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精英家教網(wǎng) > 初中地理 > 題目詳情

【題目】一船在燈塔C的正東方向8海里的A處，以20海里/時(shí)的速度沿北偏西30°方向行駛．

（1）多長時(shí)間后，船距燈塔最近?

（2）多長時(shí)間后，船到燈塔的正北方向?此時(shí)船距燈塔有多遠(yuǎn)?(其中:162-82≈13．92)

【答案】(1)12分鐘后，船距燈塔最近；（2）此時(shí)船距燈塔有13.9海里.

【解析】試題分析：（1）作出圖象在直角三角形ACD中求得AD的長后除以速度即可求得時(shí)間；

（2）在直角三角形ACB中求得AB的長后除以速度即可求得時(shí)間，由勾股定理可求出此時(shí)船距燈塔的距離．

試題解析：(1)如圖所示，由題意可知，當(dāng)船航行到D點(diǎn)時(shí)，距燈塔最近，

此時(shí)，CD⊥AB．

因?yàn)椤?/span>BAC=90°-30°=60°，所以∠ACD=30°．

所以AD==4(海里)．

又因?yàn)?/span>4÷20=0．2(小時(shí))=12(分鐘)，

所以12分鐘后，船距燈塔最近．

(2)當(dāng)船到達(dá)燈塔的正北方向的B點(diǎn)時(shí)， BC⊥AC．

此時(shí)∠B=30°，

所以AB=2AC=2×8=16(海里)．

所以16÷20=0．8(小時(shí))=48(分鐘)．

所以BC2=AB2-AC2=162-82≈13． 92．

所以BC≈13．9(海里)．

即48分鐘后，船到燈塔的正北方向，此時(shí)船距燈塔有13.9海里．

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