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一种路灯的示意图如图所示,其底部支架AB与吊线FG平行,灯杆CD与底部支架AB所成锐角$α=15^{\circ }$. 顶部支架EF与灯杆CD所成锐角$β=45^{\circ }$,则EF与FG所成锐角的度数为( ).
A. $60^{\circ }$
B. $55^{\circ }$
C. $50^{\circ }$
D. $45^{\circ }$
A. $60^{\circ }$
B. $55^{\circ }$
C. $50^{\circ }$
D. $45^{\circ }$
答案：
A 过点 $E$ 作 $EH // AB$,
$ \because AB // FG $, $EH // AB $,
$ \therefore AB // EH // FG $,
$ \therefore \angle BEH = \alpha = 15^{\circ} $,
$ \angle FEH + \angle EFG = 180^{\circ} $,
$ \because \beta = 45^{\circ} $, $ \therefore \angle FEH = 180^{\circ} - 45^{\circ} - 15^{\circ} = 120^{\circ} $,
$ \therefore \angle EFG = 180^{\circ} - \angle FEH = 180^{\circ} - 120^{\circ} = 60^{\circ} $, $ \therefore EF $ 与 $FG$ 所成锐角的度数为 $60^{\circ}$.
A 过点 $E$ 作 $EH // AB$,
$ \because AB // FG $, $EH // AB $,
$ \therefore AB // EH // FG $,
$ \therefore \angle BEH = \alpha = 15^{\circ} $,
$ \angle FEH + \angle EFG = 180^{\circ} $,
$ \because \beta = 45^{\circ} $, $ \therefore \angle FEH = 180^{\circ} - 45^{\circ} - 15^{\circ} = 120^{\circ} $,
$ \therefore \angle EFG = 180^{\circ} - \angle FEH = 180^{\circ} - 120^{\circ} = 60^{\circ} $, $ \therefore EF $ 与 $FG$ 所成锐角的度数为 $60^{\circ}$.
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