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Question Number 197499    Answers: 1   Comments: 0

Question Number 197287    Answers: 0   Comments: 3

answer to the question number 197017 AF=FI & AG=GJ⇒FG=(1/2)IJ=(1/6)BC △FGH is squilatral ⇒ △FGH≈△ABC ⇒(S_(FGH) /S_(SBC) ) =(1/(36 )) ✓

$${answer}\:{to}\:{the}\:{question}\:{number} \\ $$$$\mathrm{197017} \\ $$$${AF}={FI}\:\&\:\:{AG}={GJ}\Rightarrow{FG}=\frac{\mathrm{1}}{\mathrm{2}}{IJ}=\frac{\mathrm{1}}{\mathrm{6}}{BC} \\ $$$$\bigtriangleup{FGH}\:\:{is}\:\:{squilatral}\:\Rightarrow\:\bigtriangleup{FGH}\approx\bigtriangleup{ABC} \\ $$$$\Rightarrow\frac{{S}_{{FGH}} }{{S}_{{SBC}} }\:=\frac{\mathrm{1}}{\mathrm{36}\:}\:\checkmark \\ $$$$ \\ $$

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prove that the curve (√((x−1)^2 +y^2 ))+(√((x+1)^2 +y^2 ))=4 is an ellipse and find its semi major axis and semi minor axis.

$${prove}\:{that}\:{the}\:{curve}\: \\ $$$$\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }+\sqrt{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }=\mathrm{4}\: \\ $$$${is}\:{an}\:{ellipse}\:{and}\:{find}\:{its}\:{semi} \\ $$$${major}\:{axis}\:{and}\:{semi}\:{minor}\:{axis}. \\ $$

Question Number 196774    Answers: 1   Comments: 0

A man standing on top of Burj Khalifa. If the height of Burj Khalifa including a man is 830 m, then what is the maximum distance up to which a man can see objects on Earth? (Earth's radius 6371 km)

$$ \\ $$A man standing on top of Burj Khalifa. If the height of Burj Khalifa including a man is 830 m, then what is the maximum distance up to which a man can see objects on Earth? (Earth's radius 6371 km)

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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

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

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