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Question-65119




Question Number 65119 by Tawa1 last updated on 25/Jul/19
Commented by Tony Lin last updated on 25/Jul/19
Commented by Tony Lin last updated on 25/Jul/19
 { ((9cosθ=r→(1))),((cos(180°−2θ)=((r^2 +r^2 −4^2 )/(2r^2 ))→(2))) :}  from(1)⇒cosθ=(r/9)  from(2)⇒−cos2θ=1−(8/r^2 )  −cos2θ=1−2cos^2 θ=1−2((r/9))^2 =1−((2r^2 )/(81))  ⇒1−((2r^2 )/(81))=1−(8/r^2 )  ⇒r=3(√2)
$$\begin{cases}{\mathrm{9}{cos}\theta={r}\rightarrow\left(\mathrm{1}\right)}\\{{cos}\left(\mathrm{180}°−\mathrm{2}\theta\right)=\frac{{r}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} }{\mathrm{2}{r}^{\mathrm{2}} }\rightarrow\left(\mathrm{2}\right)}\end{cases} \\ $$$${from}\left(\mathrm{1}\right)\Rightarrow{cos}\theta=\frac{{r}}{\mathrm{9}} \\ $$$${from}\left(\mathrm{2}\right)\Rightarrow−{cos}\mathrm{2}\theta=\mathrm{1}−\frac{\mathrm{8}}{{r}^{\mathrm{2}} } \\ $$$$−{cos}\mathrm{2}\theta=\mathrm{1}−\mathrm{2}{cos}^{\mathrm{2}} \theta=\mathrm{1}−\mathrm{2}\left(\frac{{r}}{\mathrm{9}}\right)^{\mathrm{2}} =\mathrm{1}−\frac{\mathrm{2}{r}^{\mathrm{2}} }{\mathrm{81}} \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}{r}^{\mathrm{2}} }{\mathrm{81}}=\mathrm{1}−\frac{\mathrm{8}}{{r}^{\mathrm{2}} } \\ $$$$\Rightarrow{r}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 25/Jul/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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