PROBLEM X. To find the length of any arc of a circle. RULE.* 1. When the chord of the arc and the versed sine of half the arc are given. To 15 times the square of the chord, add 33 times the square of the versed sine,t and reserve the number. To the square of the chord, add 4 times the square of. the versed sine, and the square root of the sum will be twice the chord of half the arc. Multiply twice the chord of half the arc by 10 times the * Demon. Put c= =} the chord of the arc, and v= =the versed sine of half the arc, then the rule may be expressed thus : ✓ (4c2 + v2).1002 ✓(4c2 + 402)+ =2 /(c +02).(1+ 60c2 +3302 10v c2 +02 1+ c? +03 270 60c2 +3302)=216 V V 2Vdo(1+602—270 --270)=2 / 1+6d + 10v w 302 2703 =2 v dvs1 + 607 40d 40d3 +80od) 302 503 Now 2 / dv.(1+6d4002 1112d3+ &c.) is known to be the length of an arc whose diameter is d, and the versed sine of half the arc v; and this differs from the 61v3 preceding only by &c. 5600d3' + Here, as in many other places in the following part of the work, the term versed sine is used instead of versed sine of half the arc, but in all cases of the kind, it is the versed sine of half the arc that is to be understood. square of the versed sine, divide the product by the re served number, and add the quotient to twice the chord of half the arc : the sum will be the length of the arc very nearly. When the chord of the arc, and the chord of half the arc are given.—From the square of the chord of half the arc subtract the square of half the chord of the arc, the remainder will be the square of the versed sine : then pro. ceed as above. 2. When the diameter and the versed sine of half the arc are given. From 60 times the diameter subtract 27 times the versed sine, and reserve the number. Multiply the diameter by the versed sine, and the square root of the product will be the chord of half the arc. Multiply twice the chord of half the arc by 10 times the versed sine, divide the product by the reserved number, and add the quotient to twice the chord of half the arc; the sum will be the length of the arc very nearly. Note 1.-When the diameter and chord of the arc are given, the versed sine may be found thus : From the square of the diameter subtract the square of the chord, and extract the square root of the remainder. Subtract this ront froin the diameter, and half the remainder will give the versed sine of half the arc. 2. The square of the chord of half the arc being divided by the diameter will give the versed sine, or being divided oy the versed sine will give the diameter. 3. The length of the arc may also be found by multiplying together the number of degrees it contains, the radius and the number .01745329. Or, as 180 is to the number of degrees in the arc, so is 3.1416 times the radius, to the length of the arc. Or, as 3 is to the number of degrees in the arc, so is .05236 times the radius, to the length of the arc.* EXAMPLES. 1. If the chord DE be 48, and the versed sine CB 18 what is the length of the arc? Ans. 64.2959. 45252 reserved number. 482=2304=the square of the chord. 182 X 4=1296=4 times the square of the versed sine. ✓ 3600=60=twice the chord of half the arc ACB. 60 x 182 x 10 194400 Now 45252 45252 =4.2959, which added to twice the chord of half the arc gives 64.2959=the length of the arc. 2. Given the diameter CE 50, and the versed sine CD 18, what is the length of the arc? Ans. 64.2959. * When very great accuracy is required, the following theorem may be used. Let d denote the diameter of the circle, and v tie versed sine of half the arc, then the arc=2 X (1+ 302 5v3 35794 6305 + + + +, &c.) od 4002 112d3 115204 281605 + 50 X 60=3000 2514 reserved number. ACNV50 x 18=30=the chord of half the arc. 30 x 2 x 18 x 10 10800 =4.2959, which added to tuic 2514 2514 the chord of half the arc gives 64.2959=the length of the arc ACB. 3. The c! ord of the whole arc is 7, and the versed sine 2, what is t'e length of the arc ? Ans. 8.4343. 4. The ( ord of the whole arc is 40, and the versed sine 15, what i the length of the arc? Ans. 53.5800. 5. Thi: chord of the whole arc is 50, and the chord of half the cc 27, required the length of the arc. Ans. 55.3720. 6. (!ven the diameter of the circle 100, and the versed sine '., required the length of the arc. Ans. 60.9380. 7. Given the chord of the whole arc 16, and the diameter of the circle 20, required the length of the arc. Ans. 18.5439. 8. The diameter of the circle is 50, and the chord of half the arc 30, what is the length of the arc? Ans. 64.2959. 9. The chord of half the arc is 25, and the versed sine 5, required the length of the arc. Ans. 53.5800. PROBLEM XI. To find the area of a circle. RULE I. * Multiply half the circumference by half the diameter, and the product will be the area. • Demon. A circle may be considered as a regular polygon of ad infinite number of sides, the circumference being equal to the perime. ter, and the radius to the perpendicular. But the area of a regular polygon is equal to half the perimeter multiplied by the perpendicular, Or take $ of the product of the whole circumference and diameter. EXAMPLES. 1. What is the area of a circle whose diameter is 42 and circumference 131.946 ? 2)131.946 65 973=} circumference. 21=} diameter. 65973 131946 1385.433=area required. ? What is the area of a circle whose diameter is 10 feet 6 inches, and circumference 31 feet 6 inches? fe. in. 9=15.75=} circumference. 3= 5.25=} diameter. 15 7875 3150 7875 82.6875 12 8.2500 Ans. 82 feet 8 inches and consequently the area of a circle is equal to half the circum ference multiplied by the radius, or half the diameter. Q. E. D. This rule may be otherwise demonstrated by the doctrine of thus ions. |