For let DE be the sine, and CE the versed sine of the arc DC, and since ab2bc, by sim. triangles, AB 2BC, but by the property of tangents to the circle, BCB D, therefore, BDDA, and by Simp. on Max. et Min. the rectangle DE CE is a maxi

mum,

Inference. The arc DC 120°. For drawing the radius DO, since AB2BC, ABC is evidently half an equilateral triangle whose side is AB;, therefore, the B 60°, the supplement of which is the angle DOC arc DC= 120°.

Otherwise by Messrs. Baines, Brooke, Cattrall, Cummins, Gawthorp, Hirst, Nesbit, Putsey, Rylando, and Webster.

Let radius=1, and put the versed siner; then, the sine (2 x − x2), and √(2 x − x2) is to be a max.; make its fluxiono, and by reduction we have versed sine of 120°, the arc required.

10. Qu. (65) Answered by Mr. Vayheeg, the Proposer. ANALYSIS. Suppose

the thing done, and that PAC is the circle required; through P and O, the centre of the given circle, draw the right line CPEOF; then since AP, PD, PB is given, AP, PE, PF

P

is given also. For PD, PB PE, PF, and because the point P and the circle EDBF are given in position, PE, PF is given; therefore, A P 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 AP. Consequently AP, PE, PF is equal to AP, PD, 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 AP, will be that required.

Let 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

with radius SPPT PA, and centre P, describe the circle STA, 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, Butterworth, Gawthorp, Hine, Maffett, Nesbit, Putsey, Tomlinson, and Winward.

11. Qu. (66) Answered by Messrs. Eyres and Rylando.

Let DR, CS, be the radii of 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 PD2+ DR is constant, PD2 + DR2 (PR) +PC+CS2 (PS2),

R

P

that is PR PS' is constant; and because R and S are constant the locus of P will be a circle.

Cor. 1. 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 P is a 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, Nesbit, Putsey, Vayheeg, Whitley, and Winward.

12. Qu. (67) Answered by the Proposer.

Because AE. DC is, by the

question, to be a max. and EDP,

DAP, are right angles, AE =

therefore,

A D2 AP;

A D2

AP

× DC

must be a max. or AD2 x DC a

max. which is known to be the case when AD 2DC. Hence we have the following

Construction. Divide AC in D, such, that AD= 2 DC, join PD, and draw DE perp. 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 in the figure above. Put A Ex, PA=m, and AC=p; then, AD/mx, the area of the triangle C AE pr, and the area of the triangle DAE= x/mx; therefore, by the question px-xmx is a max. and its fluxion 0, whence x is found =

4p2

AD=3p, and

9 m2

To construct the

problem; make ADAC, join PD, and bisect it in m, draw mn perp. to PD, meeting PE in n, with centre n, and radius Pn, describe a circle, which will cut PA produced in E, the point required.

Answers were also given by Messrs. Gawthorp, Hine, Maffett, Nesbit, Putsey, Whitley, and Winward.

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, l, and w, and let S, s, G, g, be as in the question; then, by art. 225, SXAX G Sxaxg Marrat's Mechanics,

Lx W

χω

but

when the lengths are equal or L, the weights are as the areas, or W:w:: A: a, therefore, the

above equation, under these circumstances, becomes sx GSX g, that is, Ss :: G: g. 2. E. D. And in nearly the same manner it was answered by Messrs. Gawthorp, Harvey, Rylando, and Whitley.

14. Qu. (69) Answered by Mr. G. Harvey, jun. Plymouth.

ANALYSIS. Suppose the thing done, and that ABC is the triangle required; because NC, the diff. of the seg, of the base, and NCF, the diff. of the angles at the base are given, HF, the diam. of the circumscribing cirele is given; and because LKMC is given, ML drawn parallel to HF is given in position, and the

centre of the inscribed circle is

A

N

DM

ELK B.

F

in this line, by the question; but the centre of the inscribed circle is known also to be in the line CF, it is therefore at S, their point of intersection.

Now, by prop. 35. Mod. Geom. Student. CS x SF HF 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.111 Mod. Geom. Student. ELEK × EI; therefore, EI becomes known, and of course IK; but HD × DE EK x IK; therefore, DE becomes known, and the method of construction is evident.

Solutions were given also by Messrs. Butterworth Eyres, Hine, Webster, and Winward.

15. Qu. (70) Answered by Mr. Nesbit, Farnley Academy.

In the annexed figure let

HO denote the horizon, EQB

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" + 33" 48' 47", the semi-diameter and refraction, and add 9" for parallax, and we get 11° 56′ 22′′ 4 OC 69 = 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 4073.1141 nearly; from which take 3985, and there remains AF 88,1141 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; 816 feet, half the velocity or force at F; the velocity required, and t the time of falling from A to F; then (see Dr. Hutton's Course, vol. ii. p. 335; Marrat's Mechanics, art. 665; or Vince's Flux. art. 81)

4gr x=5411 feet per second. Again, with the radius AG GC, describe the semicircle ABC; draw B F perpendicular to AC, and join BG; then, we have BG 10753021.224, and GoF= 10287778.776, to find the angle BGF 16° 54-56, and side BF 3128751.6. We also find the arc ABC 33781691.4773; and, as 180° : 33781691.4773 :: 16° 54′ 56: 3174644.883 the arc AB; then, per AB4BF 1RO

idem, ut supra, t =

4g

CF

=173" =

253", the whole time of falling to the surface. Y