$解：如圖，?∠C=∠E=90°，??∠A=60°，??∠B=30°，??∠D=∠F=45°?$
$∴?AB=2AC，??DE=EF?$
$∴?BC=\sqrt {AB^2-AC^2}=\sqrt 3AC，??DF=\sqrt {DE^2+EF^2}=\sqrt 2DF?$
$∴?sin 60°=sinA=\frac {BC}{AB}=\frac {\sqrt 3}2，??cos 60°=cosA=\frac {AC}{AB}=\frac 12?$
$?tan 60°=tan A=\frac {BC}{AC}=\sqrt 3?$
$?sin 30°=sinB=\frac {AC}{AB}=\frac 12，??cos 30°=cosB=\frac {BC}{AB}=\frac {\sqrt 3}2?$
$?tan 30°=tanB=\frac {AC}{BC}=\frac {\sqrt 3}3?$
$?sin 45°=sinD=\frac {EF}{DF}=\frac {\sqrt 2}2，??cos 45°=cosD=\frac {DE}{DF}=\frac {\sqrt 2}2?$
$?tan 45°=tan D=\frac {EF}{DE}=1?$
