《2026年启东中学作业本九年级数学下册苏科版》

9. 在$△ ABC$中,$∠ C = 90^{\circ}$,$a$,$b$,$c$分别为$∠ A$,$∠ B$,$∠ C$的对边,若$b^{2} = ac$,则$\sin A$的值为$\frac{\sqrt{5}-1}{2}$.

答案： 9.$\frac{\sqrt{5}-1}{2}$

三、解答题(共 46 分)
10. (15 分)(1)如图①,在锐角$△ ABC$中,$AB = c$,$BC = a$,$AC = b$,$AD⊥ BC$于点$D$.
在$\mathrm{Rt}△ ABD$中,$\sin B = \frac{AD}{c}$,则$AD = c\sin B$;在$\mathrm{Rt}△ ACD$中,$\sin C =$$\frac{AD}{b}$,则$AD =$$b\sin C$;
所以,$c\sin B = b\sin C$,即$\frac{b}{\sin B} = \frac{c}{\sin C}$,
进一步即得正弦定理:$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$(此定理适合任意锐角三角形).
(2)利用正弦定理解答问题:
如图②,在$△ ABC$中,$∠ B = 75^{\circ}$,$∠ C = 45^{\circ}$,$BC = 2$,求$AB$的长.

答案： 10.
(1)$\frac{AD}{b}$　$b\sin C$
(2)解:在△ABC中,∠B = 75°,∠C = 45°,则∠A = 60°.
∵$\frac{AB}{\sin C}=\frac{BC}{\sin A}$,
∴$\frac{AB}{\sin 45°}=\frac{2}{\sin 60°}$,即$\frac{AB}{\frac{\sqrt{2}}{2}}=\frac{2}{\frac{\sqrt{3}}{2}}$,解得$AB = \frac{2\sqrt{6}}{3}$,即AB的长是$\frac{2\sqrt{6}}{3}$.

11. (15 分)在三角形纸片$ABC$(如图①)中,$∠ BAC = 78^{\circ}$,$AC = 10$.小霞用 5 张这样的三角形纸片拼成了一个内外都是正五边形的图形(如图②).
(1)$∠ ABC =$30°;
(2)求正五边形$GHMNC$的边$GC$的长.(参考数据:$\sin 78^{\circ}\approx 0.98$,$\cos 78^{\circ}\approx 0.21$,$\tan 78^{\circ}\approx 4.7$)

答案：
11.
(1)30°
(2)解:如答图，作CQ⊥AB于点Q.在Rt△AQC中,
$\sin ∠ QAC = \frac{QC}{AC}$,
∴$QC = AC · \sin ∠ QAC \approx 10 × 0.98 = 9.8$.

在Rt△BQC中,∠ABC = 30°,
∴$BC = 2QC = 19.6$,
∴$GC = BC - BG = 19.6 - 10 = 9.6$.

12. (16 分)(2023·张家界)如图,$\odot O$是$△ ABC$的外接圆,$AD$是$\odot O$的直径,$F$是$AD$延长线上一点,连接$CD$,$CF$,且$∠ DCF = ∠ CAD$.
(1)求证:$CF$是$\odot O$的切线;
(2)若$AD = 10$,$\cos B = \frac{3}{5}$,求$FD$的长.

答案：
12.
(1)证明:连接OC,如答图

∵AD是⊙O的直径,
∴∠ACD = 90°,
∴∠ADC + ∠CAD = 90°,
∵OC = OD,
∴∠ADC = ∠OCD.
∵∠DCF = ∠CAD,
∴∠DCF + ∠OCD = 90°,
即OC⊥FC,
∴FC是⊙O的切线.
(2)解:
∵∠B = ∠ADC,$\cos B = \frac{3}{5}$,
∴$\cos ∠ ADC = \frac{3}{5}$.在Rt△ACD中,
∵$\cos ∠ ADC = \frac{3}{5} = \frac{CD}{AD}$,AD = 10,
∴$CD = AD · \cos ∠ ADC = 10 × \frac{3}{5} = 6$,
∴$AC = \sqrt{AD^2 - CD^2} = 8$,
∴$\frac{CD}{AC} = \frac{3}{4}$,
∵∠FCD = ∠FAC,∠F = ∠F,
∴△FCD∽△FAC,
∴$\frac{CD}{AC} = \frac{FC}{FA} = \frac{FD}{FC} = \frac{3}{4}$.
设FD = 3x,则FC = 4x,AF = 3x + 10,
又
∵$FC^2 = FD · FA$,即$(4x)^2 = 3x(3x + 10)$,
解得$x = \frac{30}{7}$或x = 0(舍去),
∴$FD = 3x = \frac{90}{7}$.