A complete and practical solution book for the common school teacher . the MENSURATION. 186 PROBLEM 380. If 2 ft. be the radius of a circle, how far from the center must achord be drawn parallel! to divide the semicircle into two equal parts? Solution. (1) Let ADB be the semicircle, AB the diameter, GH the line thatdivides the semicircle into twoequal parts. (2) On OD, perpendicular to AB, let OF=;t, the distance required,and 0=angle HOB. r=OB, orradius. (3) Then *=rsin0, FH=rcos0. (4) Area of sector HOB=^-20. (5) Area of sector DOH=|r2(^r—0). (6) Area of triangle FOH=^r2sin0cos0. (7) We then
A complete and practical solution book for the common school teacher . the MENSURATION. 186 PROBLEM 380. If 2 ft. be the radius of a circle, how far from the center must achord be drawn parallel! to divide the semicircle into two equal parts? Solution. (1) Let ADB be the semicircle, AB the diameter, GH the line thatdivides the semicircle into twoequal parts. (2) On OD, perpendicular to AB, let OF=;t, the distance required,and 0=angle HOB. r=OB, orradius. (3) Then *=rsin0, FH=rcos0. (4) Area of sector HOB=^-20. (5) Area of sector DOH=|r2(^r—0). (6) Area of triangle FOH=^r2sin0cos0. (7) We then have r20-r-r2sin0cos0=r2(^r— 0) — FIG. 67. (8) Transposing and reducing we have sin20-|-20 2* (9) We will find the value of 0 by double position, and it isfound as follows: (10)(ID Let 0=2Oo=<r for first trial. 2tT TV =o« or 9 . sin40 .64279. (12) .-. .64279= .87267. (13) .87267—.64279=.22988llir90- si n40°=^-=. 87267,lo but sin40°= (14) Let0=22c (15) 11. (1) 2& .-. Sin44+ — = -, or si n44°=^-=. _ yu (16) .-. .69466=.80285. (17) .80285—.69466=.10819 ... (2). (18) (1)—(2)=.12169, 22°—20°=2°. (19) /.: 2° :: .22988 : .12169. (20) .\ <£=° 46 47. (21) Sin47° 3334+47°3334=90°. (22) Sin47°3334=42°2626=|||^=.74073. (23) .. .73798=74073. 186 FAIRCHIL&S SOLUTION BOOK. (24) .74073—.73798 = .00275 . . (3). (25) 23° 46 47—22°=1° 46 47=.6407. (26) (2) —(3) =.10544. (27) : .6407 :: .10819 : .10544. (28) .*. <£=.6574=1°4934. (29) .\0=23°4934, andsin47°398=42°2052=?|^=.73911. (30) .-. .73907=.73911. (31) .73911—.73907=.00004. (32) 23° 49
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