《2025年思维新观察八年级数学上册人教版湖北专版》

9. (2025·恩施)如图,将三角形纸片$ABC沿虚线剪掉两角得五边形CDEFG$,若$DE// CG$,$FG// CD$,根据所标数据,则$\angle A$的度数为(A)

A. $58^{\circ}$
B. $64^{\circ}$
C. $66^{\circ}$
D. $72^{\circ}$

答案： A

10. 求下列各图中$\alpha$的值.

“对顶型” $\alpha =$$10^{\circ}$ “互补型” $\alpha =$$80^{\circ}$($AB// CD$) “一线三等角型” $\alpha =$$20^{\circ}$

答案： $10^{\circ}$ $80^{\circ}$ $20^{\circ}$

11. (教材$P12例2$改编)如图,$B处在A处南偏西45^{\circ}$方向,$C处在A处南偏东15^{\circ}$方向,$C处在B处北偏东80^{\circ}$方向,求$\angle ACB$的度数.

解:$\angle CBE = 80^{\circ}$，$\angle EBA = 45^{\circ}$，$\angle ABC =$35°，$\because \angle BAC = 45^{\circ} + 15^{\circ} =$60°，$\therefore \angle ACB = 180^{\circ} - 35^{\circ} - 60^{\circ} =$85°。

答案： 解:$\angle CBE = 80^{\circ}$，
$\angle EBA = 45^{\circ}$，$\angle ABC = 35^{\circ}$，
$\because \angle BAC = 45^{\circ} + 15^{\circ} = 60^{\circ}$，
$\therefore \angle ACB = 180^{\circ} - 35^{\circ} - 60^{\circ} = 85^{\circ}$。

12. (教材$P17T9$改编)如图,在$\triangle ABC$中,$\angle ABC$,$\angle ACB的平分线BE$,$CD相交于F$.
(1)若$\angle ABC = 42^{\circ}$,$\angle A = 60^{\circ}$,求$\angle BFC$的度数；120°
(2)直接写出$\angle A与\angle BFC$的数量关系.$\angle BFC = \frac{1}{2}\angle A + 90^{\circ}$

答案： 解:
(1)在$\triangle ABC$中，
$\angle BCA = 180^{\circ} - 60^{\circ} - 42^{\circ} = 78^{\circ}$，
$\therefore \angle BCF = 39^{\circ}$，$\angle CBF = 21^{\circ}$，
$\therefore$在$\triangle BCF$中$\angle BFC = 120^{\circ}$；
(2)设$\angle FBC = \alpha$，$\angle FCB = \beta$，
$\therefore 2\alpha + 2\beta = 180^{\circ} - \angle A$，
$\alpha + \beta = 180^{\circ} - \angle BFC$，
$\therefore \angle BFC = \frac{1}{2}\angle A + 90^{\circ}$。

13. (1)如图1,在△ACB中,∠BCD = ∠BDC = α,∠ACE = ∠AEC = β,∠DCE = 40°,求∠ACB的大小；100°
(2)如图2,∠1 = ∠2,∠3 = ∠4,∠B = ∠BAC,求∠BAN的大小.60°

答案： 解:
(1)在$\triangle CDE$中，$\alpha + \beta = 140^{\circ}$，
在$\triangle ACB$中，$\angle A = 180^{\circ} - 2\beta$，
$\angle B = 180^{\circ} - 2\alpha$，
$\therefore \angle A + \angle B = 360^{\circ} - 280^{\circ} = 80^{\circ}$，
$\therefore \angle ACB = 100^{\circ}$；
(2)设$\angle 1 = \angle 2 = \alpha$，
$\angle 3 = \angle 4 = \beta$，
$\therefore \angle B = 2\alpha + \beta$，
$\therefore$在$\triangle BAN$中，$2\alpha + \beta + \beta + \alpha + \beta = 180^{\circ}$，
$\alpha + \beta = 60^{\circ}$，$\therefore \angle BAN = 60^{\circ}$。