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Question Number 205406 by mr W last updated on 20/Mar/24
Commented by lepuissantcedricjunior last updated on 21/Mar/24
dβ€²apreβ€²s pythagore d^2 =7^2 +k^2 (1) dβ€²apres alkashil 15^2 +15^2 =k^2 +2Γ—15^2 cos(π›‘βˆ’a) 2.15^2 =k^2 βˆ’2.15^2 cosa k^2 =2.15^2 (1+cos^2 a) (2) (k/(sin(π›‘βˆ’a)))=((15)/(sina))=((15^2 k)/(2A))=2r=d (k/(cosa))=((15)/(sina))=((15^2 .k)/(2A))=d cos(a)=(k/d) (2) remplacons dans (2) k^2 =2.15^2 (1+(k^2 /d^2 ))=(((15d(√2))^2 +(15(√2)k)^2 )/d^2 ) k^2 d^2 βˆ’(15(√2))^2 k^2 =2.15^2 d^2 k^2 =((2.15^2 d^2 )/(d^2 βˆ’2.15^2 )) (3) remplacons dans(1) d^2 =7^2 +((2.15^2 d^2 )/(d^2 βˆ’2.15^2 )) =>d^4 βˆ’d^2 (7^2 +2.15^2 )+2.(7.15)^2 =0 ⇔d^4 βˆ’d^2 (49+450)+2(105)^2 =0 posons d^2 =X=>d^4 =X^2 X^2 βˆ’499X+22050=0 β–³=249001βˆ’88200=160801=401^2 X=((499βˆ’401)/2)=49 X=((499+401)/2)=450 or X=d^2 =>d=(√(49))=7 impossible d=(√(450))=15(√2) d ou determinant (((dβ‰ 7)),((d=15(√2) (√)))) ============================= ...............le puissant Dr.........
$$\boldsymbol{{d}}'\boldsymbol{{apre}}'\boldsymbol{{s}}\:\boldsymbol{{pythagore}}\: \\ $$$$\boldsymbol{{d}}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \:\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{d}}'\boldsymbol{{apres}}\:\boldsymbol{{alkashil}} \\ $$$$\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} =\boldsymbol{{k}}^{\mathrm{2}} +\mathrm{2}Γ—\mathrm{15}^{\mathrm{2}} \boldsymbol{{cos}}\left(\boldsymbol{\pi}βˆ’\boldsymbol{{a}}\right) \\ $$$$\mathrm{2}.\mathrm{15}^{\mathrm{2}} \:\:\:\:\:\:\:=\boldsymbol{{k}}^{\mathrm{2}} βˆ’\mathrm{2}.\mathrm{15}^{\mathrm{2}} \boldsymbol{{cosa}} \\ $$$$\boldsymbol{{k}}^{\mathrm{2}} =\mathrm{2}.\mathrm{15}^{\mathrm{2}} \:\left(\mathrm{1}+\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{a}}\right)\:\left(\mathrm{2}\right) \\ $$$$\frac{\boldsymbol{{k}}}{\boldsymbol{{sin}}\left(\boldsymbol{\pi}βˆ’\boldsymbol{{a}}\right)}=\frac{\mathrm{15}}{\boldsymbol{{sina}}}=\frac{\mathrm{15}^{\mathrm{2}} \boldsymbol{{k}}}{\mathrm{2}\boldsymbol{{A}}}=\mathrm{2}{r}={d} \\ $$$$\frac{\boldsymbol{{k}}}{\boldsymbol{{cosa}}}=\frac{\mathrm{15}}{\boldsymbol{{sina}}}=\frac{\mathrm{15}^{\mathrm{2}} .\boldsymbol{{k}}}{\mathrm{2}\boldsymbol{{A}}}=\boldsymbol{{d}} \\ $$$$\boldsymbol{{cos}}\left(\boldsymbol{{a}}\right)=\frac{\boldsymbol{{k}}}{\boldsymbol{{d}}}\:\left(\mathrm{2}\right)\:\boldsymbol{{remplacons}}\:\boldsymbol{{dans}}\:\left(\mathrm{2}\right) \\ $$$$\boldsymbol{{k}}^{\mathrm{2}} =\mathrm{2}.