Bonjour Anonyme67
Exercice 3
[tex]a)\ \tan(\widehat{A})=\dfrac{KC}{CA}=\dfrac{3}{6}=\dfrac{1}{2}\\\\\widehat{A}=\tan^{-1}(\dfrac{1}{2})\\\\\boxed{\widehat{A}\approx27^o\ (arrondi\ au\ degr\acute{e}\ pr\grave{e}s)}\\\\b)\ \sin(\widehat{A})=\dfrac{TR}{TA}=\dfrac{5}{8}\\\\\widehat{A}=\sin^{-1}(\dfrac{5}{8})\\\\\boxed{\widehat{A}\approx37^o\ (arrondi\ au\ degr\acute{e}\ pr\grave{e}s)}[/tex]
Exercice 4
Par Thalès dans le triangle BKS,
[tex]\dfrac{BK}{BT}=\dfrac{KS}{TA}\\\\\dfrac{BK}{20}=\dfrac{21}{14}\\\\\dfrac{BK}{20}=1,5\\\\BK=1,5\times20\\\\\boxed{BK=30\ mm}[/tex]
Dans le triangle BKE rectangle en K,
[tex]\tan(\widehat{E})=\dfrac{BK}{EK}\\\\\tan(64^o)=\dfrac{30}{EK}\\\\EK\times\tan(64^o)=30\\\\EK=\dfrac{30}{\tan(64^o)}\\\\\boxed{EK\approx15\ mm\ (arrondi\ \grave{a}\ 10^{-1}\ cm\ pr\grave{e}s)}}[/tex]
Exercice 5
a) Figure en pièce jointe.
b) AM² + DM² = 2,4² + 3,2²
AM² + DM² = 5,76 + 10,24
AM² + DM² = 16
AM² + DM² = 4²
AM² + DM² = AD².
Par la réciproque du théorème de Pythagore, le triangle AMD est rectangle et [AD] est l'hypoténuse.
Le triangle AMD est donc rectangle en M.
[tex]c)\ \tan(\widehat{DAM})=\dfrac{DM}{DA}\\\\\tan(\widehat{DAM})=\dfrac{3,2}{2,4}\\\\\tan(\widehat{DAM})=\dfrac{32}{24}\\\\\tan(\widehat{DAM})=\dfrac{4}{3}\\\\\widehat{DAM}=\tan^{-1}(\dfrac{4}{3})\\\\\boxed{\widehat{DAM}\approx53^o\ (arrondi\ au\ degr\acute{e}\ pr\grave{e}s)}}[/tex]
d) Dans le triangle ADI,
[tex]\tan(\widehat{DAI})=\dfrac{DI}{DA}\\\\Or\ \widehat{DAI}=\widehat{DAM}\\\\Donc\ \tan(\widehat{DAM})=\dfrac{DI}{DA}\\\\\dfrac{4}{3}=\dfrac{DI}{DA}\\\\\dfrac{4}{3}=\dfrac{DI}{4}\\\\DI=4\times\dfrac{4}{3}\\\\DI=\dfrac{16}{3}\ cm\\\\\boxed{DI\approx5,3\ cm\ (arrondi\ au\ mm\ pr\grave{e}s)}[/tex]
Exercice 6
[tex]\tan(\widehat{FAD})=\dfrac{FD}{AD}\\\\\tan(65^o)=\dfrac{FD}{15}\\\\FD=15\times\tan(65^o)\\\\\\\tan(\widehat{BAD})=\dfrac{BD}{AD}\\\\\tan(30^o)=\dfrac{BD}{15}\\\\BD=15\times\tan(30^o)\\\\\\[/tex]
D'où
[tex]FB = FD - BD\\\\FB=15\times\tan(65^o)-15\times\tan(30^o)\\\\\boxed{FB\approx23,50\ m\ (arrondi\ au\ cm\ pr\grave{e}s)}}[/tex]
