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第63頁

信息發(fā)布者：
?$=\frac {\sqrt 2}2×\frac {\sqrt 2}2+\sqrt 3×\frac {\sqrt 3}2$?
?$=\frac {1}{2}+\frac {3}{2}$?
?$=2$?
?$=2+3×\frac 12-\frac 12-1$?
?$=2+\frac {3}{2}-\frac {1}{2}-1$?
?$=2$?
?$=\frac {\frac 12}{1+\frac {\sqrt 3}2}+\frac 1{\frac {\sqrt 3}3}$?
$=\frac 1{2+\sqrt 3}+\sqrt 3$
$=2-\sqrt 3+\sqrt 3$
$=2$
?$=3×\frac {\sqrt 3}3-\frac 1{\frac 12}+\sqrt 8×\frac {\sqrt 2}2+\sqrt {(1-\sqrt 3)^2}$?
?$=\sqrt 3-2+2+\sqrt 3-1$?
?$=2\sqrt 3-1$?

解：?$(1)$?如圖，點(diǎn)?$E$?即為所求
?$ (2)$?∵?$cos∠DAB=\frac {AE}{AD}$?
∴?$AE=AD · cos 30°=4× \frac {\sqrt{3}}{2}=2 \sqrt{3}$?
∴?$BE=AB-AE=6-2 \sqrt{3}$?
解：?$(1)$?∵?$BD$?平分?$∠ABC$?
∴?$∠DBC=\frac {1}{2}∠ABC=30°$?
∵?$∠ADB$?是?$△BDC$?的一個(gè)外角
∴?$∠ADB=∠DBC+∠C=75°$?
∴?$∠ADB$?的度數(shù)為?$75°$?
?$(2)$?∵?$AE⊥BC$?
∴?$∠AEB=∠AEC=90°$?
∵?$∠ABC=60°$?
∴?$∠BAE=90°-∠ABC=30°$?
∴?$AE=\sqrt {3}BE=3\sqrt {3}$?
∵?$∠C=45°，$??$tanC=\frac {AE}{EC}=1$?
∴?$AE=EC=3\sqrt {3}$?
∴?$EC$?的長(zhǎng)度為?$3\sqrt {3}$?
解：?$s in 15°= s in (45°-30°)= s in 45° · cos 30°-cos 45° · s in 30°$?
?$=\frac {\sqrt{2}}{2} ×\frac {\sqrt{3}}{2} -\frac {\sqrt{2}}{2}×\frac 12=\frac {\sqrt 6-\sqrt 2}4$?
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