A complete and practical solution book for the common school teacher . 4 FIG. 6K. k, or0=15c o (4) Angle KOT=X=90°. (5) .-. A— 0=74°3813=KOB (6) Sin^= .96426+. (7) .. AB=DT the = rd. (8) KO=rsin0= rd. .*. rd. is the length of the fence, and distance from the center. PROBLEM circle containing- one acre is cut by another whose center is onthe circumference of the given circle, and the area common to both is\ acre: find the radius of the cutting- circle. Solution. (1) Designate by O the center of thecircle, and by A the center of thecutt
A complete and practical solution book for the common school teacher . 4 FIG. 6K. k, or0=15c o (4) Angle KOT=X=90°. (5) .-. A— 0=74°3813=KOB (6) Sin^= .96426+. (7) .. AB=DT the = rd. (8) KO=rsin0= rd. .*. rd. is the length of the fence, and distance from the center. PROBLEM circle containing- one acre is cut by another whose center is onthe circumference of the given circle, and the area common to both is\ acre: find the radius of the cutting- circle. Solution. (1) Designate by O the center of thecircle, and by A the center of thecutting = R, the A as a center and radius R,draw the arc ECD, cutting the cir-cumference at E and D.(4) Join OD and AE. FIG. 69. (2)(3). 188 FAIR CHILDS SOLUTION BOOK. (5) Put r=OA = ±( — rd., radius of the 1st circle. (6) Put angle EAO = angle OEA=0. (7) Then angle EOA=tt—20. (8) Sector EOA:= ^ -, sector EAC = —~—, triangle EAO=iRrsin0. /en tu u ^2(tt-20) ,ttR20 Trr2 (9) Then we have ~ -\—5 %Krsmd=—r. (10) But R=2rco&6. (11) Substituting the value of R, we have <jrr2(7r—20) -j- 4^2cos20—wr2sin20=-£*rV2, or 20cos20—sin20=-|ir, orsin20—20cos0=-i?r ... (1). (12) Let <£=20, and the equation becomes, sin<£—<£cos£= *r • - . (2). (13) It is obvious that <£ cannot be less than \-x. Let us as- sume =\Tr, and substitute this in (2). (14) Then sin<£ = l. ^ttcos = 135°=f*r, and sin=. 70711 —f cos<fr—1 66609 , which is greaterthan \v. (16) Then, 135° ^. L 90°_ L : 45° :: .57080 : °, the correction tobe added to the 1st assumed value of <[&
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