In a square (ABCD) there is a quarter of a circle ADC (AD = DC), put a point N in the arc AC such that AN = 1 and NC = 2(√2) find BN.


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Question Number 183773 by HeferH last updated on 30/Dec/22

$I n a s q u a r e (A B C D) t h e r e i s a q u a r t e r o f$ $a c i r c l e A D C (A D = D C), p u t a p o i n t N$ $i n t h e a r c A C s u c h t h a t A N = 1 a n d N C = 2 \sqrt{2}$ $f i n d B N .$
	Answered by mr W last updated on 30/Dec/22

	Commented by mr W last updated on 30/Dec/22

	$A C = \sqrt{2} a$ ${A C}^{2} = 1^{2} + {(2 \sqrt{2})}^{2} - 2 \times 1 \times 2 \sqrt{2} \times \cos 135 °$ $2 a^{2} = 13$ $\Rightarrow a = \sqrt{\frac{13}{2}}$ $\cos α = \frac{2 \sqrt{2}}{2 a} = \frac{2}{\sqrt{13}} \Rightarrow \sin α = \frac{3}{\sqrt{13}} = \cos β$ ${B N}^{2} = {(2 \sqrt{2})}^{2} + {(\sqrt{\frac{13}{2}})}^{2} - 2 (2 \sqrt{2}) \times \sqrt{\frac{13}{2}} \times \frac{3}{\sqrt{13}} = \frac{5}{2}$ $\Rightarrow B N = \frac{\sqrt{10}}{2} ✓$
	Answered by mr W last updated on 30/Dec/22

	$M e t h o d I I$ $D a s o r i g i n (0, 0)$ $s a y N (x, y)$ $x^{2} + y^{2} = a^{2}$ $x^{2} + {(a - y)}^{2} = 1^{2} \Rightarrow a^{2} - a y = \frac{1}{2}$ ${(a - x)}^{2} + y^{2} = {(2 \sqrt{2})}^{2} \Rightarrow a^{2} - a x = 4$ ${(\frac{a^{2} - 4}{a})}^{2} + {(\frac{a^{2} - \frac{1}{2}}{a})}^{2} = a^{2}$ $a^{4} - 9 a^{2} + \frac{65}{4} = 0$ $\Rightarrow a^{2} = \frac{13}{2} \Rightarrow a = \sqrt{\frac{13}{2}}$ $\Rightarrow x = \frac{5}{\sqrt{26}}$ $\Rightarrow y = \frac{12}{\sqrt{26}}$ $B N = \sqrt{{(\sqrt{\frac{13}{2}} - \frac{5}{\sqrt{26}})}^{2} + {(\sqrt{\frac{13}{2}} - \frac{12}{\sqrt{26}})}^{2}}$ $= \frac{\sqrt{10}}{2} ✓$
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