【#第一文档网# 导语】以下是®第一文档网的小编为您整理的《已知弧弦求半径计算》,欢迎阅读!
A:B: K=A÷B Q=1 Lbl 1 C=0.0174533RQ D=2RSin(Q÷2) L=C÷D K Q=Q-0.1:Lbl 2 C=0.0174533RQ D=2RSin(Q÷2) L=C÷D K Q=Q-0.1:Lbl 3 C=0.0174533RQ D=2RSin(Q÷2) L=C÷D K>L=> Q=Q+0.01:Lbl 4 C=0.0174533RQ D=2RSin(Q÷2) L=C÷D Goto2:Goto2:Goto3:≠>Q=Q+1:Goto1 ≠>Q=Q+0.01:Goto3 ≠> Q=Q-0.001:Goto4 K>L=> Q=Q-0.001:Goto4:≠> Goto5 Lbl 5 R=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))))*Rn R0=10 R1=11.214 R2=11.684 R3=11.737 R4=11.738 R5=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n 180L/πr=n, sin(90L/πr)=7/r sin(15/2r)=7/r<=1,设m=15/2r,r=15/2m sinm=14m/15 余弦定理得;L^=2r^-2r^*cosn 222225÷2r^=1-Cosn cosn=1-225÷2r^2=cos15/2r cosm=1-225/2(15/2m)^2=1-225*4m^2/(2*225)=1-2m^2 sin^2m+cos^2m=1 代入196m^2/225+(1-2m^2)^2=1 设m^2=t 196t/225+1-4t+t^2=1 t^2-704t/225=0 t=704/225,m=8√11/15 r=15/2m=15*15/16√11=2475√11/16 2 本文来源:https://www.dywdw.cn/802fb82bb62acfc789eb172ded630b1c58ee9b42.html