Автор - mishatish06

решите пожалуйста очень нужно

Ответ

Проверено экспертом

Автор - KuOV
1. ∠АОС = ∠АОВ + ∠СОВ = 45° + 21° = 66°
2. ∠FDK = ∠EDK - ∠EDF = 36° - 21° = 15°
3.  ∠ABC = ∠ABD + ∠CBD
∠ABD = x
∠DBC = x + 26°
x  + x + 26° = 72°
2x = 46°
x = 23°
∠ABD = 23°
∠DBC = 23° + 26° = 49°

4. ∠ABD = ∠ABC + ∠CBD
∠ABC = x
∠CBD = 4x
x + 4x = 100°
5x = 100°
x = 20°
∠ABC = 20°
∠CBD = 4 · 20° = 80°

5.
1) ∠BAD = ∠BAC + ∠CAD
∠CAE = ∠DAE + ∠CAD
Так как ∠BAC = ∠DAE, то и ∠BAD = ∠CAE
2) ∠BAD = ∠BAC + ∠CAD
∠CAE = ∠DAE + ∠CAD
Так как ∠BAD = ∠CAE, то и ∠BAC = ∠DAE

6. ∠ABC = ∠ABD - ∠CBD = 85° - 40° = 45°
∠ABE = ∠ABC + ∠CBE = 45° + 45° = 90°

7. ∠POK = ∠MOK - ∠MOP = 110° - 73° = 37°
∠NOK = ∠NOP + POK = 64° + 37° = 101°

8. ∠AOD + ∠DOK = 90°
∠DOK + ∠KOE = 90°, значит ∠AOD = ∠KOE.

∠KOE + ∠EOB = 90°
∠DOK + ∠KOE = 90°, значит  ∠DOK = ∠EOB.

9.
1) ∠AOE = ∠AOC + ∠COE = 96°
∠AOC = 2∠BOC
∠COE = 2∠COD
2(∠BOC + ∠COD) = 96°
∠BOC + ∠COD = 48°
∠BOD = ∠BOC + ∠COD = 48°
2) ∠BOD = 42°
∠BOD = ∠BOC + ∠COD = 42°
∠AOC = 2∠BOC
∠COE = 2∠COD, ⇒ ∠AOC + ∠COE = 2 · 42° = 84°
∠AOE = ∠AOC + ∠COE = 84°

10. ∠ABD = ∠ABC + ∠CBD
Пусть х - коэффициент пропорциональности.
∠ABC = 3x
∠CBD = 4x
3x + 4x = 105°
7x = 105°
x = 15°
∠ABC = 3 · 15° = 45°
∠CBD = 4 · 15° = 60°

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