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An0nyme1
@An0nyme1
May 2019
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Bonjour,
J’ai besoin d’aide pour cet exercice de trigonométrie (exercice 15 page 161, collection Math’x 2nd, éditions Didier).
Merci d’avance.
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taalbabachir
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1) a) Exprimer HB et HC en fonction de PH
tan 45° = PH/HB ⇒ HB = PH/tan 45° = PH/1 = PH
⇒ HB = PH
tan 30° = PH/HC ⇒ HC = PH/tan 30° = PH/0.577 = 1.732 x PH
b) Utiliser BC = 40 m et montrer que PH = 20(√3 + 1)
HC = 1.732 x PH
HB + BC = 1.732 x PH or HB = PH ; PHB est un triangle isocèle rectangle en H
PH + 40 = 1.732 x PH or 1.732 = √3
PH + 40 = PH x √3 ⇔ PH√3 - PH = 40 ⇔ PH(√3 - 1) = 40 ⇒ PH = 40/(√3 - 1)
⇒ PH = 40 x (√3 + 1)/(√3 - 1)(√3 + 1) = 40(√3 + 1)/2 = 20(√3 + 1)
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Verified answer
1) a) Exprimer HB et HC en fonction de PHtan 45° = PH/HB ⇒ HB = PH/tan 45° = PH/1 = PH
⇒ HB = PH
tan 30° = PH/HC ⇒ HC = PH/tan 30° = PH/0.577 = 1.732 x PH
b) Utiliser BC = 40 m et montrer que PH = 20(√3 + 1)
HC = 1.732 x PH
HB + BC = 1.732 x PH or HB = PH ; PHB est un triangle isocèle rectangle en H
PH + 40 = 1.732 x PH or 1.732 = √3
PH + 40 = PH x √3 ⇔ PH√3 - PH = 40 ⇔ PH(√3 - 1) = 40 ⇒ PH = 40/(√3 - 1)
⇒ PH = 40 x (√3 + 1)/(√3 - 1)(√3 + 1) = 40(√3 + 1)/2 = 20(√3 + 1)