Correction - Exercice 01 page 46 - Rapports trigonométriques d'un angle aigu - Relations métriques dans un triangle rectangle
1ère année secondaire
Rapports trigonométriques d'un angle aigu
Relations métriques dans un triangle rectangle
Exercice 01 page 46
Déterminons dans chacun des cas suivants \(x\) et \(y\) sachant que le triangle \(ABC\) est rectangle en \(A\).
* 1er cas :
On a :
\(cos~\widehat{ABC} = \frac{AB}{BC}\)\(\Rightarrow\)
\(cos~30° = \frac{x}{1}\)\(\Rightarrow\)
\(x = cos~30°\)\(\Rightarrow\)
\(x = \frac{\sqrt{3}}{2}\)
- Cherchons \(y\)
On a :
\(AB^2+AC^2=BC^2\)\(\Rightarrow\)
\(x^2+y^2=1^2\)\(\Rightarrow\)
\((\frac{\sqrt{3}}{2})^2+y^2=1\)\(\Rightarrow\)
\(y^2=1-(\frac{\sqrt{3}}{2})^2\)\(\Rightarrow\)
\(y^2=1-\frac{3}{4}\)\(\Rightarrow\)
\(y^2=\frac{4}{4}-\frac{3}{4}\)\(\Rightarrow\)
\(y^2=\frac{1}{4}\)\(\Rightarrow\)
\(y=\sqrt{\frac{1}{4}}\)\(\Rightarrow\)
\(y^2=1-\frac{3}{4}\)\(\Rightarrow\)
\(y^2=\frac{4}{4}-\frac{3}{4}\)\(\Rightarrow\)
\(y^2=\frac{1}{4}\)\(\Rightarrow\)
\(y=\sqrt{\frac{1}{4}}\)\(\Rightarrow\)
\(y=\) \(\frac{1}{2}\)
\(x = \frac{1}{sin~30°}\)\(\Rightarrow\)
\(x = \frac{1}{\frac{1}{2}}\)\(\Rightarrow\)
\(x = \)\(2\)
- Cherchons \(y\)
On a :
\(AB^2+AC^2=BC^2\)\(\Rightarrow\)
* 2ème cas :
- Cherchons \(x\)
On a :
\(sin~\widehat{BCA} = \frac{AB}{BC}\)\(\Rightarrow\)
\(sin~30° = \frac{1}{x}\)\(\Rightarrow\)
On a :
\(sin~\widehat{BCA} = \frac{AB}{BC}\)\(\Rightarrow\)
\(sin~30° = \frac{1}{x}\)\(\Rightarrow\)
\(x = \frac{1}{sin~30°}\)\(\Rightarrow\)
\(x = \frac{1}{\frac{1}{2}}\)\(\Rightarrow\)
\(x = \)\(2\)
- Cherchons \(y\)
On a :
\(AB^2+AC^2=BC^2\)\(\Rightarrow\)
\(x^2+y^2=2^2\)\(\Rightarrow\)
\(1^2+y^2=4\)\(\Rightarrow\)
\(y^2=4-1\)\(\Rightarrow\)
\(y^2=3\)\(\Rightarrow\)
\(y=\) \(\sqrt{3}\)
\(x = \frac{1}{cos~60°}\)\(\Rightarrow\)
\(x = \frac{1}{\frac{1}{2}}\)\(\Rightarrow\)
\(x = \)\(2\)
- Cherchons \(y\)
On a :
\(AB^2+AC^2=BC^2\)\(\Rightarrow\)
\(y^2=3\)\(\Rightarrow\)
\(y=\) \(\sqrt{3}\)
* 3ème cas :
- Cherchons \(x\)
On a :
\(cos~\widehat{BCA} = \frac{AC}{BC}\)\(\Rightarrow\)
\(cos~60° = \frac{1}{x}\)\(\Rightarrow\)
On a :
\(cos~\widehat{BCA} = \frac{AC}{BC}\)\(\Rightarrow\)
\(cos~60° = \frac{1}{x}\)\(\Rightarrow\)
\(x = \frac{1}{cos~60°}\)\(\Rightarrow\)
\(x = \frac{1}{\frac{1}{2}}\)\(\Rightarrow\)
\(x = \)\(2\)
- Cherchons \(y\)
On a :
\(AB^2+AC^2=BC^2\)\(\Rightarrow\)
\(x^2+y^2=2^2\)\(\Rightarrow\)
\(1^2+y^2=4\)\(\Rightarrow\)
\(y^2=4-1\)\(\Rightarrow\)
\(y^2=3\)\(\Rightarrow\)
\(y=\) \(\sqrt{3}\)
\(y^2=3\)\(\Rightarrow\)
\(y=\) \(\sqrt{3}\)
Libellés:
1ère année secondaire
Correction
Corrigées
exercice
Le Mathématicien
manuel scolaire
Math
Mathématiques
Rapports trigonométriques d'un angle aigu
Relations métriques dans un triangle rectangle
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