Winchester Republican

MATHEMATICAL SOLUTIONS.

Mr. Editor, In your paper of 12th May last, I noticed three questions. Please to insert the following solutions, and oblige, sir,
Yours, &c.
TH. CURRY.

[The printer's unable to give the first, from a want of suitable types. The following were the second and third questions. 2d A person is offered ground at the rate of 3 half pence per foot in front and 1 penny per foot in depth. He is desirous to have (in form of a right angled parallelogram) an area 3456 feet for the least money possible. Required the dimensions thereof? 3d Required the time between 8 and 9 o'clock, when the hour and minute hands of a watch are at right-angles; with a general theorem for finding they shall many given angle in any given hour?]

Solution 2nd. Put x for the front, then will x * (3456 / x) = those in depth
3x + (3456 / x) = the rate of purchase.
By the question must be to find x, so that the fluxion &c. gives x=48 feet, width in front, and 72 feet in depth.

Solution 3rd. At 8, the hour hand is 40 minutes of time before the other, and 90°=15 minutes of time: therefore 40 - 15z = 25, the minutes of time on the watch which the minute hand must overtake the other; and in 60 minutes it overtakes the other 55 minutes: therefore as 11 : 12 :: 25 : 27 3/11ths, the minutes after 8 required.

General Theorem. Change the given angle into minutes of time; and if they be less than the minutes which the hour hand is before the minute hand, take their difference: but if greater, take their sum (but if the sum would exceed 60 minutes you must then likewise take the difference) and say, as 11 : 12 so is the sum or difference to the minutes required to form the angle. Or thus: It is evident the hands are conjoined at 12 o'clock, and their next conjunction is at 5 5/11 minutes after 1 o'clock. Therefore as 12 : 60 :: 5 5/11 : 27 3/11, the time required.