Question-185953

Question Number 185953 by ajfour last updated on 30/Jan/23

Commented by ajfour last updated on 30/Jan/23

$${Would}\:{it}\:{be}\:{r}\approx\mathrm{0}.\mathrm{432}\:? \\ $$

Commented by TUN last updated on 30/Jan/23

S(triangle)=p×r <=>(1/2)(√(1+r^2 ))×(√((1/x^2 )+1))=(((√(1+r^2 ))+(√((1/r^2 )+1))+r+(1/r))/2)×r <=>r=0.4320408003

$${S}\left({triangle}\right)={p}×{r} \\ $$$$<=>\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+{r}^{\mathrm{2}} }×\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}}=\frac{\sqrt{\mathrm{1}+{r}^{\mathrm{2}} }+\sqrt{\frac{\mathrm{1}}{{r}^{\mathrm{2}} }+\mathrm{1}}+{r}+\frac{\mathrm{1}}{{r}}}{\mathrm{2}}×{r} \\ $$$$<=>{r}=\mathrm{0}.\mathrm{4320408003} \\ $$$$ \\ $$

Commented by ajfour last updated on 30/Jan/23

$${thanks},\:{I}\:{understand}.\:{wow}! \\ $$

Answered by mr W last updated on 30/Jan/23

Commented by mr W last updated on 30/Jan/23

tan θ=(r/1)=r cos θ=((r+(r/(tan (θ/2))))/((r/(tan (θ/2)))+(√(1+r^2 ))−r)) cos θ=((1+(1/(tan (θ/2))))/((1/(tan (θ/2)))+(√(1+(1/(tan^2 θ))))−1)) let t=tan (θ/2) ((1−t^2 )/(1+t^2 ))=((1+t)/(1−t+t(√(1+(((1−t^2 )/(2t)))^2 )))) t^3 −t^2 +5t−1=0 (s+(1/3))^3 −(s+(1/3))^2 +5(s+(1/3))−1=0 s^3 +((14s)/3)+((16)/(27))=0 s=((((2(√(78)))/9)−(8/(27))))^(1/3) −((((2(√(78)))/9)+(8/(27))))^(1/3) t=(1/3)(1+((6(√(78))−8))^(1/3) −((6(√(78))+8))^(1/3) ) ≈0.206783 ⇒r=((2t)/(1−t^2 ))≈0.432

$$\mathrm{tan}\:\theta=\frac{{r}}{\mathrm{1}}={r} \\ $$$$\mathrm{cos}\:\theta=\frac{{r}+\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}}{\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\sqrt{\mathrm{1}+{r}^{\mathrm{2}} }−{r}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}}{\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\theta}}−\mathrm{1}} \\ $$$${let}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }=\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}+{t}\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}\right)^{\mathrm{2}} }} \\ $$$${t}^{\mathrm{3}} −{t}^{\mathrm{2}} +\mathrm{5}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\left({s}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} −\left({s}+\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{5}\left({s}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\mathrm{1}=\mathrm{0} \\ $$$${s}^{\mathrm{3}} +\frac{\mathrm{14}{s}}{\mathrm{3}}+\frac{\mathrm{16}}{\mathrm{27}}=\mathrm{0} \\ $$$${s}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{\mathrm{78}}}{\mathrm{9}}−\frac{\mathrm{8}}{\mathrm{27}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{2}\sqrt{\mathrm{78}}}{\mathrm{9}}+\frac{\mathrm{8}}{\mathrm{27}}} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{78}}−\mathrm{8}}−\sqrt[{\mathrm{3}}]{\mathrm{6}\sqrt{\mathrm{78}}+\mathrm{8}}\right) \\ $$$$\:\:\approx\mathrm{0}.\mathrm{206783} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\approx\mathrm{0}.\mathrm{432} \\ $$

Commented by ajfour last updated on 30/Jan/23

$${Thank}\:{you}\:{Sir}.\:{All}\:{correct}! \\ $$

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