Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com