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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
${ ((9cosθ=r→(1))),((cos(180°−2θ)=((r^2 +r^2 −4^2 )/(2r^2 ))→(2))) :} from(1)⇒cosθ=(r/9) from(2)⇒−cos2θ=1−(8/r^2 ) −cos2θ=1−2cos^2 θ=1−2((r/9))^2 =1−((2r^2 )/(81)) ⇒1−((2r^2 )/(81))=1−(8/r^2 ) ⇒r=3(√2)$
${\begin{cases} 9 c o s θ = r \to (1) \\ c o s (180 ° - 2 θ) = \frac{r^{2} + r^{2} - 4^{2}}{2 r^{2}} \to (2) \end{cases}$ $f r o m (1) \Rightarrow c o s θ = \frac{r}{9}$ $f r o m (2) \Rightarrow - c o s 2 θ = 1 - \frac{8}{r^{2}}$ $- c o s 2 θ = 1 - 2 {c o s}^{2} θ = 1 - 2 {(\frac{r}{9})}^{2} = 1 - \frac{2 r^{2}}{81}$ $\Rightarrow 1 - \frac{2 r^{2}}{81} = 1 - \frac{8}{r^{2}}$ $\Rightarrow r = 3 \sqrt{2}$
Commented by Tawa1 last updated on 25/Jul/19

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