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3. Qu. (77) By Mr. J. Cattrall, Fifer, and R. L. Militia,
Plymouth. A frustum of a cone whose diameters and altitude are 6, 10, and 12, is standing on its less end; at what distance from the bottom must it be cut by a plane parallel to its base, so that the solidity of each part may be equal ?
4. Qu. (78) By Mr. Baines, jun. Horbury Bridge.
On the 26th of June, 1808, I observed when the sun's shadow was parallel to the front of a house, that the true altitude of his centre was 41° 55', and when his shadow was perpendicular to the front of the same house, his altitude was 58° 8';-required the declination of the front wall, and the latitude of the place where it stands.
5. Qu. (79) By Mr. Dunn, Broughton. From a glass in form of the frustum of a cone, the top diameter of which is 4, bottom diameter 2, and depth 3 inches; A. drank till the surface of the liquor, and top diameter made an angle of 43°, and B. drank the remainder;-how much must each of them pay, if the liquor cost sixpence; and what is the length required to complete the frustum into a cone? 6. Qu. (80.) By Mr. Gawthrop, Master of the Lancas
terian School, Leeds. Determine a point in the side of a cylinder whose length is 10, and the diameter of its base 2 feet, to which if a string be fastened, and the cylinder suspended by it, the axis of the cylinder may make an angle of 25° 15', with a plane parallel to the horizon. 7. Qu. (81) By Mr. R. Maffett, Plymouth.
s Required, in finite terms, the fluent of
8. Qu. (82) By Mr. J. Whitley, Masbro'.
d rated while % increases from a to d = 10.
9. Qu. (83) By Mr. A. Nesbit, Furnley, Leeds.
2-2+ys+x2-xy) 3 formulæ become squares. 10. Qu. (84) By Mr. J. Butterworth, Haggate, near
Oldham. If to one end of a given rod, a string of a given length be fastened, and to the other end of the string a given weight; if then, the rod and weight thus connected revolve in a vertical plane about the other end of the rod; what will be the locus of the centre of gravity of the rod and weight, supposing the string always to.remain perp. to the horizon?
11. Qu. (85) By Mr. J. Hine, 2nd R. L. Militia: ACB is a given semicircle, CD any ordinate perpendicular to the diameter AB, and P a point in DC produced, such that DP x DC = AC?; required the equation and quadrature of the curve which defines the locus
12. Qu. (86) By the same. ABC is a given quadrant of a circle whose centre is A, and D any point in the curve; draw DE perp. to the radius AB; in AB take AF = DE, and draw FG parallel to DE meeting FG in P, and the circle in G; to determine the equation and quadrature of the curve which is the locus of P.
13. Qu. (87) By Mr. J. Whitley: In a horizontal dial, the angle included between the hour lines of 4 and 5 was observed to contain just 21 degrees; to find, geometrically, the latitude of the place it was inade for. 14. Qu. (88) By Mr. J. Winward, and R. L. Militia.
Given, the base and vertical angle, to construct the plane triangle, when the area has a given ratio to the square of the sum of the sides,
15. Qu. (89) By the same. AB, AC, are two right lines given by position, and
B is a given point in AB; it is required to determine geometrically, a point C in AC, such, that drawing CD perp. to A B, n times DB minus in times DC
may equal to a given line.
16. Qu. (90) By Mr. R. Majfelt. It is required to construct a triangle, given in species, such, that its three sides and the line bisecting its vert. angle may pass through four given points in the same
17. Qui (91) By Mr.J. Kay, Royton, near Oldham.
Through a given point P, to draw a right line, so that if right lines KD, SR, be drawn from two given points K, S, to make given angles therewith, the rectangle PD X PR may be equal to a given space.
18. Qu. (92) By Mr. A. Hirst, Mursden. If the outer diameter of a spur-wheel containing 1000 teeth be 60 feet, the depth to which the teeth work being 1.3 inches; it is required to determine the outer diameters of two other wheels, one of 100, and the other of 10 teeth, so that they may either drive the great wheel, or be driven by it without shaking: also to infer from hence, whether it be possible, upon true mechanical principles, to make a complete set of what are called geer-wheels of different diameters, which may all work together freely.
19. Qu. (93) By Rylando. Given, the vertical angle, the sum or difference of the sides, and the length of a line drawn from the vertical angle dividing the base in a given ratio; to determine the plane triangle.
20. Qu. (94) By Mr. G. Harvey jun. Plymouth. Three points, A, B, C, are given in position; it is required to determine the position of a fourth, so that n X CP2, minus a certain given space, may be equal to P X AP9 * BP?.
21. Qu. (95) By Mr. Rt. Maar, Holland-Fen. P is a given point in the diameter AB produced of a given circle AE BA; draw DE perp. to AB, meeting
the circumference of the circle in E; join PE, and let S be the centre of the circle. In ED take DN always DE2 x PS
; to determine the equation and area of PE the curve which is the locus of the point N. (Newton's Princip. Prop. 80.)
22. Qu. (96) By Mr. J. Whitley. From four given points in the circumference of a given circle, to draw four right lines to meet in a point in the circumference, so that their continual product may be equal to a given space, or a maximum.
INVESTIGATION OF FORMULA FOR THE VALUES OF THE
TANGENT AND COTANGENT OF AN ARC.
By Mr. G. Harvey, jun. Plymouth. Let x represent any arc or variable angle, radius being unity; then, by the principles of finite differences, we shall have a sin. x = sin.(x + Ax) - sin. x; but, by trigonometry, sin. (x + x) = sin. x cos. A x + cos. I sin. A x; therefore, A sin. x = sin. x cos. A 2 + cos. I sin. Ax—sin. x = cos. x sin. A x cos. A x) sin. x; and, by a similar process, we obtain, Acos=sin. x sin. Ax— (1 cos. A x) cos. x.
Let us now, to simplify, put sin. Ar=m, and 1-cos. Ar=n; then the preceding formulæ for the value of a sin. x = n cos. x -insin, x, and Acos. x =
msin, X - n cos. x, and the first of these equations multiplied by -- n, and the second by-m, produces
n A sin. x = mn cos. x + na sin. x; and — mA cos. x = m2 sin. r + mn cos. x; the sum of these gives -n Asin.x— m Acos. x = m2 sin. x + n’sin. x, whence
n A sin.x — m Acos. x sin, x = m2 + na
, and by a nearly si
m A sin. 3 cos. I milar method, we find cos. x =
m2 + n? sin. x
- Asin. Am A cos. x hence = tang. I =
and m A sin, x n A cos. I
m A sin. 2n A cos. cotang. =
#m Acos. I
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HEN clouds obscure the lunar
rays, And well-known stars no longer shine, To hear'n th' affrighted seaman prays,
In sweet retirement to recline. The warlike Thracians ease require;
The quiverd Parthians' martial train With frequent pray’rs that peace desire,
Which gold and gems can never gain. For neither wealth nor potent sway
Can soothe the tumults of the mind; Nor can the Lictor-chase away
The cares 'round Consuls' hearts entwin'd. That man is bless'd in humble life,
Whose simple meals no pomp display; Whose sweet repose no cares, 'nor strife,
Nor base desires can snatch away. VOL. II.