Find the number of degrees at the head or foot of the table; and the minutes in the left or right-hand column, and in the common angle of meeting, formed by the title placed at the head or foot of the column, and the minutes, you will find the fine, tangent or fecant required. But if the given arch contains any parts of a minute, intermediate to thofe found in the tables; take the difference between the fines, &c. of the given degrees and minutes, and of the minute next greater. Then as I minute is to that difference, fo is the given intermediate part of a minute in decimals to a fourth proportional, Which being added to the fine, &c. first found will give the fine, &c. required. Example. The Log. fine of 1°. 48'. 28". 12", is required. The Log. fine of 1°. 49'. is The Log. fine of 1°. 48′. is 8.5010798 8.4970784 Their difference is 40014 Then as 1 : 40014:: 0.47 (28′′. 12#.): 18807 nearly. Therefore to the Log. fine of 1°. 48′. Add the difference found The fum is the Log. fine of 1°. 48′. 28′′. 12′′ 8.4970784 40014 8.5010798 2. To find the arch anfwering to any given fine, tangent, or fecant. Take the next less found in the tables, and fubtract it from that given, obferving the degrees and minutes answering to it, and divide this remainder by the difference between it and the next greater, adding as many cyphers as are neceffary to the dividend, and the quotient will be the decimal part of a minute to be added to the degrees and minutes before found. Example 1. Suppofe it were required to find the arch anfwering to the Log, fine. 9.8393859 The next lefs is the Log. fine of 43°. 41'.-9.8392719 The difference is 1140 Alfo the difference between the Log. fine of 43°. 41′, and 43°. 42'. is 1322. Then 1322)1140.000(0.86251.43, which being added to 43°. 41′, gives 43°. 41'. 51". 43", for the arch required. Example ponding to the Log. tangent Example 2. Let it be required to find the arch corref The tang. of 25°. 24', the next lefs is The difference is 9.6766687 9.6765426 1261 Also the difference between the Log, tangent of 25°. 24′• and the Log. tangent of 25°. 25. is 3260. Then, 3260)1261.0000(0′. 387— 23". 13′′". Therefore 25°. 24. 23". 13" is the arch anfwering to the given tangent. F IN I S |