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第82頁(yè)

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$解：如圖，過(guò)點(diǎn)?A?作?AE⊥BC?于點(diǎn)?E?$

$∵?AB=DC$

$?∴梯形?ABCD?是等腰梯形$

$∴?BC=AD+2BE?$

$在?Rt△ABE?中，?BE=AE÷tan B?$

$∴?BC=AD+2BE=16+12÷tan 55°≈24.4(\mathrm {cm})$

$解：∵?AB?的坡度為?1∶\sqrt 3，?即?tan∠ABE=\frac {\sqrt 3}3?$

$∴?∠ABE=30° ?∴?AE=\frac 12AB=100m?$

$?CE=100-20=80m，??ED=4CE=320m?$

$?CD=\sqrt {CE^2+ED^2}=80\sqrt {17}m?$

$解：過(guò)點(diǎn)?B?作?BE⊥AD?于點(diǎn)?E，??BF⊥CD?于點(diǎn)?F?$

$由題意可得?AB=120m，??BC=160m?$

$∵?sin 10°=\frac {BE}{AB}，??sin 15°=\frac {CF}{BC}?$

$∴?BE=AB · sin 10°=120×0.17=20.4m?$

$?CF=BC · sin 15°=160×0.26=41.6m?$

$則點(diǎn)?C?相對(duì)于起點(diǎn)?A?升高了?BE+CF=62.0m?$

$解：過(guò)點(diǎn)?D?作?DE⊥BC?于點(diǎn)?E?$

$?sin∠C=\frac {DE}{CD}?$

$∴?DE=CD · sin 18°≈20×0.31=6.2m?$

$∴?AF=DE=6.2m?$

$則?\frac {AF}{BF}=3：??4，??BF=\frac {4×AF}3≈8.3m?$

$∴?AB=\sqrt {AF^2+BF^2}≈10.3m?$

$答：斜坡?AB?的長(zhǎng)為?10.3m。?$

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