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Question Number 185915 by ajfour last updated on 29/Jan/23

Commented by ajfour last updated on 29/Jan/23

$I f b o t h b l u e s e g m e n t s h a v e l e n g t h$ $x, f i n d x .$
	Answered by ajfour last updated on 29/Jan/23

	$S a y x = \tan θ$ $x^{2} = {(1 + \sin^{2} θ)}^{2} + \sin^{2} θ \cos^{2} θ$ $\Rightarrow \sin^{2} θ = \cos^{2} θ {(1 + \sin^{2} θ)}^{2}$ $+ \sin^{2} θ \cos^{4} θ$ $1 - s = s {(2 - s)}^{2} + (1 - s) s^{2}$ $\Rightarrow 1 - s = 4 s - 3 s^{2}$ $3 s^{2} - 5 s + 1 = 0$ $s = \frac{5 \pm \sqrt{13}}{6} = \cos^{2} θ$ $\frac{1}{s} = \frac{1}{\cos^{2} θ} = \frac{5 \pm \sqrt{13}}{2}$ $x = \tan θ = \sqrt{\frac{1}{\cos^{2} θ} - 1}$ $= \sqrt{\frac{3 \pm \sqrt{13}}{2}}$ $t h e r e f o r e x = \sqrt{\frac{3 + \sqrt{13}}{2}} \approx 1.8174$
	Commented by ajfour last updated on 29/Jan/23

	Answered by mr W last updated on 29/Jan/23

	Commented by mr W last updated on 29/Jan/23

	$m = \tan θ = \frac{1}{h}$ $e q n . o f A C :$ $y = (x - 1) m$ $e q n . o f B C :$ $y = h - \frac{(x - 1)}{m}$ $y_{C} = h - \frac{(x_{C} - 1)}{m} = (x_{C} - 1) m$ $h + m + \frac{1}{m} = (m + \frac{1}{m}) x_{C}$ $x_{C} = 1 + \frac{h}{m + \frac{1}{m}} = 1 + \frac{h}{h + \frac{1}{h}} = 1 + \frac{h^{2}}{h^{2} + 1}$ $y_{C} = \frac{h}{h^{2} + 1}$ $x_{C}^{2} + y_{C}^{2} = h^{2}$ ${(1 + \frac{h^{2}}{h^{2} + 1})}^{2} + \frac{h^{2}}{{(h^{2} + 1)}^{2}} = h^{2}$ $h^{6} - 2 h^{4} - 4 h^{2} - 1 = 0$ $(h^{2} + 1) (h^{4} - 3 h^{2} - 1) = 0$ $\Rightarrow h^{2} = \frac{3 + \sqrt{13}}{2}, (\frac{3 - \sqrt{13}}{2}, - 1 r e j e c t e d)$ $\Rightarrow h = \sqrt{\frac{3 + \sqrt{13}}{2}} \approx 1.817 ✓$
	Commented by ajfour last updated on 29/Jan/23

	$T h a n k y o u s i r .$
	Answered by HeferH last updated on 29/Jan/23

	Commented by HeferH last updated on 29/Jan/23

	$a = \frac{x^{2}}{x^{2} + 1}; b = \frac{x}{x^{2} + 1}$ ${(1 + \frac{x^{2}}{x^{2} + 1})}^{2} + {(\frac{x}{x^{2} + 1})}^{2} = x^{2}$ ${(\frac{2 x^{2} + 1}{x^{2} + 1})}^{2} + {(\frac{x}{x^{2} + 1})}^{2} = x^{2}$ ${(2 x^{2} + 1)}^{2} + x^{2} = x^{2} {(x^{2} + 1)}^{2}$ $l e t x^{2} = q$ $4 q^{2} + 5 q + 1 = q {(q + 1)}^{2}$ $(4 q + 1) (q + 1) = q {(q + 1)}^{2}$ $q^{2} - 3 q - 1 = 0$ $q = \frac{3 \pm \sqrt{13}}{2} \Rightarrow x = \sqrt{\frac{3 + \sqrt{13}}{2}} \approx 1.817$
	Commented by ajfour last updated on 29/Jan/23

	$t h a n k s s i r .$
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