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ID:38114050
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时间:2019-06-06
《已知弧弦求半径计算》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、A:B:K=A÷BQ=1Lbl1C=0.0174533RQD=2RSin(Q÷2)L=C÷DKQ=Q-0.1:Goto2:≠>Q=Q+1:Goto1Lbl2C=0.0174533RQD=2RSin(Q÷2)L=C÷DKQ=Q-0.1:Goto2:≠>Q=Q+0.01:Goto3Lbl3C=0.0174533RQD=2RSin(Q÷2)L=C÷DK>L=>Q=Q+0.01:Goto3:≠>Q=Q-0.001:Goto4Lbl4C=0.0174533RQD=2RSin(Q÷2)L=C÷DK>L=>Q=Q-0.001:Goto4:≠>Goto5Lbl
2、5R=A÷0.0174533Q已知弧长C=15,弦长L=14求圆半径R!!!Rn+1=(1+(L-2*Rn*SIN(C/(2*Rn)))/(L-C*COS(C/(2*Rn))))*RnR0=10R1=11.214R2=11.684R3=11.737R4=11.738R5=11.738圆半径R=11.738弧长公式:弧长L=2πr*n/360,(r=圆半径,n=弧长所对圆心角度数)根据三角形三边得sinn/2=7/r消去n180L/πr=n,sin(90L/πr)=7/rsin(15/2r)=7/r<=1,设m=15/2r,r=15/2msinm=14m/15余
3、弦定理得;L^2=2r^2-2r^2*cosn225÷2r^2=1-Cosncosn=1-225÷2r^2=cos15/2rcosm=1-225/2(15/2m)^2=1-225*4m^2/(2*225)=1-2m^2sin^2m+cos^2m=1代入196m^2/225+(1-2m^2)^2=1设m^2=t196t/225+1-4t+t^2=1t^2-704t/225=0t=704/225,m=8√11/15r=15/2m=15*15/16√11=2475√11/16
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