题目内容
如图,AB是⊙O的直径,弦CD⊥AB,垂足为E,连接AC、BC,若∠BAC=30º,CD=6cm.
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小题1:求∠BCD的度数;
小题2:求⊙O的直径.
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小题1:求∠BCD的度数;
小题2:求⊙O的直径.
小题1:∵直径AB⊥CD,
∴⌒BC=⌒BD
∴∠DCB=∠CAB=30°
小题2:∵直径AB⊥CD,CD=6,
∴CE=3……………………………………………… 4分
在Rt△ACE中,∠A=30°,
∴AC=6 ………………………………………………………………6分
∵AB是直径,∴∠ACB=90°
在Rt△ACB中,AB===4(cm)
(1)由垂径定理知,
=
,∴∠DCB=∠CAB=30°;
(2)由垂径定理知,点E是CD的中点,有CE=
CD=3,AB是直径,∴∠ACB=90°,再求出AC的长,利用∠A的余弦即可求解.
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(2)由垂径定理知,点E是CD的中点,有CE=
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