在直角△ABC中,∠C=90°,∠ABC=30°,D是邊AC的中點,則sin∠DBA= .
【答案】
分析:過點D作DE⊥AB于點E,將求sin∠DBA的問題轉(zhuǎn)化到Rt△BDE中求解,即求
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的值,設(shè)AB=2x,則AC=x,BC=
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,又△ABC,△ADE都是30°的直角三角形,可求DE,用勾股定理可求BD.
解答:
解:過點D作DE⊥AB于點E,
∵在Rt△ABC中,∠C=90°,∠ABC=30°,
∴sinA=
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,
設(shè)AB=2x,則AC=x,BC=
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,
又∵D是邊AC的中點,
∴AD=CD=
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,
在Rt△DBC中,BD
2=BC
2+CD
2=
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,
∴BD=
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,
在Rt△ADE中,DE=AD•sinA=
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,
在Rt△BDE中,sin∠DBA=
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.
故本題答案為:
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.
點評:求銳角的三角函數(shù)值的方法:利用銳角三角函數(shù)的定義,通過設(shè)參數(shù)的方法求三角函數(shù)值.