frre 
Hs cf 
268 Miscellaneous. [May, 
parts being bound and suspended by strong twisted withes. The bridges were as 
may be supposed very vibratory, but were sufficiently strong to admit of the hill 
men carrying their loads of iron or charcoal across them with safety. 
If this short notice of a somewhat ingenious and picturesque object is worth 
publishing in the JouRNAL oF THE Astatic Socrery, it is very much at your ser- 
vice. 
Notre.—We are always happy to give insertion to notices of this nature, and 
especially of the simple inventions and processes of the natives. In the present case, 
we regret that our correspondent has not given us the dimensions and span of his 
rustic bridge. 
2.—Remarks on the Paper on the Trisection of an Angle in No. 14 of the “‘ Jour- 
NAL OF THE ASIATIC SOCIETY.” 
The difficulty of the problem is touched on in the second proposition of the 
paper in question, which is as follows: “To draw the base ofa triangle so that of 
the interior angles at the base, one shall be double of the other, the vertical angle 
of the triangle being a given rectilinear angle, greater than half a right angle.” 
The construction is, to take B F=2 B A, inflect A C=3 A B, from the point 
A on B & and make B C=BF3F C. The writer has failed, as he admits, 
in his first attempted demonstration of this construction, nor in his supplementary 
emendation of it is he more successful. The phrase ‘‘ which it does not,” in line 
12th, and repeated in line 14th, is mere assertion ; the eighteen following lines are 
superfluous; for if the angle K D G is a right angle, the question is settled. 
A numerical example or two will perhaps be the easiest way of convincing Mr. 
Morrieson of his failure. 
Suppose then B=90° B G A is by hypothesis 309 and calling A B=1 AG 
will be equal to 2, and B G=4/4—1 =v 3-1°7320508 ; but A C=3, BC=/9—1 
= /8=2°8284271 and B F=2 #e B C—B F='8284271, § of which is 2761423, 
and this taken from 2, leaves 1°723577, the length of B G by Lt. Morrieson’s shew- 
ing ; but it ought to be 1°7320508. 
If B be taken=45, the limit of Mr. M.’s problem B G will come out by Mr. M.’s 
construction=1°65363908, but it ought to be 1°4142136. 
If 60 be taken, the difference is smaller again ; B G ought to be 1°53207, where- 
as Lt. Morrieson’s construction makes it 1°542579, &c. On the whole the method is 
a very good mechanical rule for trisecting an angle; mathematical solution it is not. 
Mr. M. has hit on the difficulty in the problem which is “ to draw the base of a 
triangle, so that of the angles at the base one shall be double of the otherg.”’—In 
different words, the problem comes to this, ‘‘ To draw a line GA such that G D 
shall be equal to the radius of the circle which has Basa centre and BA asa 
radius,” and this rule will answer in all cases where B is equal to 45. D and A 
coincide when greater D falls between A and G, when less D falls beyond A, and 
farther from G'. Low 
But the problem is not to be solved by straight lines and circles : if a conchoid 
(pl. ix. fig. 2) having A for its pole and B C for its assymptote be described, it will 
cut the circle A D Ein the points D, D’ and D” and straight lines joining these points 
with A, or their extensions will form with B G triangles B G A, B G’ A, BG” 
A, &c. of the species required. This follows from the nature of the curve, in 
which G D is a constant quantity, and here equal to B D the radius of AD E. 
D E drawn parallel to { H gives I E an arc which measures § of AB G’ D’ E! 
