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Determiner-r-et-R-voir-figure-Sachant-aue-C-120-AB-12-




Question Number 200761 by a.lgnaoui last updated on 23/Nov/23
Determiner  r  et  R  (voir figure )   Sachant aue:∡C=120  :AB=12
$$\boldsymbol{\mathrm{Determiner}}\:\:\boldsymbol{\mathrm{r}}\:\:\mathrm{et}\:\:\boldsymbol{\mathrm{R}}\:\:\left({voir}\:{figure}\:\right) \\ $$$$\:\mathrm{Sachant}\:\mathrm{aue}:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{120}\:\::\boldsymbol{{AB}}=\mathrm{12} \\ $$
Commented by a.lgnaoui last updated on 23/Nov/23
Commented by mr W last updated on 23/Nov/23
do you mean “r=? and R=?” or  “r+R=?” ?
$${do}\:{you}\:{mean}\:“{r}=?\:{and}\:{R}=?''\:{or} \\ $$$$“{r}+{R}=?''\:? \\ $$
Commented by a.lgnaoui last updated on 23/Nov/23
find r  and R
$$\mathrm{find}\:\boldsymbol{\mathrm{r}}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{R}} \\ $$
Commented by a.lgnaoui last updated on 23/Nov/23
(A and B are the centre of circle :radius r)
$$\left(\boldsymbol{\mathrm{A}}\:\mathrm{and}\:\boldsymbol{\mathrm{B}}\:\mathrm{are}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{circle}\::\mathrm{radius}\:\boldsymbol{\mathrm{r}}\right)\: \\ $$
Commented by mr W last updated on 23/Nov/23
i think you can not determine r and  R. you can only determine the sum  of r and R, i.e. r+R.
$${i}\:{think}\:{you}\:{can}\:{not}\:{determine}\:{r}\:{and} \\ $$$${R}.\:{you}\:{can}\:{only}\:{determine}\:{the}\:{sum} \\ $$$${of}\:{r}\:{and}\:{R},\:{i}.{e}.\:{r}+{R}. \\ $$
Answered by mr W last updated on 23/Nov/23
AC=r+R−5  BC=r+R−3  say s=r+R  (s−5)^2 +(s−3)^2 +(s−5)(s−3)=12^2   3s^2 −24s−95=0  s=((12+(√(429)))/3)≈10.9
$${AC}={r}+{R}−\mathrm{5} \\ $$$${BC}={r}+{R}−\mathrm{3} \\ $$$${say}\:{s}={r}+{R} \\ $$$$\left({s}−\mathrm{5}\right)^{\mathrm{2}} +\left({s}−\mathrm{3}\right)^{\mathrm{2}} +\left({s}−\mathrm{5}\right)\left({s}−\mathrm{3}\right)=\mathrm{12}^{\mathrm{2}} \\ $$$$\mathrm{3}{s}^{\mathrm{2}} −\mathrm{24}{s}−\mathrm{95}=\mathrm{0} \\ $$$${s}=\frac{\mathrm{12}+\sqrt{\mathrm{429}}}{\mathrm{3}}\approx\mathrm{10}.\mathrm{9} \\ $$
Commented by a.lgnaoui last updated on 26/Nov/23
thanks
$$\mathrm{thanks} \\ $$

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