Question-214350

Question Number 214350 by ajfour last updated on 06/Dec/24

Answered by mr W last updated on 06/Dec/24

tan α=(1/2) ⇒tan (α/2)=(√5)−2 2+r tan (α/2)=(√((r+1)^2 −(r−1)^2 )) 2+((√5)−2)r=2(√r) ((√5)−2)r−2(√r)+2=0 (√r)=((1−(√(5−2(√5))))/( (√5)−2)) ⇒r=(((1−(√(5−2(√5))))/( (√5)−2)))^2 ≈1.342

$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\mathrm{5}}−\mathrm{2} \\ $$$$\mathrm{2}+{r}\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\sqrt{\left({r}+\mathrm{1}\right)^{\mathrm{2}} −\left({r}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}+\left(\sqrt{\mathrm{5}}−\mathrm{2}\right){r}=\mathrm{2}\sqrt{{r}} \\ $$$$\left(\sqrt{\mathrm{5}}−\mathrm{2}\right){r}−\mathrm{2}\sqrt{{r}}+\mathrm{2}=\mathrm{0} \\ $$$$\sqrt{{r}}=\frac{\mathrm{1}−\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$$\Rightarrow{r}=\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{5}}−\mathrm{2}}\right)^{\mathrm{2}} \approx\mathrm{1}.\mathrm{342} \\ $$

Commented by ajfour last updated on 06/Dec/24

https://youtu.be/6veUu8wBV_8?si=aJGNsb2DtsOBFYbJ

Commented by ajfour last updated on 06/Dec/24

sir could you check the question in the video for me?

Commented by mr W last updated on 06/Dec/24

$${for}\:{some}\:{reason}\:{i}\:{can}'{t}\:{open}\:{this} \\ $$$${video}. \\ $$

Commented by ajfour last updated on 06/Dec/24

$${try}\:{again}\:{now}\:{please} \\ $$

Commented by mr W last updated on 06/Dec/24

$${i}\:{can}\:{view}\:{it}\:{now}.\:{great}! \\ $$

Answered by A5T last updated on 06/Dec/24

Commented by A5T last updated on 06/Dec/24

AF=(√((1+r)^2 −r^2 ))=(√(1+2r)) ⇒FE=(√(1^2 +2^2 ))−(√(1+2r))=(√5)−(√(1+2r))=EC (1+r)^2 =(r−1)^2 +(2+(√5)−(√(1+2r)))^2 ⇒r=14+6(√5)−2(√(85+38(√5)))≈1.3419

$${AF}=\sqrt{\left(\mathrm{1}+{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }=\sqrt{\mathrm{1}+\mathrm{2}{r}} \\ $$$$\Rightarrow{FE}=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }−\sqrt{\mathrm{1}+\mathrm{2}{r}}=\sqrt{\mathrm{5}}−\sqrt{\mathrm{1}+\mathrm{2}{r}}={EC} \\ $$$$\left(\mathrm{1}+{r}\right)^{\mathrm{2}} =\left({r}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{1}+\mathrm{2}{r}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{14}+\mathrm{6}\sqrt{\mathrm{5}}−\mathrm{2}\sqrt{\mathrm{85}+\mathrm{38}\sqrt{\mathrm{5}}}\approx\mathrm{1}.\mathrm{3419} \\ $$

Commented by ajfour last updated on 06/Dec/24

$${Thanks}.\:{I}\:{see}. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply