如图,将含30角的直角三角形ABC(角A=30)绕其顶点C顺势针旋转A度(0-90)得到RT三角形A'B'C,A'C与AB交与点D过点D作DE平行A'B',交C'D'于点E,连接BE,已知,在旋转过程中,三角形BDE为直角三角形,设BE=1,AD=X,
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如图,将含30角的直角三角形ABC(角A=30)绕其顶点C顺势针旋转A度(0-90)得到RT三角形A'B'C,A'C与AB交与点D过点D作DE平行A'B',交C'D'于点E,连接BE,已知,在旋转过程中,三角形BDE为直角三角形,设BE=1,AD=X,
如图,将含30角的直角三角形ABC(角A=30)绕其顶点C顺势针旋转A度(0-90)得到RT三角形A'B'C,A'C与AB交与点D
过点D作DE平行A'B',交C'D'于点E,连接BE,已知,在旋转过程中,三角形BDE为直角三角形,设BE=1,AD=X,三角形BDE的面积为S.求
1.当A=30°时求X的值
2.求S与X的函数关系式,并写出X的取值范围.
3.以点E为圆心,BE为半径做圆E,当S=1/4S三角形ABC时,判断圆E与A'C的位置关系,并求相应的tanA的值
如图,将含30角的直角三角形ABC(角A=30)绕其顶点C顺势针旋转A度(0-90)得到RT三角形A'B'C,A'C与AB交与点D过点D作DE平行A'B',交C'D'于点E,连接BE,已知,在旋转过程中,三角形BDE为直角三角形,设BE=1,AD=X,
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NEW YORK, July 18, 2008 (AFP) - Oscar-nominated Mexican actress Salma Hayek and her French billionaire businessman fiance Fran?ois-Henri Pinault have called off their engagement, a spokeswoman s...
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NEW YORK, July 18, 2008 (AFP) - Oscar-nominated Mexican actress Salma Hayek and her French billionaire businessman fiance Fran?ois-Henri Pinault have called off their engagement, a spokeswoman said Friday, July 18th.\x0d
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Salma Hayek\x0d
We are sad to announce the engagement of Salma Hayek and Francois-Henri Pinault has been canceled. There will be no further comment," her representative in New York, Cari Ross, said.
The couple announced plans to wed back in March 2007.
They met in Italy in 2006,wedding dresses, and have a daughter, Valentina, who was born in September 2007.
Hayek was nominated for an Oscar in 2003 for her portayal of Mexican painter Frida Kahlo in "Frida." Pinault heads the luxury group PPR.
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没图怎么说
(1)∵∠α=30°,∠DAC=30°∴DA=DC
∵∠BCD=∠DBC=60°∴DC=DB=BC
∴AD=BC ∵BC=1∴AD=1 ∴x=1
(2)S=负根号3除以6+根号3除以3x(0<x<2)