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15. 如图,$A$,$B$,$C$是$\odot O$上的三点,若$△ OBC$是等边三角形,则$\cos ∠ A =$
$ \frac{\sqrt{3}}{2} $
.答案
15. $ \frac{\sqrt{3}}{2} $
16. 如图,正五边形$ABCDE$内接于$\odot O$,$P$为$\overset{\frown}{DE}$上的一点(点$P$不与点$D$重合),则$∠ CPD$的度数为
$ 36^{\circ} $
.答案
16. $ 36^{\circ} $
17. (8分)计算:$2\tan 45^{\circ} - \dfrac{1}{\sin 30^{\circ}} - 2\sin^{2}60^{\circ} + (\dfrac{2}{3})^{-1}$.
答案
17. 解: 原式 $ = 2 × 1 - 2 - 2 × \frac{3}{4} + \frac{3}{2} = 0 $.
18. (10分)如图,在$△ ABC$中,$∠ ACB = 90^{\circ}$,$CD ⊥ AB$交$AB$于点$D$,$AD = 4$,$BD = 9$,求$\tan A$.
答案
18. 解: $ \because ∠ ACB = 90^{\circ}, CD ⊥ AB $,
$ \therefore ∠ B + ∠ A = ∠ A + ∠ ACD = ∠ B + ∠ BCD = 90^{\circ} $.
$ \therefore ∠ B = ∠ ACD, ∠ BCD = ∠ A $.
$ \therefore △ BCD ∽ △ CAD $.
$ \therefore CD^{2} = AD · BD = 36 $.
$ \therefore CD = 6 $.
$ \therefore \tan A = \frac{CD}{AD} = \frac{6}{4} = \frac{3}{2} $.
$ \therefore ∠ B + ∠ A = ∠ A + ∠ ACD = ∠ B + ∠ BCD = 90^{\circ} $.
$ \therefore ∠ B = ∠ ACD, ∠ BCD = ∠ A $.
$ \therefore △ BCD ∽ △ CAD $.
$ \therefore CD^{2} = AD · BD = 36 $.
$ \therefore CD = 6 $.
$ \therefore \tan A = \frac{CD}{AD} = \frac{6}{4} = \frac{3}{2} $.
