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Leahharold62800
@Leahharold62800
January 2021
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je n'y arrive pas a mon dm aidez moi svp
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aymanemaysae
Bonjour ;
1)
En considérant le triangle HZA rectangle en H , on a :
sin(HZA) = HA/HZ = 16,5/24 = 0,6875 ;
donc : HZA ≈ 43,43° .
En considérant le triangle HZB rectangle en H , on a :
sin(HZB) = HB/HZ = (HA + AB)/HZ = (16,5 + 7,3)/24 ≈ 0,9917 ;
donc : HZB ≈ 82,60° .
b)
L'angle AZB = HZB - HZA ≈ 82,60° - 43,43° = 39,17° .
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Leahharold62800
merci mais lexercice 2 ?
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Leahharold62800
January 2021 | 0 Respostas
Rebonjour pouvez vous maidez a l'exercice 2 svp
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May 2019 | 0 Respostas
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1)
En considérant le triangle HZA rectangle en H , on a :
sin(HZA) = HA/HZ = 16,5/24 = 0,6875 ;
donc : HZA ≈ 43,43° .
En considérant le triangle HZB rectangle en H , on a :
sin(HZB) = HB/HZ = (HA + AB)/HZ = (16,5 + 7,3)/24 ≈ 0,9917 ;
donc : HZB ≈ 82,60° .
b)
L'angle AZB = HZB - HZA ≈ 82,60° - 43,43° = 39,17° .