Again, by Messrs. Baines, Brooke, Butterworth, Eyres) Putsey, Webster, and Whitley:

In the annexed fig. let AC≈x, 16x =g, and the diameter of the inscribed to circle 4. Draw A D perp. to A C, meeting BC produced in D, then by art. 301 Marrat's Mechanics the time down AC time down CD, therefore the time down CD 1" and conse

quently C D = 16+ feet. Now BC=,

and AB =√(x2

g

also by a known property of right angled triangles AB+BC AC+4; therefore, AB

=

+ 4.

——✔ (a2 —— x2) as above, and, by reduction, x+— g x3

— 4g x+8 · 4 g x2 + 4 g2 x + 8 g 0, or in numbers - 16

643 r2 x9.90725

7 80445.

10343x+206970; whence AC; then BC = 6.1028 and AB=

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 correspondents 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 -x, the two first conditions are answered, and we have only to make x2+(4−x)2 = 7− {x+2x2 a square; equate it with (na) = {} = \ n x + n2x2, and we have r where n is indeterminate.

n 1

2n 4

Let n 3, then, and the required squares are ra and t

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

VOL. 11.

Gawthorp, Hine, Maffett, Nesbit, Putsey, Rylando, Webster, Tomlinson, and Winward..

5. Qu. (60) Answered by M. M.

Let ABC be the required triangle, and AD the bisecting line. Put A Ba, AD = b, and AC; then, CD= (b2x2), and by Leslie's

a

niz-bas

= DB; also 6.21. ibid. — × (b2 — x3) + b2= a *,

=

x= b2, or, in numbers, a2 - 32 x =

74.78775, and the area — 248a.

Again, by Messrs. Gawthorp and Hirst. Put A B 100= a, A D = 80=b, and cos. 4 CAD X, then AC=bx; and Em. Trig. p. 105, edit. 2d, 2 a b x: a b + b2 x : : 1 : x ; therefore, 2 a b x2 = a = ab+ b2x, from whence x = +√37 = .9348469, then br 74.7877, and the area of the field 248a. Or. 37 p. The same, by Messrs. Hine, Maffett, and Winward.. Put AB=a, AD=c, and AC; then, by tri

gonometry, sin. B. to rad. 1 = cos. C AB, and

a

sin. CDA cos, CAD; also sin. DAC=√(c2x2)

32 x 3200;

duced, in numbers we shall have 2.

whence x=74.78775, and the area as above,

Again, by Messrs. Baines, Cattrall, and Cummins.

Put a sin. CAD

sin. A D C; also 2

DAB, then, (122)

(12) = sin. CAB, and

√ (1-4x+4)=sin. ABC; therefore, rad. AB. 100: sin. ABC=√(1) 4x + 4x2): A B≈ 100 √ (1-4x+4). Again rad.: AD 80:: sin. ADC=√(1): AB 80 (1-x), whence 100 (1-4x+4x2) = 80(1); therefore, ✔ by squaring and reducing we get x4.84 x4.09, and x.35505; consequently sin. CAB 66383 and sin. A B C .74788, therefore, A C➡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. (61) Answered by Rylando. Let h height of the gnomon, b and c sine and cosine of the given latitude 53° 45′, d and p = sine and cosine of the sun's declination. Then, by art. 11, Simpson's Dissertations, and the nature of the question 2hp d 2 ph

9:11 ::

9

√ (c2-d2)

:

; therefore,

c2-d2 √(c2 — d2)

√202

11d

d

.3744392 the

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, Putsey, 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 CF parallel to AB, cutting the arc m n in E; join ED, and parallel to it, draw A F meeting C Fin F;

m

E

n

D

B

then will the parallelogram A DEF be equal in area

and perimeter to the triangle ABC. For since A D is =AB the parellelogram is evidently equal to the triangle, and because DE (AC+BC), therefore AF+DE+2AD=AC+BC+AB.

Otherwise by Mr. E. S. Eyres, Liverpool. Make AE the given I perimeter, and perp. thereto draw A I, and make AI = A E, and join IE; on AI set off A H the side of a square equal in area to the given triangle, draw HG parallel to D A E, meeting a semicircle on AE in G, and on A E let fall the perp. GB, cutting IE in

BE

H

A

E

B

C lastly, draw CD parallel to A B, and ABCD is the rectangle required. For GB' AHAB X ABX BC the given area; and A B+BC +CD+DA2AB+2BC-2AE. 2. E.D... Mr. Whitley's answer is the same in effect as the above.

Again, by Messrs. Baines and Putsey.

On A E the semi-perimeter of the given triangle, describe the semicircle A G E, and, from A, erect AH perp. to A E, and to a mean proportional between half the base and the perpendicular, draw H G parallel. to A E, meeting the semicircle in G, and from G let fall the perp. GB, so shall AB be the length, and BEBC 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 ABC represent the inclined t plane; put g 16, then the velocity acquired by descending from C to A is 2√ (g x C B) = 50.728 (arts. 287 and 313, Marrat's Mechanics). Now if a body be projected up the plane with this A

03

1

B

velocity, it will, in the time x, describe the space v z

CB

-gfr2 (art. 292 idem) ƒ being = AC .8; but if it

were let fall from C, it would, in the same time, describe the space gr2C D. Let A E be the space described by the ascending body, then CE=50-vx+ƒg x2. .8=cosine of ACB,

By trig. AC: rad.:: BC:

BC

AC

which put = c; then, by p. 124, Em. Trig. 2d edit. DE CE+CD-2c.CE.CD= (50-vx+fgx2): +g2x2cgr2x (50-vs+fg) which is to be a minimum, by the question. Make its fluxion = 0, and, by reduction, we get the following equation ☛3 + 13.8169 x 13.6186, from whence = .9278', and DE 8.5291 feet, is the distance required.

This is the only true answer which we have received to the question.

9 Qu. (64) Answered by Messrs. Butterworth, Eyres, Hine, Maffett, and Winward.

Let AB be the diameter of the circle AD BA, and let DB be the required arc; draw DC perp. to AB, and by the question DC x CB, A or DC x CB', or (because ACX. CB=DC2) ACX BC3 is to be a maxinium; which is known to be

D

B

the case when 3 AC CB, whence AC is = { A B

and the arc D B is evidently

120°.

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 point, and to CF apply ba=2bc; parallel to ab draw the tangent A B, touching the curve in C D, and C D will be the arc required.

D

F