数学证明题,α = arc tan (2h / l ) R cos β = R cos α + h R sin β =
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数学证明题,α = arc tan (2h / l ) R cos β = R cos α + h R sin β = l - R sin α
已知α,β < 90° ,
α = arc tan (2h / l )
R cos β = R cos α + h
R sin β = l - R sin α
求证: R = ((h*h + l*l) / 2hl) * √( 4h*h + l*l)
![](http://img.wesiedu.com/upload/b/15/b1509b66b3f18683ffd1dcfb3fd85f8b.jpg)
已知α,β < 90° ,
α = arc tan (2h / l )
R cos β = R cos α + h
R sin β = l - R sin α
求证: R = ((h*h + l*l) / 2hl) * √( 4h*h + l*l)
![](http://img.wesiedu.com/upload/b/15/b1509b66b3f18683ffd1dcfb3fd85f8b.jpg)
![数学证明题,α = arc tan (2h / l ) R cos β = R cos α + h R sin β =](/uploads/image/z/4473027-27-7.jpg?t=%E6%95%B0%E5%AD%A6%E8%AF%81%E6%98%8E%E9%A2%98%2C%CE%B1+%3D+arc+tan+%282h+%2F+l+%29+R+cos+%CE%B2+%3D+R+cos+%CE%B1+%2B+h+R+sin+%CE%B2+%3D)
α=arctan(2h/l)
tnaα=2h/l
cos²α=1/(1+tan²α)=l²/(4h²+l²)
cosα=l/√(4h²+l²)
sinα=tanαcosα=2h/√(4h²+l²)
又
Rcosβ=Rcosα+h (1)
Rsinβ=l-Rsinα (2)
(1)²+(2)²,得,
R²=R²cos²α+2hRcosα +h²+l² -2lRsinα +R²sin²α
2(lsinα-hcosα)R=h²+l²
即2lh/√(4h²+l²)R=h²+l²
R=[(h²+l²)/(2hl)]·√(4h²+l²)
tnaα=2h/l
cos²α=1/(1+tan²α)=l²/(4h²+l²)
cosα=l/√(4h²+l²)
sinα=tanαcosα=2h/√(4h²+l²)
又
Rcosβ=Rcosα+h (1)
Rsinβ=l-Rsinα (2)
(1)²+(2)²,得,
R²=R²cos²α+2hRcosα +h²+l² -2lRsinα +R²sin²α
2(lsinα-hcosα)R=h²+l²
即2lh/√(4h²+l²)R=h²+l²
R=[(h²+l²)/(2hl)]·√(4h²+l²)
数学证明题,α = arc tan (2h / l ) R cos β = R cos α + h R sin β =
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