Question-208493

Question Number 208493 by Tawa11 last updated on 17/Jun/24

Answered by A5T last updated on 17/Jun/24

a(3a)=x^2 ⇒x=a(√3) (4a)^2 =2(a(√3)+y)^2 =16a^2 ⇒r=y=2a(√2)−a(√3) ⇒r=y=a(2(√2)−(√3)) a^2 (2(√2)−(√3))^2 =a^2 (2(√2)−(√3))^2 +4a^2 −4a^2 (2(√2)−(√3))cosθ ⇒θ=cos^(−1) ((1/(2(√2)−(√3))))≈24.2°

$${a}\left(\mathrm{3}{a}\right)={x}^{\mathrm{2}} \Rightarrow{x}={a}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}\sqrt{\mathrm{3}}+{y}\right)^{\mathrm{2}} =\mathrm{16}{a}^{\mathrm{2}} \Rightarrow{r}={y}=\mathrm{2}{a}\sqrt{\mathrm{2}}−{a}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{r}={y}={a}\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right) \\ $$$${a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){cos}\theta \\ $$$$\Rightarrow\theta={cos}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}}\right)\approx\mathrm{24}.\mathrm{2}° \\ $$

Commented by Tawa11 last updated on 17/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 17/Jun/24

Am still trying to understand your workings sir. It will be appreciated if you can expanciate sir.

$$\mathrm{Am}\:\mathrm{still}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{workings}\:\mathrm{sir}. \\ $$$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{appreciated}\:\mathrm{if}\:\mathrm{you}\:\mathrm{can}\:\mathrm{expanciate}\:\mathrm{sir}. \\ $$

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^2 =a(a+2a)=a(3a) ⇒x=a(√3); y=r and Pythagoras theorem gives: (a+2a+a)^2 =(x+y)^2 +(x+y)^2 =2(x+y)^2 ⇒(4a)^2 =2(a(√3)+y)^2 ⇒r=y=a(2(√2)−(√3)) Cosine rule on the triangle with sides r,r,2a gives the last equation.

$$'{Power}\:{of}\:{a}\:{point}'\:{will}\:{give}\:{x}^{\mathrm{2}} ={a}\left({a}+\mathrm{2}{a}\right)={a}\left(\mathrm{3}{a}\right) \\ $$$$\Rightarrow{x}={a}\sqrt{\mathrm{3}};\:\:\:\:\:\: \\ $$$$\:{y}={r}\:{and}\:{Pythagoras}\:{theorem}\:{gives}:\: \\ $$$$\left({a}+\mathrm{2}{a}+{a}\right)^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} +\left({x}+{y}\right)^{\mathrm{2}} =\mathrm{2}\left({x}+{y}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{2}\left({a}\sqrt{\mathrm{3}}+{y}\right)^{\mathrm{2}} \Rightarrow{r}={y}={a}\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right) \\ $$$${Cosine}\:{rule}\:{on}\:{the}\:{triangle}\:{with}\:{sides}\:{r},{r},\mathrm{2}{a} \\ $$$${gives}\:{the}\:{last}\:{equation}. \\ $$

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=2a DE=(√(r^2 −a^2 )) AD=(√2)r (√2)r+(√(r^2 −a^2 ))=2a r^2 −a^2 =(2a−(√2)r)^2 =4a^2 +2r^2 −4(√2)ar r^2 −4(√2)ar+5a^2 =0 r=(2(√2)−(√3))a cos θ=(a/((2(√2)−(√3))a))=((2(√2)+(√3))/5) ⇒θ=cos^(−1) ((2(√2)+(√3))/5)≈24.203°

$${AE}=\mathrm{2}{a} \\ $$$${DE}=\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${AD}=\sqrt{\mathrm{2}}{r} \\ $$$$\sqrt{\mathrm{2}}{r}+\sqrt{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\mathrm{2}{a} \\ $$$${r}^{\mathrm{2}} −{a}^{\mathrm{2}} =\left(\mathrm{2}{a}−\sqrt{\mathrm{2}}{r}\right)^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{r}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}{ar} \\ $$$${r}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}{ar}+\mathrm{5}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${r}=\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){a} \\ $$$$\mathrm{cos}\:\theta=\frac{{a}}{\left(\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right){a}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}{\mathrm{5}} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}}{\mathrm{5}}\approx\mathrm{24}.\mathrm{203}° \\ $$

Commented by Tawa11 last updated on 17/Jun/24

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by Tawa11 last updated on 17/Jun/24

Please, how is AD = (√2)r I got every other steps sir.

$$\mathrm{Please},\:\mathrm{how}\:\mathrm{is}\:\mathrm{AD}\:\:=\:\:\sqrt{\mathrm{2}}\mathrm{r} \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{every}\:\mathrm{other}\:\mathrm{steps}\:\mathrm{sir}. \\ $$

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