Question and Answers Forum

All Questions   Topic List

GeometryQuestion and Answers: Page 25

Question Number 196275    Answers: 1   Comments: 0

Question Number 197582    Answers: 1   Comments: 0

Question Number 196225    Answers: 2   Comments: 1

Question Number 196021    Answers: 1   Comments: 0

Question Number 195896    Answers: 2   Comments: 0

Question Number 195892    Answers: 1   Comments: 0

Question Number 195880    Answers: 1   Comments: 1

Question Number 195806    Answers: 1   Comments: 0

Question Number 195740    Answers: 2   Comments: 0

Question Number 195688    Answers: 1   Comments: 0

Question Number 195344    Answers: 2   Comments: 0

Question Number 195287    Answers: 2   Comments: 0

Question Number 195289    Answers: 1   Comments: 0

Question Number 195273    Answers: 0   Comments: 1

Question Number 195046    Answers: 2   Comments: 1

Question Number 195043    Answers: 2   Comments: 1

Question Number 194998    Answers: 0   Comments: 2

Resolution du probldme pose par sonukgindia (16.7.2023) voir Q194819 △ABC AM=AN=ADcos (𝛂/2) { ((AC=AM+MC=17 (1))),((AB=AN+NB =18 (2))) :} AB−AC=1=NB−MC (3) △CDE cos (C/2)=((CM)/(CD))=((CE)/(CD)) ⇒CM=CE △BDE cos (B/2)=((BE)/(BD))=((BN)/(BD))⇒BN=BE ⇒BE−CE=1 BC=BE+CE=2BE−1 △BEF BF=9 cos B=((BE)/9) BE=9cos B ⇒BC=18cos B−1 Posons BC=x x=18cos B−1 d apres triangle ABC AC^2 =AB^2 +BC^2 −2AB.BCcos B ⇒17^2 =18^2 +x^2 −36(((x+1)/(18))) x^2 +2x−35=0 alors x=5

$$\mathrm{Resolution}\:\mathrm{du}\:\mathrm{probldme}\:\mathrm{pose}\:\mathrm{par}\: \\ $$$$\mathrm{sonukgindia}\:\left(\mathrm{16}.\mathrm{7}.\mathrm{2023}\right) \\ $$$$\mathrm{voir}\:\:\mathrm{Q194819} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{ABC}}\:\:\boldsymbol{\mathrm{AM}}=\boldsymbol{\mathrm{AN}}=\boldsymbol{\mathrm{AD}}\mathrm{cos}\:\frac{\boldsymbol{\alpha}}{\mathrm{2}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{AC}}=\boldsymbol{\mathrm{AM}}+\boldsymbol{\mathrm{MC}}=\mathrm{17}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AN}}+\boldsymbol{\mathrm{NB}}\:\:=\mathrm{18}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\:\:\boldsymbol{\mathrm{AB}}−\boldsymbol{\mathrm{AC}}=\mathrm{1}=\boldsymbol{\mathrm{NB}}−\boldsymbol{\mathrm{MC}}\:\:\:\left(\mathrm{3}\right) \\ $$$$\bigtriangleup\boldsymbol{\mathrm{CDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{CM}}}{\boldsymbol{\mathrm{CD}}}=\frac{\boldsymbol{\mathrm{CE}}}{\boldsymbol{\mathrm{CD}}}\:\Rightarrow\boldsymbol{\mathrm{CM}}=\boldsymbol{\mathrm{CE}} \\ $$$$\bigtriangleup\boldsymbol{\mathrm{BDE}}\:\:\:\mathrm{cos}\:\frac{\boldsymbol{\mathrm{B}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{BE}}}{\boldsymbol{\mathrm{BD}}}=\frac{\boldsymbol{\mathrm{BN}}}{\boldsymbol{\mathrm{BD}}}\Rightarrow\boldsymbol{\mathrm{BN}}=\boldsymbol{\mathrm{BE}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{BE}}−\boldsymbol{\mathrm{CE}}=\mathrm{1} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{BC}}=\boldsymbol{\mathrm{BE}}+\boldsymbol{\mathrm{CE}}=\mathrm{2}\boldsymbol{\mathrm{BE}}−\mathrm{1} \\ $$$$\:\bigtriangleup\boldsymbol{\mathrm{BEF}}\:\:\:\boldsymbol{\mathrm{BF}}=\mathrm{9}\:\:\:\:\mathrm{cos}\:\boldsymbol{\mathrm{B}}=\frac{\boldsymbol{\mathrm{BE}}}{\mathrm{9}}\:\: \\ $$$$\:\:\boldsymbol{\mathrm{BE}}=\mathrm{9cos}\:\boldsymbol{\mathrm{B}}\:\:\:\:\Rightarrow\boldsymbol{\mathrm{BC}}=\mathrm{18cos}\:\boldsymbol{\mathrm{B}}−\mathrm{1} \\ $$$$\:\:\boldsymbol{\mathrm{Posons}}\:\:\boldsymbol{\mathrm{BC}}=\boldsymbol{\mathrm{x}}\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{18cos}\:\boldsymbol{\mathrm{B}}−\mathrm{1} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{d}}\:\boldsymbol{\mathrm{apres}}\:\boldsymbol{\mathrm{triangle}}\:\:\boldsymbol{\mathrm{ABC}} \\ $$$$\:\boldsymbol{\mathrm{AC}}^{\mathrm{2}} =\boldsymbol{\mathrm{AB}}^{\mathrm{2}} +\boldsymbol{\mathrm{BC}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{AB}}.\boldsymbol{\mathrm{BC}}\mathrm{cos}\:\boldsymbol{\mathrm{B}} \\ $$$$\:\:\:\: \\ $$$$\Rightarrow\mathrm{17}^{\mathrm{2}} =\mathrm{18}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{36}\left(\frac{\boldsymbol{\mathrm{x}}+\mathrm{1}}{\mathrm{18}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{35}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{alors}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{5} \\ $$

