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lidity of K EF, and by sim. solids, &c. as in the other solutions.

This question was answered also by Messrs. Baines, Bamford, Brewer, Brooke, Bruster, Cummins, Ely, Eyres, Gawthorp, Harrison, Hine, Jones, Macann, Maffett, Nesbit, Rylando, Webster, and Winward.

4. Qu. (78) Answered by Mr. A. Nesbit, Farnley Academy

In the annexed figure, let HO denote the horizon, EQ the equinoctial, m m the parallel of the sun's declination, S the sun's place when his rays H were parallel to the front of the house, T his place when they were perpendicular to it, Z the zenith, and P the north pole. Then, in the spherical

E

m

Z

P

Ο

m

triangle S ZT, we have the first zenith distance SZ= 48° 5, the second zenith distance TZ 31° 52′, and the SZT 90°, to find ST = 55° 26′, and the TSZ 39° 52′ 32".

Again, in the triangle SPT, we have the polar distance SP TP 66° 37′, and ST, to find the TSP 76° 52′ 2′′, from which take the 2 TSZ, and we obtain the PSZ = 36° 59′ 30′′.

Lastly, in the triangle PSZ, we have the sides SP and SZ, and their included angle, to find PZ = 35° 50′ 16", the co-latitude of the place, and the 4 SZP= 109° 23′ 43′′, the sun's azimuth from the north, at the time of the first observation. Hence, the latitude of the place is 54° 9′ 44′′ N. and the declination of the front wall 19° 23′ 43′′ from the east or west.

Again, (by the terrestrial globe) by Mr. S. Jones, Liverpool,

Bring any place through which a meridian passes, as London, to the brazen meridian, and under the first given altitude, on that meridian, make a mark upon the globe; turn the globe on its axis eastward 90°, (by the question) and under the second given altitude, on the

brazen meridian, make another mark on the globe; bring the north pole towards the southern edge of the horizon, and at the same time turn the globe on its axis till the two marks appear in the horizon, the horizon will then cut the brazen meridian at 59 degrees, the sun's meridian altitude on the given day, whence the zenith distance 30 degrees, added to the declination 23 N. gives the latitude of the place = 54o nearly. The globe still remaining in the same position, thrust a quill between the globe and the brazen meridian, to keep the globe from turning on its axis, and elevate the north pole 90°; the meridian (of London) which passes through the first mark, will cut the horizon in S.S. E. & E. nearly, which is the declination of the wall.

In the same manner exactly as the first solution, it was answered by Messrs. Haines, Cattrall, Gawthorp, Hine, Kay, Maffett, Rylando, Smith, and Winward.

5 Qu. (79) Answered by Mr. Baines, and Mr. Bruster, Donington.

Let the lines be drawn as in the annexed figure, where AJ represents the surface of the liquor before coming to rest, and ▲ BAJ 43° per question. Then BH (1) HC (3) :: BG (2): GE6 the altitude of the complete cone. Again, by trigonometry, EG: radius: BG: tan. 18° 20′ 5" 49""▲ BEG=▲ BJI; moreover AJI=90°-43° 47°, and AJB 47° +18° 26′ 5′′′ 49′′ — 65° 26′ 5′′, 49""; hence, S. 4 AJB; AB:: S. 4 BAJ: BJ 2,99947. Then /BG+GE2 BE 6.324555, and BE- · BJ = 3,325085 = JE. Now by similar triangles EG: AB :: EJ: JK 2,102968 diameter of the hoof at the section, also BE: EG: BJ: JI = 2,845547 = perp. height of the hoof. Then, by Hutton or Bonnycastle, AB-JKAB. JK)

=

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× IJ X AB

GHI

F

AB-JK

A

16-6,09928

X,2618 =

4-2,102968

E

(5,21905) × 4 × 2,845547 × ,2618 15,552 cubic

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inches content of the hoof AJB, which was the part A drank. But (4a + 22 + 4×2) ×,2618 × 3=21,9912 cubic inches the capacity of the glass.

Therefore we have

share, and

As 21,992: 6d. ::

21,9912-15,5526,4392-B's

́15,552 : 4d. 243 + or about 43d. A must pay.

6,4392: ld. 756 + or about 13d. B must pay.

