3.如图6,直线AB,CD互相垂直,垂足为O,直线EF过点O,∠DOF = 32°,求∠AOE的度数.

答案
3.解:
∵直线CD与EF交于O,
∴∠EOC = ∠DOF.
∵∠DOF = 32°,
∴∠EOC = 32°.
∵AB,CD互相垂直,
∴∠AOC = 90°.
∴∠AOE + ∠EOC = 90°.
∴∠AOE = 90°−∠EOC = 90°−32° = 58°.
∵直线CD与EF交于O,
∴∠EOC = ∠DOF.
∵∠DOF = 32°,
∴∠EOC = 32°.
∵AB,CD互相垂直,
∴∠AOC = 90°.
∴∠AOE + ∠EOC = 90°.
∴∠AOE = 90°−∠EOC = 90°−32° = 58°.
4.如图7,在$△ ABC$中,$∠ ABC = 90°, ∠ A = 50°, BD // AC$,求$∠ CBD$的度数.

答案
4.解法一 由题意知,BD // AC,
∴∠ABD + ∠A = ∠CBD + ∠ABC + ∠A = 180°.
∴∠CBD = 180°−50°−90° = 40°.
解法二 由题意知,∠C = 90°−∠A = 90°−50° = 40°.
又
∵BD // AC,
∴∠CBD = ∠C = 40°.
∴∠ABD + ∠A = ∠CBD + ∠ABC + ∠A = 180°.
∴∠CBD = 180°−50°−90° = 40°.
解法二 由题意知,∠C = 90°−∠A = 90°−50° = 40°.
又
∵BD // AC,
∴∠CBD = ∠C = 40°.
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