Question Number 208493 by Tawa11 last updated on 17/Jun/24
Answered by A5T last updated on 17/Jun/24
$${a}\left(\mathrm{3}{a}\right)={x}^{\mathrm{2}} \Rightarrow{x}={a}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}\sqrt{\mathrm{3}}+{y}\right)^{\mathrm{2}} =\mathrm{16}{a}^{\mathrm{2}} \Rightarrow{r}={y}=\mathrm{2}{a}\sqrt{\mathrm{2}}−{a}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{r}={y}={a}\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right) \\ $$$${a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){cos}\theta \\ $$$$\Rightarrow\theta={cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}\right)\approx\mathrm{24}.\mathrm{2}° \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{Am}\:\mathrm{still}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{workings}\:\mathrm{sir}. \\ $$$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{appreciated}\:\mathrm{if}\:\mathrm{you}\:\mathrm{can}\:\mathrm{expanciate}\:\mathrm{sir}. \\ $$
Commented by A5T last updated on 17/Jun/24
Commented by A5T last updated on 17/Jun/24
$$'{Power}\:{of}\:{a}\:{point}'\:{will}\:{give}\:{x}^{\mathrm{2}} ={a}\left({a}+\mathrm{2}{a}\right)={a}\left(\mathrm{3}{a}\right) \\ $$$$\Rightarrow{x}={a}\sqrt{\mathrm{3}};\:\:\:\:\:\: \\ $$$$\:{y}={r}\:{and}\:{Pythagoras}\:{theorem}\:{gives}:\: \\ $$$$\left({a}+\mathrm{2}{a}+{a}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{2}\left({x}+{y}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}\sqrt{\mathrm{3}}+{y}\right)^{\mathrm{2}} \Rightarrow{r}={y}={a}\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right) \\ $$$${Cosine}\:{rule}\:{on}\:{the}\:{triangle}\:{with}\:{sides}\:{r},{r},\mathrm{2}{a} \\ $$$${gives}\:{the}\:{last}\:{equation}. \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 17/Jun/24
Commented by mr W last updated on 17/Jun/24
$${AE}=\mathrm{2}{a} \\ $$$${DE}=\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${AD}=\sqrt{\mathrm{2}}{r} \\ $$$$\sqrt{\mathrm{2}}{r}+\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\mathrm{2}{a} \\ $$$${r}^{\mathrm{2}} −{a}^{\mathrm{2}} =\left(\mathrm{2}{a}−\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}{ar} \\ $$$${r}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}{ar}+\mathrm{5}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${r}=\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){a} \\ $$$$\mathrm{cos}\:\theta=\frac{{a}}{\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){a}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}{\mathrm{5}}\approx\mathrm{24}.\mathrm{203}° \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$