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Again, by Messrs. Baines, Brooke, Butterworth, Eyres,

* Putsey, Webster, and Whitley: "..? ! In the annexed fig. let AC3%, 165: = g, and the diameter of the inscribed circle = 4. Draw A D perp. to A C, meeting B C produced in D, then by art. 301 Marrat's Mechanics the time AT down AC = time down CD, therefore: the time down CD = 1" and conse.. t quently CD= 164 feet. Now BC= Vi and AB=(- ) (0 ); 1.72 also by a known property of right angled triangles AB+BC=AC+4; therefore, AB = =+ 4

V (22— 5*) as above, and, by reduction, tt-gra -482? +4g? x+8 g = 0, or in numbers - 167 m 644 m2 = 103445 x + 206978 = 0; whence x = 9.90725 = AC; then BC=6.1028 and AB 7 80445.

The numbers in the first of the above solutions appear to be correct, but we have not time to examine those in the second. The opinion of our correspond. ents is requested for the next Number of the Enquirer.

4. Qu. (59) Answered by Messrs. Hirst and Whitley.

Denoting the roots of the required squares by x and 1-X, the two first conditions are answered, and we have only to make x + (1-x) = -1x + 2x2 a square; equate it with (1-nx)2 = M ina + 12x?,

un-1 and we have r=0 where n is indeterminate. Let n= 3, then r= 4, and the required squares are

and gr Nearly in the same manner it was answered by Messrs. Baines, Brooke, Butterworth, Cummins, Eyres,

VOL. II.

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Gawthorp, Hine, Maffett, Nesbit, Putsey, Rylando, Web ster, Tomlinson, and Winward. : i . ,.

- 5. Qu. (60) Answered by M. M. Let ABC be the rea..

001 quired triangle, and AD

en
C

yriteupaya the bisecting line. Put ...

bris AB = a, AD = b, and :

a ns AC =*; then, CD=

9

9 v (b?x*), and by Leslie's

Gibson Geom. 6.11. (64--12)

A

7 3 219B = DB; also 6.21. ibid. x 162 —- *•) + b=ar, that is 2x mor =bz, or, in numbers, ? — 32 r = 3200; whence r = 74.78775, and the area = 2A8Q. Or. 371p. .

. Again, by Messrs. Gawthorp and Hirst. Put A B = 100 = 0, AD=80=b, and cos. ¿CAD = x, then AC=bx; and Em. Trig. p. 105, edit. 2d, 2 abx : ab + b2 x ::1:x; therefore, 2 a b 2 = ah+ b2 x, from whence x= + 3=.9348469, then bs = 74.7877, and the area of the field = 248a.Or. 371p.

The same, by Messrs. Hinc, Maffett, and Winward. • Put-AB=a, AD=c, and AC = x; then, by trigonometry, = sin. B. to rad. 1 = cos: C A B, and = sin. C D A = cos. C A D; also sin. DACEV (0.3) and cos. 2 DAC_ la Croix, p. 29), therefore the = 2 *** duced, in numbers we shall have je — 32 r = 3200; whence x = 74.78775, and the area as above, i

Again, by Messrs. Baines, Cattrall, and Cummins. ** Put a sin. CAD-DAB, then, v (1 = z)= kin. ADC; also 2 x (1-2) = sin. C AB, and

=2= (Trig. de

; which re

W(1.-4* 7.434) = sin. ABC; therefore, rada: AB, =100 :: sin. ABC=>(

1 4x +4*): A B = 100 (

1 4 * +4x). Again rad. : AD= 80 :: sin. ADC=N(1 *): AB = 80Ň (1 — X+), whence 100 V (1 —41 + 4 x*) = 80 (1 x); therefore, by squaring and reducing we get 24 .84 x3 ,09, and x = .35505; consequently sin. C A B = .66383 and sin. ABC=.74788, therefore, AC = 74.788, and BC= 66.383, and &c.

