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Question Number 158543 by ajfour last updated on 06/Nov/21

Commented by ajfour last updated on 06/Nov/21

$T h e q u a d r i l a t e r a l i s a s q u a r e .$
Commented by mr W last updated on 06/Nov/21

Commented by mr W last updated on 06/Nov/21

$\tan θ = a$ $φ = \frac{π}{4} + \frac{θ}{2}$ $R a d i u s b i g c i r c l e = a + \frac{a}{\sin φ} = \frac{a}{\sin θ \cos θ}$ $1 + \frac{1}{\sin (\frac{π}{4} + \frac{θ}{2})} = \frac{1}{\sin θ \cos θ}$ $\Rightarrow θ \approx 35.0446 °$ $\Rightarrow a \approx 0.7014$
Commented by Tawa11 last updated on 06/Nov/21

$Weldone sir$
	Answered by ajfour last updated on 06/Nov/21
	$let a=tan θ 1+a^2 =R (R−a)^2 =a^2 +b^2 a^2 +a^4 ={(a^2 +b)sin θ+b}^2 ⇒ (1+a^2 −a)^2 =a^2 +b^2 a^2 +a^4 ={((a(a^2 +b)+b(√(1+a^2 )))/( (√(1+a^2 ))))}^2 a(1+a^2 )=a^3 + (a+(√(1+a^2 )))(1−a)(√(1+a^2 )) ⇒ a((√(1+a^2 ))−a)=(1−a)(√(1+a^2 )) ⇒ (2a−1)(√(1+a^2 ))=a^2 ⇒ (2−(1/a))(√(1+(1/a^2 )))=1 ⇒ a≈0.70136 (>(1/2))$
	$l e t a = \tan θ$ $1 + a^{2} = R$ ${(R - a)}^{2} = a^{2} + b^{2}$ $a^{2} + a^{4} = {(a^{2} + b) \sin θ + b}^{2}$ $\Rightarrow$ ${(1 + a^{2} - a)}^{2} = a^{2} + b^{2}$ $a^{2} + a^{4} = {\frac{a (a^{2} + b) + b \sqrt{1 + a^{2}}}{\sqrt{1 + a^{2}}}}^{2}$ $a (1 + a^{2}) = a^{3} +$ $(a + \sqrt{1 + a^{2}}) (1 - a) \sqrt{1 + a^{2}}$ $\Rightarrow$ $a (\sqrt{1 + a^{2}} - a) = (1 - a) \sqrt{1 + a^{2}}$ $\Rightarrow (2 a - 1) \sqrt{1 + a^{2}} = a^{2}$ $\Rightarrow (2 - \frac{1}{a}) \sqrt{1 + \frac{1}{a^{2}}} = 1$ $\Rightarrow a \approx 0.70136 (> \frac{1}{2})$
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