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-

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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