Question Number 194939    Answers: 0   Comments: 0

Question Number 194938    Answers: 0   Comments: 0

Question Number 194861    Answers: 1   Comments: 0

Question Number 194847    Answers: 0   Comments: 2

$$\:\:\:\:\:\:\underbrace{\:} \\ $$

Question Number 194815    Answers: 1   Comments: 0

Question Number 194624    Answers: 2   Comments: 2

Question Number 194564    Answers: 2   Comments: 0

soit A(2,1) B(3,2) C(4,3) points du plan( ox,oy) 1)Determiner l ′ equation du cercle qui passe par A; B; C ? 2) points d intersection du cercle avec l axe(ox,oy)?

$$\mathrm{soit}\:\:\:\boldsymbol{\mathrm{A}}\left(\mathrm{2},\mathrm{1}\right)\:\:\:\:\:\boldsymbol{\mathrm{B}}\left(\mathrm{3},\mathrm{2}\right)\:\:\:\:\:\boldsymbol{\mathrm{C}}\left(\mathrm{4},\mathrm{3}\right)\:\:\mathrm{points}\:\mathrm{du}\: \\ $$$$\:\:\mathrm{plan}\left(\:\mathrm{ox},\mathrm{oy}\right)\:\: \\ $$$$ \\ $$$$\left.\:\mathrm{1}\right)\mathrm{Determiner}\:\:\mathrm{l}\:'\:\mathrm{equation}\:\mathrm{du}\:\mathrm{cercle}\:\mathrm{qui}\: \\ $$$$\:\:\:\:\mathrm{passe}\:\mathrm{par}\:\boldsymbol{\mathrm{A}};\:\boldsymbol{\mathrm{B}};\:\boldsymbol{\mathrm{C}}\:\:\:? \\ $$$$\left.\:\:\mathrm{2}\right)\:\mathrm{points}\:\mathrm{d}\:\mathrm{intersection}\:\mathrm{du}\:\mathrm{cercle}\:\mathrm{avec} \\ $$$$\:\:\:\:\:\mathrm{l}\:\mathrm{axe}\left(\mathrm{ox},\mathrm{oy}\right)? \\ $$

Question Number 194410    Answers: 0   Comments: 1

Question Number 194270    Answers: 1   Comments: 1

  Pg 20      Pg 21      Pg 22      Pg 23      Pg 24      Pg 25      Pg 26      Pg 27      Pg 28      Pg 29   

Terms of Service

Privacy Policy

Contact: info@tinkutara.com