And thus very nearly it was answered by Messrs. Armitage, Brewer, Brooke, Cattrall, Cummins, Dunn, Ely, Gawthorp, Hine, Jones, Macann, Maffet, Nesbit, Rylando, Smith, Webster, and Winward.

6. Qu. (80) Answered by Mr. Bruster, Donington.

The cylinder being supposed of equal density throughout, the centre of gravity will of course be the middle point of its axis; and at any given elévation, the line of suspension continued will always pass through the centre of

H

P

gravity, and make an angle with the middle section equal to the given angle of elevation.

In the annexed scheme P is the point of suspension, HO and ho are parallels to the horizon, and the angle of elevation Aho 4 BAC➡ 25° 15'. As rad. : AB (1) tan. (25° 15′) ▲ BAC:,47163 = BC.

Hence the point C =

10

2

,471635,47163 feet from

the bottom, or 4,52837 feet from the top end.

Again, by Messrs. Baines, Brewer, Cattrall, Ely, and
Smith.

Let A be the centre of gravity, and P C the string; then, when the cylinder is at rest, PC will pass through

A; therefore, in the triangle ABC, we have all the angles, (for the angle BAC is angle A ho) and the side A B, to find BC,47163; whence FC = 5,47163, and CF 4,52837 feet, as required.

This question was answered also by Messrs. Armitage, Brooke, Cummins, Eyres, Gawthorp, Hine, Jones, Kay, Maffett, Nesbit, Rylando, Webster, and Winward.

7. 2v. (81) Answered by Mr. Gawthorp, Mr. Nesbit, and Rylando.

2,

Put zx- 1; then, and 2% = 2xtherefore, 2% +1=2x· 1, and (2%+ 1) × % = 2x2 3x+1; hence the given fluxion will be transformed

ż

to

=

282+%

[ocr errors]

; whose fluent is

2,30258

× log. 2+8=-1 x hyp. log. 2+; and by restoring

1

x we have — 1 × h. l. 2 + for the fluent required.

Nearly thus it was answered by Messrs. Hine, Jones, Maffett, Whitley, and Winward.

8. Qu. (82) Answered by Mr. Gawthorp and Mr. J,

Whitley.

and√(d

Let 2 = 1 (d † v), then, ✯ = 1, and 3⁄4 √ (ď' — 4 dz

+82 2) = √√√√√({ d2 + { v1) = × v √ (ď2 + v2);

√2

4d

8 d

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whose fluent, by prop. 73, p. 178, Vince's Fluxions, is

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+ 1}+C. This expression ought to vanish when z=

fore, the correct fluent, generated while z increases from

d d√2

o; hence C=

xh.l. (d/2-d); there

8

16

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(2 + √ 10 ) = {(3√/5 + 1) + √ √ 2x 8.4. (3+√/10)

√2-1

[ocr errors]

d

2

h.l.(3+✔10)

× (1+ √2)} × = 1,20215 × d= 12,0215 as

required.

And thus it was answered by Messrs. Hine, Maffett, Nesbit, Rylando, and Winward.

Otherwise by Mr. S. Jones, Richmond Academy, Liverpool.

With semiaxes CD, CV, each

d. describe the equilateral hyperbola HVB, draw the ordinates DH, AB, and join AV; by a known property of the hyperbola, ABAV

H

B

(CD+C A'); put DA = z, then,

1

CA=% d. and AB= x (d2-4 dz +8%';)

-

8

but the fluxion of the area ADHBAAB x fluxion

1

[ocr errors]

AD = * × — — ̧ × √ (d3 — 4 dz + 8 2'), consequently

V8
ż

the fluent of (d'— 4 dx + 8 x2) = v' (d2.

ADHVBA; as required.

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9. Qu. (83) Answered by Mr. Elias Webster, Armley Mills.

Take yx-r, and z = x+2r; then, the first becomes a square, and it only remains to make x2 — 6rx -6r2, and r2 2rx-52 squares; to effect which, let r6rx 6 r2 = (x — nr)', and we get r

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Χ Τ.

-

·

n2 + 6· 2n-6 latter formula, we obtain, by reduction, n 4 n3 + 4 n2 +96 n-72; which must be made a square. Assume p+5 for its root, and we shall find p 8, and n9;

This value of x substituted for it in the

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Exam. Taker 4, then y = 25,x=29, and z = 37; three numbers that answer the conditions of the ques

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