« PreviousContinue »
For let D E be the sine, and CE the versed sine of the arc DC, and since ah — 2 be, by sim. triangles, AB = 2BC, but by the property of tangents to the circle, B C = B D, therefore, BD = DA, and by Simp, on Max. et Min. the rectangle DE x CE is a maximum, - •, . • .
Inference. The arc DC =120°. For drawing the radius DO, since AB=2BC, ABC is evidently half an equilateral triangle whose side is AB; therefore, the /.B=:60o, the supplement of which is the angle DOC = arc DC = 120°.
Otherwise by Messrs. Baines, Brooke, Cattrall, Cummins, Gauithorp, Hirst, Nesbit, Putsey, Rylando, and Webstef'.
Let radius — \, and put the versed sine = x; then, the sine = ^/(2 x — x'), andxv/'(2a; —.»') is to be a max.; make its fluxion = a, and by reduction we have x — versed sine of 120°, the arc required.
10. Qu. (65) Answered by Mr. Vayheeg, the Proposer.
Analysis. Suppose As~~~**\ the thing done, and CT that PACis the circle required; through P and O, the centre of the given circle, draw the right lineCPEOF; then since AP,PD, PB is given, AP, PE,PF is given also. For PD, PB = PE, PF, and because the pointPand the circle EDBF are given in position, PE, PF is given; therefore, AP is equal to a given line, but one extremity P of it is given, consequently, the other extremity A must be in the circumference of a circle whose radius is A P. Consequently AP, PE, PF is equal to AP, P D, PB, the given solid.
Again, by Mr. Whitley.
The point P and the circle EDBF being given, the rectangle PD, PB is given; and since the solid AP, PD, PB is given, AP is given; therefore, a circle described about the centre P, with the given radius A P, will be that required.
n *;" Otherwise by Rylando.
L«t P be the given point, and QBD the" given circle; draw the tangent PQ, and produce it to S, so that QP'. PS may be equal to the given solid; then 6 with radius SP = PT = PA, and centre P, describe the circle ST A, and the thing is done. For the given solid is QP'. PS = QP' x PT = PA. PD. PB, by a known property of the circle.
In the same elegant manner it was answered also by Messrs. Brooke, Buttemorth, Gawthorp, Hint, Maffett, Nesbit, Putsey, Tomlmson, and WinWord.
tU Qu. (66) Answered by Messrs. Eyres and Rylando.
Let DR., CS, be the radii of P, the given circles, and P a point in the required locus; from P draw the tangents PD, PC, and join PR, PS, RS; then, D/ it is evident that when P D2 + I>R2 is constant, PF + DR' (P§2>.+. PC'+ CS* (PS2), that is P R2 + PS2 is constant; and because R and S are constant the locus of P will be a circle.
Cor. I. If the difference DP' —CP' be constant, the locus of P will be a right line perp. to RS.
Cor. 2. By Mr. Whitley. When R and S coincide, RS vanishes, so that, in this case, the locus of Pisa given circle, concentric with the given circles; hence question the 18th (72) is only a particular case of the above.
This question was answered also by Messrs. Brooke, Butterworth, Gawthorp, Hine, Maffett, Hesb.it* Putsey, Kiyheeg, Whitley, and Winward..
-• ~ • ** . i-> ■ ^ -. •. ,-. i K
12: fto. (67) Answered by the Proposer.
Because AE. DC is, by the
question, to be a max. and EDP,
DAP, are right angles, AE =
AD' , AD2
^p-; therefore, ^-p- X DC
must be a max. or AD1 x DC a B
max. which is known to be the case when AD= 2DC.
Hence we have the following '.*
Construction. Divide AC in D, such, that AD S 2 DC, join PD, and draw DEperp. to PD, intersecting BA produced in E, join EC, and the triangle EDC will be a max. as required.