According to one or other of the above methods, answers were given by. Messrs. Brooke, Eyres, Hirst, Nesbit, Putsey, Rylando, Smith (Alton Park), and Webster; and geometrical constructions by Messrs. Hine, Maffett, Winward, and Whitley. - -- 6. Qu. (01) Answered by Rylando. *Let h = height of the gnomon, b and c =sine ånd cosine of the given latitude 53° 4.5', d and p = sine and cosine of the sun's declination. Then, by art. 1l, Simpson's Dissertations, and the nature of the question

2hpd. 2 ph; therefore, we do 9:11 :: dai vad?)',

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V(c?—d?); therefore,

9c ; from whenced= = .3744392 = nat. sine of 21° 59', which answers in the tables to the 31st of May, or the 12th of July.

It was also answered by Messrs. Baines, Nesbit, Pete sey, and Tomlinson. 7 Qu. (62) Answered by Messrs. Brooke, Cattrall, Hine,

Maffett, Nesbit, and Winward. . .? Let ABC be the given triangle; bisect A B in D, and on D as a centre, with radius equal to half AC + BC, describe the arc m, n, draw C F parallel to AB, cutting the arc m n in E; join É D, and parallel to it, draw A F meeting C F in F;

D. then will the parallelogram A DEF be equal in area

and perimeter to the triangle A BC. For since A D is ={A B the parellelogram is évidently.equal to the triangle, and because DE=I (AC + BC), therefore AF+DE+2 AD= AC + BC + AB. ::

Otherwise by Mr. E. S. Eyres, Liverpool..... Make A E= the given in perimeter, and perp. thereto draw A I, and make AI= A E, and join IE; on A I set off AH = the side of a square, equal in area to the given triangle, draw HG parallel to DE A E, meeting a semicircle on AE in G, and on A E let fall AL the perp. G B, cutting I E in C; lastly, draw C D parallel to A B, and ABCD is the rectangle required. For, GB’ = AH? FABX BE=AB X BC= the given area ; and AB TBC +CD+DA2AB+ 2 BC = 2 A E. 2. ED.

Mr. Whitley's answer is the same in effect as the above.

Again, by Messrs. Baines and Putsey. On AE = the semi-perimeter of the given triangle, describe the semicircle AGE, and, from A, erect AH perp. to A E, and = to a mean proportional between half the base and the perpendicular, draw HG parallel to A E, meeting the semicircle in G, and from G let fall the perp. G B, so shall AB be the length, and BE=BC the breadth of the parallelogram required ; as is evident by the known property of the circle.

Other answers were sent by Messrs. Gawthorp, Hirst, Rylando, Tomlinson, and Webster. ,

8 Qu. (63) Answered by Mr. J. Gawthorp, Leeds. · Let A BC represent the inelined ! ost olyan so pred nc plane; put g = 161, then the ve- !? La compito ! locity acquired by descending from C to A is = 2 lex CB) = .,' disis 50.728. (arts. 287 and 313, Marrat's try Mechanics). Now if a body bee n for projected up the plane with this AT

T A

velocity, it will, in the time x, describe the space or

mini_CB gfx? (art. 292 idem) f being = =.8; but if it were let fall from C, it would, in the same time, describe the space g.x=CD. Let A E be the space described by the ascending body, then CE=50 0rtfg**. By trig. AC: rad. :: BC:ě = .8=cosine of ACB, · which put = c; then, by p. 124, Em. Trig. 2d edit. DE CES +CD2c. CE.CD= (500x tf g 32)

+ gê x4 – 2cg ra x (30—08 + f g .za) which is to be a minimum, by the question. Make its fluxion = 0, and, by reduction, we get the following equation 737 13.8169 * = 13.6186, from whence r = .9278', and DE=8.5291 feet, is the distance required. * This is the only true answer which we have received to the question. ; (9

9 Qu. (64) Answered by Messrs. Butterworth, Eyres,

Hine, Maffett, and Winward.

-B

Let A B be the diameter of the D circle ADB A, and let DB be the required arc; draw DC perp. to AB, and by the question DCXCB, AF or D-C" X CB, or (because ACX. CB=DC), AC x B C is to be a maxinium ; which is known to be the case when 3 AC=CB, whence A C is and the arc D B is evidently = 120°.

AB

Again, by Mr. J. Whitley.. Let CB be an indefinite Br tangent to the semicircle CDF, at the point C; in this tangent take any pointb, and to ČF apply = 26c; parallel to ab draw the tangent 1 AB, touching the curve in C aO E F D, and C D will be the arc required. ii,

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