\mathrm{15}^{\mathrm{2}} \left(\mathrm{1}+\frac{\boldsymbol{{k}}^{\mathrm{2}} }{\boldsymbol{{d}}^{\mathrm{2}} }\right)=\frac{\left(\mathrm{15}\boldsymbol{{d}}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{15}\sqrt{\mathrm{2}}\boldsymbol{{k}}\right)^{\mathrm{2}} }{\boldsymbol{{d}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{k}}^{\mathrm{2}} \boldsymbol{{d}}^{\mathrm{2}} βˆ’\left(\mathrm{15}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \boldsymbol{{k}}^{\mathrm{2}} =\mathrm{2}.\mathrm{15}^{\mathrm{2}} \boldsymbol{{d}}^{\mathrm{2}} \\ $$$$\boldsymbol{{k}}^{\mathrm{2}} =\frac{\mathrm{2}.\mathrm{15}^{\mathrm{2}} \boldsymbol{{d}}^{\mathrm{2}} }{\boldsymbol{{d}}^{\mathrm{2}} βˆ’\mathrm{2}.\mathrm{15}^{\mathrm{2}} }\:\:\:\left(\mathrm{3}\right)\:\boldsymbol{{remplacons}}\:\boldsymbol{{dans}}\left(\mathrm{1}\right) \\ $$$$\boldsymbol{{d}}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +\frac{\mathrm{2}.\mathrm{15}^{\mathrm{2}} \boldsymbol{{d}}^{\mathrm{2}} }{\boldsymbol{{d}}^{\mathrm{2}} βˆ’\mathrm{2}.\mathrm{15}^{\mathrm{2}} } \\ $$$$=>\boldsymbol{{d}}^{\mathrm{4}} βˆ’\boldsymbol{{d}}^{\mathrm{2}} \left(\mathrm{7}^{\mathrm{2}} +\mathrm{2}.\mathrm{15}^{\mathrm{2}} \right)+\mathrm{2}.\left(\mathrm{7}.\mathrm{15}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{d}}^{\mathrm{4}} βˆ’\boldsymbol{\mathrm{d}}^{\mathrm{2}} \left(\mathrm{49}+\mathrm{450}\right)+\mathrm{2}\left(\mathrm{105}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\boldsymbol{\mathrm{posons}}\:\boldsymbol{\mathrm{d}}^{\mathrm{2}} =\boldsymbol{\mathrm{X}}=>\boldsymbol{{d}}^{\mathrm{4}} =\boldsymbol{{X}}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{X}}^{\mathrm{2}} βˆ’\mathrm{499}\boldsymbol{\mathrm{X}}+\mathrm{22050}=\mathrm{0} \\ $$$$\bigtriangleup=\mathrm{249001}βˆ’\mathrm{88200}=\mathrm{160801}=\mathrm{401}^{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{X}}=\frac{\mathrm{499}βˆ’\mathrm{401}}{\mathrm{2}}=\mathrm{49} \\ $$$$\mathrm{X}=\frac{\mathrm{499}+\mathrm{401}}{\mathrm{2}}=\mathrm{450} \\ $$$${or}\:\boldsymbol{\mathrm{X}}=\boldsymbol{{d}}^{\mathrm{2}} =>\boldsymbol{{d}}=\sqrt{\mathrm{49}}=\mathrm{7}\:\boldsymbol{{impossible}} \\ $$$$\boldsymbol{{d}}=\sqrt{\mathrm{450}}=\mathrm{15}\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{{d}}\:\boldsymbol{{ou}} \\ $$$$\begin{array}{|c|c|}{\boldsymbol{{d}}\neq\mathrm{7}}\\{\boldsymbol{{d}}=\mathrm{15}\sqrt{\mathrm{2}}\:\sqrt{}}\\\hline\end{array} \\ $$$$============================= \\ $$$$……………{le}\:{puissant}\:{D}\boldsymbol{{r}}……… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 21/Mar/24
please post your answer to a question as β€œanswer”, not as β€œcomment”! thanks! btw, your answer 15(√2) is wrong.
$${please}\:{post}\:{your}\:{answer}\:{to}\:{a}\:{question} \\ $$$${as}\:“{answer}'',\:{not}\:{as}\:“{comment}''!\:{thanks}! \\ $$$${btw},\:{your}\:{answer}\:\mathrm{15}\sqrt{\mathrm{2}}\:{is}\:{wrong}. \\ $$
Answered by A5T last updated on 20/Mar/24
(√(15^2 +15^2 βˆ’2Γ—15^2 cosΞΈ))=(√(7^2 +d^2 +14dcosΞΈ)) β‡’401βˆ’450cosΞΈ=d^2 +14dcosΞΈβ‡’cosΞΈ=((d^2 βˆ’401)/(βˆ’450βˆ’14d)) r^2 =7^2 +r^2 +2Γ—7rcosΞΈβ‡’dcosΞΈ=βˆ’7β‡’cosΞΈ=((βˆ’7)/d) β‡’((d^2 βˆ’401)/(βˆ’450βˆ’14d))=((βˆ’7)/d)β‡’d=25
$$\sqrt{\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} βˆ’\mathrm{2}Γ—\mathrm{15}^{\mathrm{2}} {cos}\theta}=\sqrt{\mathrm{7}^{\mathrm{2}} +{d}^{\mathrm{2}} +\mathrm{14}{dcos}\theta} \\ $$$$\Rightarrow\mathrm{401}βˆ’\mathrm{450}{cos}\theta={d}^{\mathrm{2}} +\mathrm{14}{dcos}\theta\Rightarrow{cos}\theta=\frac{{d}^{\mathrm{2}} βˆ’\mathrm{401}}{βˆ’\mathrm{450}βˆ’\mathrm{14}{d}} \\ $$$${r}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}Γ—\mathrm{7}{rcos}\theta\Rightarrow{dcos}\theta=βˆ’\mathrm{7}\Rightarrow{cos}\theta=\frac{βˆ’\mathrm{7}}{{d}} \\ $$$$\Rightarrow\frac{{d}^{\mathrm{2}} βˆ’\mathrm{401}}{βˆ’\mathrm{450}βˆ’\mathrm{14}{d}}=\frac{βˆ’\mathrm{7}}{{d}}\Rightarrow{d}=\mathrm{25} \\ $$
Commented by A5T last updated on 20/Mar/24
Commented by A5T last updated on 20/Mar/24
General formula, take a=7,b=15,c=15,d=2R
$${General}\:{formula},\:{take}\:{a}=\mathrm{7},{b}=\mathrm{15},{c}=\mathrm{15},{d}=\mathrm{2}{R} \\ $$
Commented by mr W last updated on 20/Mar/24
thanks!
$${thanks}! \\ $$
Answered by mr W last updated on 20/Mar/24
Commented by mr W last updated on 20/Mar/24
R^2 βˆ’((7/2))^2 =15^2 βˆ’(Rβˆ’(7/2))^2 2R^2 βˆ’7Rβˆ’15^2 =0 R=((7+(√(49+8Γ—15^2 )))/4)=12.5 β‡’d=2R=25 βœ“
$${R}^{\mathrm{2}} βˆ’\left(\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{15}^{\mathrm{2}} βˆ’\left({R}βˆ’\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{R}^{\mathrm{2}} βˆ’\mathrm{7}{R}βˆ’\mathrm{15}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}=\frac{\mathrm{7}+\sqrt{\mathrm{49}+\mathrm{8}Γ—\mathrm{15}^{\mathrm{2}} }}{\mathrm{4}}=\mathrm{12}.\mathrm{5} \\ $$$$\Rightarrow{d}=\mathrm{2}{R}=\mathrm{25}\:\checkmark \\ $$
Answered by cortano12 last updated on 20/Mar/24
β‡’ 7^2 +15^2 +15^2 +((7.15.15)/R)=4R^2 β‡’499+((1575)/R)= 4R^2 β‡’4R^3 βˆ’499Rβˆ’1575=0 β‡’(2Rβˆ’25)(R+9)(2R+7)=0 β‡’ d = 2R = 25
$$\:\:\Rightarrow\:\mathrm{7}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} +\frac{\mathrm{7}.\mathrm{15}.\mathrm{15}}{\mathrm{R}}=\mathrm{4R}^{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{499}+\frac{\mathrm{1575}}{\mathrm{R}}=\:\mathrm{4R}^{\mathrm{2}} \\ $$$$\:\Rightarrow\mathrm{4R}^{\mathrm{3}} βˆ’\mathrm{499R}βˆ’\mathrm{1575}=\mathrm{0} \\ $$$$\:\Rightarrow\left(\mathrm{2R}βˆ’\mathrm{25}\right)\left(\mathrm{R}+\mathrm{9}\right)\left(\mathrm{2R}+\mathrm{7}\right)=\mathrm{0} \\ $$$$\:\:\Rightarrow\:\mathrm{d}\:=\:\mathrm{2R}\:=\:\mathrm{25}\: \\ $$
Commented by mr W last updated on 20/Mar/24
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