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1. 在$\mathrm{Rt}△ ABC$中，$∠ C=90°$，$∠ A - ∠ B=70°$，则$∠ A$的度数为（
A.$80°$
B.$70°$
C.$60°$
D.$50°$
A
）.A.$80°$
B.$70°$
C.$60°$
D.$50°$
答案
1.A
2.如图，BD平分∠ABC，CD⊥BD，D为垂足，∠C=55°，则∠ABC的度数是(
A.35°
B.55°
C.60°
D.70°
D
).A.35°
B.55°
C.60°
D.70°
答案
2.D
3. 下列选项中，不能构成直角三角形的是(
A.$∠ A = ∠ B = 3∠ C$
B.$∠ A + ∠ B = ∠ C$
C.$∠ A = ∠ B = \frac{1}{2}∠ C$
D.$∠ A : ∠ B : ∠ C = 1:2:3$
A
).A.$∠ A = ∠ B = 3∠ C$
B.$∠ A + ∠ B = ∠ C$
C.$∠ A = ∠ B = \frac{1}{2}∠ C$
D.$∠ A : ∠ B : ∠ C = 1:2:3$
答案
3.A
4.如图，$CE ⊥ AF$，垂足为$E$，$CE$与$BF$相交于点$D$，$∠ F = 40°$，$∠ C = 30°$，求$∠ EDF$和$∠ DBC$的度数.
答案
4.解 $\because CE⊥AF$,
$\therefore ∠DEF=90°$,
$\therefore ∠EDF=90°-∠F=90°-40°=50°$.
$\because ∠C+∠DBC=∠F+∠DEF$,
$\therefore 30°+∠DBC=40°+90°$,
$\therefore ∠DBC=100°$.
$\therefore ∠DEF=90°$,
$\therefore ∠EDF=90°-∠F=90°-40°=50°$.
$\because ∠C+∠DBC=∠F+∠DEF$,
$\therefore 30°+∠DBC=40°+90°$,
$\therefore ∠DBC=100°$.
5. 如图，在$△ ABC$中，$∠ B = ∠ C$，$FD ⊥ BC$，$DE ⊥ AB$，$∠ AFD = 152°$，求$∠ EDF$的度数. 
答案
5.解 $\because ∠AFD=152°$,
$\therefore ∠DFC=28°$.
$\because ∠B=∠C,FD⊥BC,DE⊥AB$,
$\therefore ∠EDB=∠DFC=28°$,
$\therefore ∠EDF = 180° - ∠EDB - ∠FDC=180°-28°-90°=62°$.
$\therefore ∠DFC=28°$.
$\because ∠B=∠C,FD⊥BC,DE⊥AB$,
$\therefore ∠EDB=∠DFC=28°$,
$\therefore ∠EDF = 180° - ∠EDB - ∠FDC=180°-28°-90°=62°$.
6.如图,在$△ ABC$中,$CD ⊥ BA$,交$BA$的延长线于点$D$,$DE ⊥ AC$于点$E$.
(1)如图①,若$∠ B＝35°$,$∠ CDE＝60°$,求$∠ ACB$的度数.
(2)如图②,若$AC$平分$∠ BCD$,$BF ⊥ AC$交$CA$的延长线于点$F$,直接写出与$∠ ACB$相等的角($∠ ACB$除外).
(1)如图①,若$∠ B＝35°$,$∠ CDE＝60°$,求$∠ ACB$的度数.
(2)如图②,若$AC$平分$∠ BCD$,$BF ⊥ AC$交$CA$的延长线于点$F$,直接写出与$∠ ACB$相等的角($∠ ACB$除外).
答案
6.解 (1)如题图①,
$\because CD⊥BD,\therefore ∠BDC=90°$.
$\because ∠B=35°$,
$\therefore ∠BCD=90°-∠B=55°$.
$\because DE⊥AC$于点$E$,
$\therefore ∠DEC=90°$.
$\because ∠EDC=60°$,
$\therefore ∠DCE=90°-∠EDC=30°$,
$\therefore ∠ACB = ∠BCD - ∠DCE = 55°-30°=25°$.
(2)$∠DCA,∠ADE,∠FBA$.
$\because CD⊥BD,\therefore ∠BDC=90°$.
$\because ∠B=35°$,
$\therefore ∠BCD=90°-∠B=55°$.
$\because DE⊥AC$于点$E$,
$\therefore ∠DEC=90°$.
$\because ∠EDC=60°$,
$\therefore ∠DCE=90°-∠EDC=30°$,
$\therefore ∠ACB = ∠BCD - ∠DCE = 55°-30°=25°$.
(2)$∠DCA,∠ADE,∠FBA$.
