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13. 如图,在△ABC中,CD⊥AB,垂足为D.若AB=12,CD=6,tan A = $\frac{3}{2}$,求sin B + cos B的值.
答案
解:在 $\mathrm{Rt} △ A C D$ 中, $\because ∠ A D C=90°,$
$\therefore \tan A=\dfrac{C D}{A D}=\dfrac{6}{A D}=\dfrac{3}{2},$
$\therefore A D=4, \therefore B D=A B-A D=12-4=8.$
在 $\mathrm{Rt} △ B C D$ 中,
$\because ∠ B D C=90°, B D=8, C D=6$, 根据勾股定理,
得 $B C=\sqrt{B D^{2}+C D^{2}}=10,$
$\therefore \sin B=\dfrac{C D}{B C}=\dfrac{3}{5}, \cos B=\dfrac{B D}{B C}=\dfrac{4}{5},$
$\therefore \sin B+\cos B=\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}.$
$\therefore \tan A=\dfrac{C D}{A D}=\dfrac{6}{A D}=\dfrac{3}{2},$
$\therefore A D=4, \therefore B D=A B-A D=12-4=8.$
在 $\mathrm{Rt} △ B C D$ 中,
$\because ∠ B D C=90°, B D=8, C D=6$, 根据勾股定理,
得 $B C=\sqrt{B D^{2}+C D^{2}}=10,$
$\therefore \sin B=\dfrac{C D}{B C}=\dfrac{3}{5}, \cos B=\dfrac{B D}{B C}=\dfrac{4}{5},$
$\therefore \sin B+\cos B=\dfrac{3}{5}+\dfrac{4}{5}=\dfrac{7}{5}.$
14. 如图,在$△ ABC$中,$AD$是$BC$边上的高,$AE$是$BC$边上的中线,$∠ C=45°,\sin B=\dfrac{1}{3},AD=1$.
(1)求$BC$的长.
(2)求$\tan∠ DAE$的值.
(1)求$BC$的长.
(2)求$\tan∠ DAE$的值.
答案
解:(1)在 $△ A B C$ 中, $\because A D$ 是 $B C$ 边上的高,
$\therefore ∠ A D B=∠ A D C=90°.$
在 $△ A D C$ 中, $\because ∠ A D C=90°, ∠ C=45°, A D=1,$
$\therefore D C=A D=1.$
在 $△ A D B$ 中, $\because ∠ A D B=90°, \sin B=\dfrac{1}{3}, A D=1,$
$\therefore A B=\dfrac{A D}{\sin B}=3,$
$\therefore B D=\sqrt{A B^{2}-A D^{2}}=2 \sqrt{2},$
$\therefore B C=B D+D C=2 \sqrt{2}+1.$
(2) $\because A E$ 是 $B C$ 边上的中线,
$\therefore C E=\dfrac{1}{2} B C=\sqrt{2}+\dfrac{1}{2},$
$\therefore D E=C E-C D=\sqrt{2}+\dfrac{1}{2}-1=\sqrt{2}-\dfrac{1}{2},$
$\therefore \tan ∠ D A E=\dfrac{D E}{A D}=\dfrac{\sqrt{2}-\dfrac{1}{2}}{1}=\sqrt{2}-\dfrac{1}{2}.$
$\therefore ∠ A D B=∠ A D C=90°.$
在 $△ A D C$ 中, $\because ∠ A D C=90°, ∠ C=45°, A D=1,$
$\therefore D C=A D=1.$
在 $△ A D B$ 中, $\because ∠ A D B=90°, \sin B=\dfrac{1}{3}, A D=1,$
$\therefore A B=\dfrac{A D}{\sin B}=3,$
$\therefore B D=\sqrt{A B^{2}-A D^{2}}=2 \sqrt{2},$
$\therefore B C=B D+D C=2 \sqrt{2}+1.$
(2) $\because A E$ 是 $B C$ 边上的中线,
$\therefore C E=\dfrac{1}{2} B C=\sqrt{2}+\dfrac{1}{2},$
$\therefore D E=C E-C D=\sqrt{2}+\dfrac{1}{2}-1=\sqrt{2}-\dfrac{1}{2},$
$\therefore \tan ∠ D A E=\dfrac{D E}{A D}=\dfrac{\sqrt{2}-\dfrac{1}{2}}{1}=\sqrt{2}-\dfrac{1}{2}.$
15. 如图,在$△ ABC$中,AD是边BC上的高,$\sin C=\dfrac{\sqrt{2}}{2}$,$\tan B=\dfrac{1}{2}$,$AD=2$.
(1)求$\cos∠ BAD$的值.
(2)求$△ ABC$的面积.
(1)求$\cos∠ BAD$的值.
(2)求$△ ABC$的面积.
答案
解:(1)在 $\mathrm{Rt} △ A B D$ 中, $\tan B=\dfrac{A D}{B D}=\dfrac{1}{2}, A D=2,$
$\therefore B D=4,$
$\therefore A B=\sqrt{A D^{2}+B D^{2}}=2 \sqrt{5},$
$\therefore \cos ∠ B A D=\dfrac{A D}{A B}=\dfrac{\sqrt{5}}{5}.$
(2) $\because \sin C=\dfrac{\sqrt{2}}{2}, \therefore ∠ C=45°.$
$\because \tan C=\dfrac{A D}{C D}, A D=2,$
$\therefore C D=2, \therefore B C=B D+C D=6,$
$\therefore S_{△ A B C}=\dfrac{1}{2} × A D × B C=6.$
$\therefore B D=4,$
$\therefore A B=\sqrt{A D^{2}+B D^{2}}=2 \sqrt{5},$
$\therefore \cos ∠ B A D=\dfrac{A D}{A B}=\dfrac{\sqrt{5}}{5}.$
(2) $\because \sin C=\dfrac{\sqrt{2}}{2}, \therefore ∠ C=45°.$
$\because \tan C=\dfrac{A D}{C D}, A D=2,$
$\therefore C D=2, \therefore B C=B D+C D=6,$
$\therefore S_{△ A B C}=\dfrac{1}{2} × A D × B C=6.$
16. [数学文化]第二十四届国际数学家大会会徽的设计基础是1 700多年前中国古代数学家赵爽的“弦图”.如图,在由四个全等的直角三角形（$△ DAE$，$△ ABF$，$△ BCG$，$△ CDH$）和中间一个小正方形$EFGH$拼成的大正方形$ABCD$中，$∠ ABF ＞ ∠ BAF$，连接$BE$.设$∠ BAF = α$，$∠ BEF = β$，若正方形$EF-GH$与正方形$ABCD$的面积之比为$1:n$，$\tan α = \tan^2 β$，则$n=$（
A.5
B.4
C.3
D.2
C
）A.5
B.4
C.3
D.2
答案
16. C