Otherwise by Mr. Rylando. Let ABC be the given triangle, P the given point, in the side AB, and draw the rest of the lines as ia thefigure above. Put AE =: x, PA = m, and AC=/>; thenjAD^^njjr, the area of the triangle C AE =z.£px, and the area of the triangle D AE= \x,Jmx; therefore, by the question \px—\x </mx is a max. and its
fluxion = 0, whence x is found = -~, AD = \p, and
the area of the triangle =r ,-f—. To construct the ° 27 m
problem; make AD = f-AC, join PD, and bisect it
in m, draw mn perp. to PD, meeting PE in n, with
centre n, and radius P», describe a circle, which will
cut PA produced in E, the point required.
Answers were also given by Messrs. Gawthorp, Hint,
Maffett, Nesbit, Putsey, Whitley, and Wihward.
13. Qu. (68) Answered by Mr. Nesbit, Farnley Academy.
Put the area, length and weight of one beam respectively A, L, and W, that of the other a, I, and ;:, and let S, s, G, g, be as i n the question; then, by art. 225,
Marrat's Mechanics, , „ tl. = —; —j but
L x W I x w ■
when the lengths are equal or L'= I, the weights
are as the areas, or W : w :: A : a, therefore, the
above equation, under these circumstances, becomes * x G = S x g, that is, S : s :: G : g. St.E:D.
And in nearly the same manner it was answered by Messrs. Gawthorp, Harvey, Rylando, and Whitley.
14s. Qv. (69) Answered by Mr. G. Harvey, jun, Plymouth.
Analysis. Suppose the thing done, and that ABC is the triangle required; because N C, the diff. of the seg, of the base, and NCF, the diff. of the angles at the base are given, HF, the diairt-oftbecircumscribingrcircle
is given; and because LK = MC ^ is given, ML drawn parallel to HF is given in position, and the centre of the inscribed circle is in this line, by the question; but the centre of the inscribed circle is known also to be in the lins CF, itis therefore at S, their point of intersection.
Now, by prop. 35. Mod. Geom. Student. CS x SF = HF x SL, and CS, SF, HF are all given; therefore LS is given; hence, if a circle be described from centre S with radius SL thus found, and CA, CB, be drawn from C, touching it, and cutting the circle ACB "in A and B, and A,B, be joined, ABC will be the triangle required; as is evident from the analysis.
The same, by Messrs. Rylando and Whitley.
Analysis. Suppose the thing done, and that ABC is the triangle required; then, since the diff. of the angles at the base, and the diff. of the seg. of the base are given, the triangle CDF is given; draw CH perp. to CF, meeting FD produced in H, then HD, and the diam. HF are given.
", But KL is given, and from thence EL; also by prop. II. Mod. Geom. Student. EL2==EKx EI; therefore, EI becomes known, and of course IK; but PxDE = EKxlK; therefore, D E become* known, and the method of construction is evident.
Solutions were given also by Messrs. Butterworthj^ Eyres, Hine, Webster, and Winward.
15. Qu. (70) Answered by Mr. Westi&
In the annexed figure let H O denote the horizon, E Q. the equator, 69 the tropic of Cancer, N the north and S the south pole, F the town of Farnley, and D the point in which the sun's upper limb would appear to a person elevated to A, on June 21st, at midnight.
Now, from the co-latitude = 36° 13', substract the declination =23° 28', and we obtain 12* 45', the depression of the sun's centre below the horizon at midnight, from which take 15' 47"+ 35' = 48' 47", the semi-diameter and refraction, and add 9" for parallax, and we get 11° 56' 22" = z. O C 69 = C ACD. Then, in the right-angled triangle ADC, right-angled at D; as the nat. co-sine of 11° 56^ 22" ; CD = 3985 :: rad. : AC = 4073.114-1 nearly; from which take 3985, and there remains AFz=88,ll41 miles, the height to which a person must be elevated, to see the sun's upper limb at the given time.
Put CF = r = 21040800 feet, the radius of the earth; AC = a= 21506042.448 feet, the distance fallen from; g — 16-jSj feet, half the velocity or force at F; v — the velocity required, and t = the time of falling from A to F; then (see Dr. Hutton's Course, vol. ii. p. 835 ; Marrat's Mechanics, art. 665; or Vince's Flux. art. 81)