

== 2.82112 = half the given Diff. of the parallel Sides ; Let a 160 Poles the given Area, and HG FL=%;

=

x

a+dx

a-dx

- FH=

and

BH AF√√√2+x, whence the Perimeter is 2+x

za

+ which is to be a Minimum; in Fluxions

XX

xx

√d2 + x2]
'd2+x2

o, Reduced 6 — azx2 = a2 d2; Put zx, then 23 -axad, where z 163,84 nearly. 12.8: Hence AB 15.321112, FH = 9.678888 and AF BH 13.1072.

In like Manner this Question was anfwered by Mr. J. Holden, Mr. Geo. Godhelp, Mr. Abr. Lord, Rimfide, Mr. Rob. Butler, Mr. John Nichols, Mr. Steph. Hartley, Mr. John Watchorne, Mr. Philip George, Mr. Geo. Stapley, Mr. William Enefer, and others, as may be feen in the Catalogue.

(8) Queft. 108 anfwer'd by Mr Tho. Garrard.

Conftruction. With 60° of Chords defcribe the Circle SBRNES (fee the Figure) and draw SN for the Meridian, and CB (making the Angle BCS 65° 59′ 53′′) for the Sun's Azimuth, alfo make the Angle BCA 54°, the Elevation of the Object CA, upon B erect the Perpend. BA, from the Point of Interfection A draw

AD making the Angle BAD 45° 50' 31" the Compt of the Sun's Altitude, with the Radius DPCS de

feribe the Circle

DEO; from the

Centre C draw the

S

A

Tang CE to touch

the Circle DEO in the Point E, and from E draw ER parallel to PB; laftly, thro' the Centre C and Point of Interfection R, draw CR, which will be the true Pofition of the

Object when the Angle included between the Meridian and Shadow of the Object is a Minimum.

For if we conceive the Triangle ABC to and upon the Bafe BC, perpendicular to the Plane of this Paper (which here represents the Plane of the Horizon) then will the Shadow of the Top of the Object A fall upon the Point D, and DC will be the Length of the Shadow when the Object is directed towards the Sun: Now if we imagine the Plane of the Triangle ABC to revolve round the Centre C upon the Bafe BC, always keeping perpendicular to the Plane of this Paper, then will A, the Top of the Object always be perpendicular to fome Point in the Periphery of the Circle SBRNES, and the Length of the Shadow (reckoned from that Point) will always be equal and parallel to BD; confequently the Shadow of A, the Top of the Object, will always fall in the Periphery of the Circle DEO. Thus when the Top of the Object A, becomes perpènd. to the Point R, then will the Shadow of the Object CA fall upon CE, making RE equal and parallel to BDCP, and fince (by Construction) PE is equal, and confequently parallel to CR, we have a very easy Method of

Calculation. Put p and q for the Sine and Cofine of the Angle of Elevation of the Object, n and m for those of the Sun's Altitude, and call the Length of the Object Radius or pm Unity; then by plain Trigonom. 2: p::m:=BD=PC.

Again,

pm

n

: 1 :: q:

gn pm

n

Sine of the Angle PCE = BCR

44° 52′ 47′′. Hence the Angle RCN 69° 07′ 20′′, and NCE 21° 07′ 06′′, whence we have this

Theorem. Multiply the Tangent of the Sun's Altitude by the Cotangent of the Elevation of the Object, the Product is the Sine of the Angle included by the Line of the Sun's Azimuth and the Direction of the Object when the Shadow of the Object and the Meridian Line make the least Angle poffible.

N. B. When the Altitude of the Sun is equal to, or greater than, the Elevation of the Object, then this admits of no Minimum.

The fame anfwer'd by Mr. Wm. Kingstone.

Let AC the Length of the Staff, then will BC= ,809017, AB,5877536, and BD,821642 = a:

Pur

71537944° 19′ 52", and

the Angle D = 45° 40′ 27′′; whence the required Angle DAE = 21° 13' 04", and the Angle BAE 68° 46′ 56′′

More curious Answers might have been

A

B

E

exbibited to this Question from Harland Widd, Mr. J. Holden, Mr. Abr. Lord, Mr. Wm Spencer, Tabularius, Mr. Tho. Mofs, and others, but we bave purposely omitted them, to make Room for what follows.

(9) Question 119 answer'd by Mr. Abr. Lord.

For the Distance of the Candle at the 2d Obfervation put x, for 60 puta, then the Distance at the 1ft Obfervation will be ax; but it is proved by the Writers on Optics, that Light decreafes in the reciprocal duplicate Ratio of the Squares of the Distances. Therefore x2: 2 :: x2+2ax+x2: 3, then multiplying Extreams and Means, and reducing, we have x= 2a√6a2 266,9693845 Yards, Therefore the 1st Distance is had 326,9693845 Yards, the Anfwer required.

The fame answer'd by Mr William Enefer.

Whether the ingenious Propofer meant that the apparent Diam. or the Effects of the Light of the Candle should be in the given Proportion, does not from the Data of the Question to me feem determined, and therefore fhall give a Solution upon both Suppofitions.

ift Supposing the apparent Diam. of the Candle's Blaze be

meant.

Put DC the given Distance = 60a, AB the Semi-Diam. of the Candle's Blazev, Sine <DCB=2x, then (by Qu.) Sine <AD B= 3x, and Rad. 1. By plain Trigonom.

C 3

B

C

E

BD, whence it is evident that BD will Dalways be to BC, in the Ratio of the <DCB to ABD, viz. (in this particular Case) of 2 to 3. Hence (fee Mr. Simpfon's Algebra, p. 310.) this Geometrical Conftruction.

A

H

Draw the Right Line CH of any convenient Length, in which make CD the given Distance; and make CE to DE as 3 to 2, then take EO to EC as DE to CE-DE, and F upon O as a Centre describe (with the Rad. OE) the Semi-circle EBF; then if the two Lines CB and DB be drawn to meet any where in the Periphery thereof, the Angles C and D will always be in the conftant Ratio of 2 to 3. Confequently the Queftion is capable of innumerable Anfwers, according to the Bignefs or Smallnefs of the Blaze; for it might be at any Distance not lefs than CD, nor greater than FC.

2dly, by fuppofing the Effects of the Light to be meant.

Put 60 d, and x the Dift. of the Obferver from the Candle at the 1ft Obfervation; then x-d will exprefs his Dist. therefrom at his 24; and fince the Effects are reciprocally as the Squares of the Distances x2: 3 :: x2—2dx2+d2: 2.. 2x2 =3×x2-2dx+d2, whence by proper Reduction, completing the Square,&c.x=√6dd +3d=326,964 Yards.

The fame answer'd by Mr. Tho. Garrard.

From the Data of this Queft. it don't clearly appear (to me) whether the ingenious Propofer's Meaning was, That the Degrees or Quantity of Light which he received from the Candle at the 1ft Obfervation, was to that which he received at the 2d as 2 to 3; or whether he means, That the apparent Diam. of the Candle (or Angle under which the fame was feen) at the firft Obfervation, was to that at the second in the fame Ratio ; I fhall therefore confider it under both Cafes.

=

Put the requir'd Distance at the first Obfervation, and d60 Yards, then will x-d the Distance at the 2d Obfervation. Now if we consider it under Case the first (because the Degrees of Light, Heat, and Attraction are reciprocally proportional to the Squares of the Distances from the Centres of their primary Caufes) we have 2xx-3x2-5dx+