Menu Close

Question-202093

Tinku Tara 0 Comments
Pin




Question Number 202093 by sonukgindia last updated on 20/Dec/23
Commented by a.lgnaoui last updated on 22/Dec/23
Commented by a.lgnaoui last updated on 22/Dec/23
ΔAMN ∡ MNA=((3π)/4)−α ∡ANM=(π/4)+α ⇒ ∡ACE=α Calcul de AB AC en finction de x; 𝛂: ∡ABC ((sin(45+α))/(AC))=((cos α)/(AB))=((sin 45)/( (√2))) AC=(((√2) sin (45+α))/(sin 45))=(√2) (sin α+cos α) AC=(√2) (sin 𝛂+cos𝛂) (1) AB=(((√2) cos α)/(sin 45))=2cos α AB=2cos 𝛂 (2) Camcul de AB AB=AM+MB ((sin 45)/x)=((sin (((3π)/4)−α))/(AM)) ;AM=((xsin (((3π)/4)−α))/(sin (π/4))) ⇒ AM=x(cos 𝛂+sin 𝛂) (3) ΔMBD ((sin (π/4))/(MB))=((sin α)/a) ⇒ MB=((asin 45)/(sin 𝛂))=((a(√2))/(2sin 𝛂)) (4) akora AB=AM+MB AB =x(sin 𝛂+cos 𝛂)+((a(√2))/(2sin 𝛂)) (5) •Calcul de AC AC=AN+NC ((sin α)/(AN))=((sin (((3π)/4)−α))/(AM))=((√2)/(2x)) ⇒ AN=(√2) xsin 𝛂 (6) •Calcul de ND ∡ANE ((cos α)/(NE))=((sin 45)/(AN)) ⇒ NE=((ANcos α)/(sin 45))=((2xsin αcos α)/ ) NE=xsin 2α ΔNEC ((sin α)/(NE))=((sin 45)/(ND)) ⇒ ND=((xsin 2𝛂sin 45)/(sin α))=((2(√2)xcos α)/2) ND=(√2) xcos 𝛂 alors AC=(√2) xsin 𝛂+(√2) xcos α AC=x(√2) (sin 𝛂+cos 𝛂) (7) (1) (2)⇒ { ((((a(√2))/(2sin 𝛂))+x(sin 𝛂+cos𝛂)=2cos 𝛂 (8))),((x(√2) (sin 𝛂+cos 𝛂) =(√2) (cos 𝛂+sin 𝛂)(9))) :} (9)⇒ x(√2) =(√2) donc x=1
$$\Delta\mathrm{AMN}\:\:\measuredangle\:\mathrm{MNA}=\frac{\mathrm{3}\pi}{\mathrm{4}}−\alpha \\ $$$$\:\:\measuredangle\mathrm{ANM}=\frac{\pi}{\mathrm{4}}+\alpha\:\:\Rightarrow\:\measuredangle\mathrm{ACE}=\alpha \\ $$$$\boldsymbol{\mathrm{Calcul}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{AB}}\:\:\boldsymbol{\mathrm{AC}}\:\boldsymbol{\mathrm{en}}\:\boldsymbol{\mathrm{finction}}\:\boldsymbol{\mathrm{de}}\:\boldsymbol{\mathrm{x}};\:\boldsymbol{\alpha}: \\ $$$$\measuredangle\mathrm{ABC}\:\:\:\:\frac{\mathrm{sin}\left(\mathrm{45}+\alpha\right)}{\mathrm{AC}}=\frac{\mathrm{cos}\:\alpha}{\mathrm{AB}}=\frac{\mathrm{sin}\:\mathrm{45}}{\:\sqrt{\mathrm{2}}}\:\:\: \\ $$$$\mathrm{AC}=\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\mathrm{45}+\alpha\right)}{\mathrm{sin}\:\mathrm{45}}=\sqrt{\mathrm{2}}\:\left(\mathrm{sin}\:\alpha+\mathrm{cos}\:\alpha\right) \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{AC}}=\sqrt{\mathrm{2}}\:\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\boldsymbol{\alpha}\right)\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{AB}=\frac{\sqrt{\mathrm{2}}\:\mathrm{cos}\:\alpha}{\mathrm{sin}\:\mathrm{45}}=\mathrm{2cos}\:\alpha \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{AB}}=\mathrm{2cos}\:\boldsymbol{\alpha}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{Camcul}\:\mathrm{de}\:\mathrm{AB} \\ $$$$\:\mathrm{AB}=\mathrm{AM}+\mathrm{MB} \\ $$$$\:\:\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{x}}=\frac{\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{4}}−\alpha\right)}{\mathrm{AM}}\:;\mathrm{AM}=\frac{\mathrm{xsin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{4}}−\alpha\right)}{\mathrm{sin}\:\frac{\pi}{\mathrm{4}}} \\ $$$$\Rightarrow\:\:\:\boldsymbol{\mathrm{AM}}=\boldsymbol{\mathrm{x}}\left(\mathrm{cos}\:\boldsymbol{\alpha}+\mathrm{sin}\:\boldsymbol{\alpha}\right)\:\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\Delta\mathrm{MBD}\:\:\:\frac{\mathrm{sin}\:\frac{\pi}{\mathrm{4}}}{\mathrm{MB}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{a}} \\ $$$$\Rightarrow\:\:\:\boldsymbol{\mathrm{MB}}=\frac{\boldsymbol{\mathrm{a}}\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\boldsymbol{\alpha}}=\frac{\boldsymbol{\mathrm{a}}\sqrt{\mathrm{2}}}{\mathrm{2sin}\:\boldsymbol{\alpha}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\mathrm{akora}\:\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{AM}}+\boldsymbol{\mathrm{MB}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{AB}}\:\:\:=\boldsymbol{\mathrm{x}}\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\boldsymbol{\alpha}\right)+\frac{\mathrm{a}\sqrt{\mathrm{2}}}{\mathrm{2sin}\:\boldsymbol{\alpha}}\:\left(\mathrm{5}\right) \\ $$$$\bullet\mathrm{Calcul}\:\mathrm{de}\:\boldsymbol{\mathrm{AC}} \\ $$$$\:\mathrm{AC}=\mathrm{AN}+\mathrm{NC} \\ $$$$\:\:\:\frac{\mathrm{sin}\:\alpha}{\mathrm{AN}}=\frac{\mathrm{sin}\:\left(\frac{\mathrm{3}\pi}{\mathrm{4}}−\alpha\right)}{\mathrm{AM}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2x}} \\ $$$$\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{AN}}=\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{x}}\mathrm{sin}\:\boldsymbol{\alpha}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{6}\right) \\ $$$$\bullet\mathrm{Calcul}\:\mathrm{de}\:\mathrm{ND} \\ $$$$\:\:\:\measuredangle\mathrm{ANE}\:\:\:\frac{\mathrm{cos}\:\alpha}{\mathrm{NE}}=\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{AN}} \\ $$$$\Rightarrow\:\:\mathrm{NE}=\frac{\mathrm{ANcos}\:\alpha}{\mathrm{sin}\:\mathrm{45}}=\frac{\mathrm{2xsin}\:\alpha\mathrm{cos}\:\alpha}{\:} \\ $$$$\:\:\:\:\:\mathrm{NE}=\mathrm{xsin}\:\mathrm{2}\alpha\:\:\: \\ $$$$\Delta\mathrm{NEC}\:\:\frac{\mathrm{sin}\:\alpha}{\mathrm{NE}}=\frac{\mathrm{sin}\:\mathrm{45}}{\mathrm{ND}} \\ $$$$\Rightarrow\:\:\:\mathrm{ND}=\frac{\boldsymbol{\mathrm{x}}\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}\mathrm{sin}\:\mathrm{45}}{\mathrm{sin}\:\alpha}=\frac{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{xcos}\:\alpha}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{ND}}=\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{x}}\mathrm{cos}\:\boldsymbol{\alpha} \\ $$$$\mathrm{alors}\:\:\mathrm{AC}=\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{x}}\mathrm{sin}\:\boldsymbol{\alpha}+\sqrt{\mathrm{2}}\:\boldsymbol{\mathrm{x}}\mathrm{cos}\:\alpha \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{AC}}=\boldsymbol{\mathrm{x}}\sqrt{\mathrm{2}}\:\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\boldsymbol{\alpha}\right)\:\:\:\:\:\left(\mathrm{7}\right) \\ $$$$\left(\mathrm{1}\right)\:\:\left(\mathrm{2}\right)\Rightarrow \\ $$$$\begin{cases}{\frac{\boldsymbol{\mathrm{a}}\sqrt{\mathrm{2}}}{\mathrm{2sin}\:\boldsymbol{\alpha}}+\boldsymbol{\mathrm{x}}\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\boldsymbol{\alpha}\right)=\mathrm{2cos}\:\boldsymbol{\alpha}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{8}\right)}\\{\boldsymbol{\mathrm{x}}\sqrt{\mathrm{2}}\:\left(\mathrm{sin}\:\boldsymbol{\alpha}+\mathrm{cos}\:\boldsymbol{\alpha}\right)\:=\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\:\boldsymbol{\alpha}+\mathrm{sin}\:\boldsymbol{\alpha}\right)\left(\mathrm{9}\right)}\end{cases} \\ $$$$\left(\mathrm{9}\right)\Rightarrow\:\:\:\boldsymbol{\mathrm{x}}\sqrt{\mathrm{2}}\:=\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{donc}}\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{1} \\ $$$$\:\:\: \\ $$$$ \\ $$
Answered by MathematicalUser2357 last updated on 22/Dec/23
Is the size of the yellow angle 45°?
$$\mathrm{Is}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of}\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{angle}\:\mathrm{45}°? \\ $$
Commented by a.lgnaoui last updated on 22/Dec/23
from picture yellow angle=45°
$$\mathrm{from}\:\mathrm{picture}\:\:\mathrm{yellow}\:\mathrm{angle}=\mathrm{45}° \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *