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Question Number 203569 by cherokeesay last updated on 22/Jan/24
Commented by a.lgnaoui last updated on 22/Jan/24
2(a+b+c+d)=360 (a+b+c+d)=180 (a+b)=180−(c+d) tan (a+b)=−tan (c+d) ((t1+t2)/(1−t1.t2))=((t3+t4)/(t3t4−1)) t1=(R/(20)) t2=(R/(15)) t3=(R/(12)) t4 =(R/(10)) ⇒ 20t1=15t2 =12t3=10t4 t2=(4/3) t1 t4=(6/5)t3 ((7t1)/(3−4(t1)^2 ))=((11t3)/(6(t3)^2 −5)) (t3=(5/3)t1) ⇒ ((7t1)/(3−4(t1)^2 ))=((11×(5/3)t1)/(6(((25)/9))t1^2 −5))=((55t1)/(50(t1^2 )−15)) 7[50(t1)^2 −15]=55(3−4(t1)^2 350(t1^2 )−105=165−220(t1)^2 [350+220](t1)^2 =270 t1=(3/( (√(19)))) ⇒ (R/(20))=(3/( (√(19)))) R=((60)/( (√(19))))=13,764
$$\mathrm{2}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right)=\mathrm{360} \\ $$$$\:\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right)=\mathrm{180} \\ $$$$\:\:\left(\mathrm{a}+\mathrm{b}\right)=\mathrm{180}−\left(\mathrm{c}+\mathrm{d}\right) \\ $$$$ \\ $$$$\:\:\:\:\mathrm{tan}\:\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)=−\mathrm{tan}\:\left(\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{d}}\right) \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{t}}\mathrm{1}+\boldsymbol{\mathrm{t}}\mathrm{2}}{\mathrm{1}−\boldsymbol{\mathrm{t}}\mathrm{1}.\boldsymbol{\mathrm{t}}\mathrm{2}}=\frac{\boldsymbol{\mathrm{t}}\mathrm{3}+\boldsymbol{\mathrm{t}}\mathrm{4}}{\boldsymbol{\mathrm{t}}\mathrm{3}\boldsymbol{\mathrm{t}}\mathrm{4}−\mathrm{1}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{1}=\frac{\boldsymbol{\mathrm{R}}}{\mathrm{20}}\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{\boldsymbol{\mathrm{R}}}{\mathrm{15}}\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{3}=\frac{\boldsymbol{\mathrm{R}}}{\mathrm{12}}\:\:\:\boldsymbol{\mathrm{t}}\mathrm{4}\:=\frac{\boldsymbol{\mathrm{R}}}{\mathrm{10}} \\ $$$$\:\:\:\Rightarrow\:\:\:\mathrm{20}\boldsymbol{\mathrm{t}}\mathrm{1}=\mathrm{15}\boldsymbol{\mathrm{t}}\mathrm{2}\:\:\:=\mathrm{12}\boldsymbol{\mathrm{t}}\mathrm{3}=\mathrm{10}\boldsymbol{\mathrm{t}}\mathrm{4} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{\mathrm{4}}{\mathrm{3}}\:\boldsymbol{\mathrm{t}}\mathrm{1}\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{4}=\frac{\mathrm{6}}{\mathrm{5}}\boldsymbol{\mathrm{t}}\mathrm{3} \\ $$$$\:\:\:\frac{\mathrm{7}\boldsymbol{\mathrm{t}}\mathrm{1}}{\mathrm{3}−\mathrm{4}\left(\boldsymbol{\mathrm{t}}\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{11}\boldsymbol{\mathrm{t}}\mathrm{3}}{\mathrm{6}\left(\boldsymbol{\mathrm{t}}\mathrm{3}\right)^{\mathrm{2}} −\mathrm{5}}\:\:\:\left(\boldsymbol{\mathrm{t}}\mathrm{3}=\frac{\mathrm{5}}{\mathrm{3}}\boldsymbol{\mathrm{t}}\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{7t1}}{\mathrm{3}−\mathrm{4}\left(\mathrm{t1}\right)^{\mathrm{2}} }=\frac{\mathrm{11}×\frac{\mathrm{5}}{\mathrm{3}}\mathrm{t1}}{\mathrm{6}\left(\frac{\mathrm{25}}{\mathrm{9}}\right)\mathrm{t1}^{\mathrm{2}} −\mathrm{5}}=\frac{\mathrm{55t1}}{\mathrm{50}\left(\mathrm{t1}^{\mathrm{2}} \right)−\mathrm{15}} \\ $$$$\mathrm{7}\left[\mathrm{50}\left(\mathrm{t1}\right)^{\mathrm{2}} −\mathrm{15}\right]=\mathrm{55}\left(\mathrm{3}−\mathrm{4}\left(\mathrm{t1}\right)^{\mathrm{2}} \right. \\ $$$$\:\:\:\mathrm{350}\left(\mathrm{t1}^{\mathrm{2}} \right)−\mathrm{105}=\mathrm{165}−\mathrm{220}\left(\mathrm{t1}\right)^{\mathrm{2}} \\ $$$$\left[\mathrm{350}+\mathrm{220}\right]\left(\mathrm{t1}\right)^{\mathrm{2}} =\mathrm{270} \\ $$$$\:\:\:\mathrm{t1}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{19}}}\:\:\:\:\:\Rightarrow\:\frac{\boldsymbol{\mathrm{R}}}{\mathrm{20}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{19}}} \\ $$$$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{R}}=\frac{\mathrm{60}}{\:\sqrt{\mathrm{19}}}=\mathrm{13},\mathrm{764} \\ $$$$ \\ $$$$\: \\ $$
Commented by a.lgnaoui last updated on 22/Jan/24
Commented by cherokeesay last updated on 22/Jan/24
perfect ! thank you sir.
$${perfect}\:! \\ $$$${thank}\:{you}\:{sir}. \\ $$
Answered by mr W last updated on 22/Jan/24
tan^(−1) ((10)/R)+tan^(−1) ((12)/R)+tan^(−1) ((15)/R)+tan^(−1) ((20)/R)=π tan (tan^(−1) ((10)/R)+tan^(−1) ((12)/R))=−tan (tan^(−1) ((15)/R)+tan^(−1) ((20)/R)) ((((10)/R)+((12)/R))/(1−((10)/R)×((12)/R)))=−((((15)/R)+((20)/R))/(1−((15)/R)×((20)/R))) ((22)/(R^2 −120))=−((35)/(R^2 −300)) R^2 =((22×300+35×120)/(22+35))=((3600)/(19)) ⇒R=((60)/( (√(19))))≈13.765
$$\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{10}}{{R}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{12}}{{R}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{15}}{{R}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{20}}{{R}}=\pi \\ $$$$\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{10}}{{R}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{12}}{{R}}\right)=−\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{15}}{{R}}+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{20}}{{R}}\right) \\ $$$$\frac{\frac{\mathrm{10}}{{R}}+\frac{\mathrm{12}}{{R}}}{\mathrm{1}−\frac{\mathrm{10}}{{R}}×\frac{\mathrm{12}}{{R}}}=−\frac{\frac{\mathrm{15}}{{R}}+\frac{\mathrm{20}}{{R}}}{\mathrm{1}−\frac{\mathrm{15}}{{R}}×\frac{\mathrm{20}}{{R}}} \\ $$$$\frac{\mathrm{22}}{{R}^{\mathrm{2}} −\mathrm{120}}=−\frac{\mathrm{35}}{{R}^{\mathrm{2}} −\mathrm{300}} \\ $$$${R}^{\mathrm{2}} =\frac{\mathrm{22}×\mathrm{300}+\mathrm{35}×\mathrm{120}}{\mathrm{22}+\mathrm{35}}=\frac{\mathrm{3600}}{\mathrm{19}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{60}}{\:\sqrt{\mathrm{19}}}\approx\mathrm{13}.\mathrm{765} \\ $$
Commented by cherokeesay last updated on 22/Jan/24
So nice ! thank you master !
$${So}\:{nice}\:! \\ $$$${thank}\:{you}\:{master}\:! \\ $$

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